Monty Hall Problem

[Rolfe]

I hope you meant to say "the answer is 1/3" ;)


I assumed this was "department of we know what he meant!
























In marked contrast to the originator of the "Monty Hall" puzzle.

Rolfe.

[Skeptical Greg]

I just found something interesting in Wikipedia. Marylin vos Savant, who may be considered to be one of the world's experts on the Monty Hall problem, wrote about the version where Monty chooses the door at random in her column in November 2006:

This difference can be demonstrated by contrasting the original problem with a variation that appeared in vos Savant's column in November 2006. In this version, Monty Hall forgets which door hides the car. He opens one of the doors at random and is relieved when a goat is revealed. Asked whether the contestant should switch, vos Savant correctly replied, "If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch" (vos Savant, 2006).



That explanation seems non-sensical ..
If Monty forgets which door hides the car, how can him ' knowing ' be a factor ?

[Michael C]

That explanation seems non-sensical ..
If Monty forgets which door hides the car, how can him ' knowing ' be a factor ?

But that's the whole point: of course if Monty forgets which door hides the car, he no longer knows where the car is! In the first version Monty knows where the car is, in the second (random choice) he doesn't know where the car is. The important point is that whether he knows or not makes a difference to the outcome.

[Skeptical Greg]

Your reference clearly says, vos Savant is asked if the contestant should switch when the host has forgotten... She says ' switch if he knows ' ...

The example is self contradictory ..

[AntiTelharsic]

YOU assume the problem to be identical if repeated. However, there is no basis for that assumption. As written in the OP, it only describes a single instance.

If "The Monty Hall Problem" is "Monty presents 3 doors; behind one is a car, and behind the other two are goats. You pick a door. But before Monty opens that door, he reveals behind another door that there is a goat. He then gives you the opportunity to switch your choice. What is the probability that you will get the car by switching, or by staying?" then that's what happens the first time I take part in "The Monty Hall Problem". It's also what happens the next time I take part in "The Monty Hall Problem" unless someone re-defined what "The Monty Hall Problem" is in the meantime. You're free to assume that it might change, and that Monty might reveal a car, but if you do you must change the statement of "The Monty Hall Problem", removing the bit where he reveals a goat, and you are then no longer talking about "The Monty Hall Problem" as it was originally stated.

The statement of the problem was not "he reveals what's behind a random door, which turns out to be a goat". The statement of the problem was "he reveals behind another door that there is a goat." That's how the problem works -- he reveals a goat. Change that, and you change the problem.

You say there's no information either way, but there's an explicit statement that he reveals a goat, and it's still there every time someone looks at the problem. You're adding information if you're assuming there are hidden processes, such as random determination of the revealed door, in operation here.

This thread looks like the Monty Python problem. "I'd like to have an argument, please." "Do you want to have the full argument, or just 13 pages?"

But seriously.. discussing variants of the problem is great; let's just call them variants.

[CurtC]

If "The Monty Hall Problem" is "Monty presents 3 doors; behind one is a car, and behind the other two are goats. You pick a door. But before Monty opens that door, he reveals behind another door that there is a goat. He then gives you the opportunity to switch your choice. What is the probability that you will get the car by switching, or by staying?" then that's what happens the first time I take part in "The Monty Hall Problem". It's also what happens the next time I take part in "The Monty Hall Problem" unless someone re-defined what "The Monty Hall Problem" is in the meantime.

I think the misunderstanding here has to be around the interpretation of the words used in the OP's statement of the puzzle. When he said "But before Monty opens that door, he reveals behind another door that there is a goat. He then gives you the opportunity to switch your choice.," you are taking that to mean that you are playing a game where Monty will always give you three doors to choose from, then after you make a choice, will always show you a non-prize door and give you an option to switch.

Again (and again), what we're saying is that the OP can be interpreted to not be specific enough - a game where Monty gives you a choice of three doors, then if you picked the prize on your first guess will then show a non-prize door and give you the chance to switch, this game is consistent with the wording in the OP if you read his words only as a description of a one-time situation that you found yourself in.

Here's the description of this issue found at the Wikipedia page:Some of the controversy was because the Parade version of the problem is technically ambiguous since it leaves certain aspects of the host's behavior unstated, for example whether the host must open a door and must make the offer to switch. Variants of the problem involving these and other assumptions have been published in the mathematical literature.

The issue is that the wording of the OP in this thread also leaves aspects of the host's behavior unstated, or at best can be interpreted ambiguously about the host's behavior.

[AntiTelharsic]

Oh, yeah, it certainly can be interpreted ambiguously, but the simplest interpretation is the standard one. Again, if you want to assume that the host has decision-making processes which aren't specified, you're adding information to the specification of the problem. If the host's methodology is unstated, it's simplest to assume he doesn't have one and does precisely what the problem said: reveals a goat.

[jsfisher]

If "The Monty Hall Problem" is "Monty presents 3 doors; behind one is a car, and behind the other two are goats. You pick a door. But before Monty opens that door, he reveals behind another door that there is a goat. He then gives you the opportunity to switch your choice. What is the probability that you will get the car by switching, or by staying?" then that's what happens the first time I take part in "The Monty Hall Problem". It's also what happens the next time I take part in "The Monty Hall Problem" unless someone re-defined what "The Monty Hall Problem" is in the meantime.

Yes, and yes again.

And if on trial number three, Monty doesn't bother opening any door, then that is not "The Monty Hall Problem". True enough. Then, on the forth trial, Monty again opens a door to reveal a goat and gives you an option to change your choice of door. Then this is "The Monty Hall Problem" again.

"The Monty Hall Problem" is whether you should change your choice when faced with the sequence of events you stated. It is a conditional probability problem: What's the probability the remaining door is a better choice given that Monty has shown us a goat.

Unfortunately, all the Monty Hall Problem statement provided was a sequence of events, not any binding guarantee that that sequence had to happen. We do not have enough information to compute the conditional probability.


You're free to assume that it might change, and that Monty might reveal a car, but if you do you must change the statement of "The Monty Hall Problem", removing the bit where he reveals a goat, and you are then no longer talking about "The Monty Hall Problem" as it was originally stated.

No, not at all. We were only told in the opening post what did happen, not what could have happened. Had a different sequence of events occurred, then, no, it would not have been "The Monty Hall Problem", but that doesn't change that it still may have been a possible outcome.

The statement of the problem was not "he reveals what's behind a random door, which turns out to be a goat". The statement of the problem was "he reveals behind another door that there is a goat." That's how the problem works -- he reveals a goat. Change that, and you change the problem.

Ok, so he revealed a goat. You have no information to indicate he intended to reveal a goat. You have no information to indicate he would have opened any door at all had we selected a different door. We only know he did open a door after we chose a door, and behind the door was a goat.

You say there's no information either way, but there's an explicit statement that he reveals a goat, and it's still there every time someone looks at the problem. You're adding information if you're assuming there are hidden processes, such as random determination of the revealed door, in operation here.

The explicit statement is that a goat was revealed. It was not that you pick a door then Monty will reveal a goat. You only know what has happened, not what had to happen.

[AntiTelharsic]

If I were posing the problem to someone, and Monty's behavior were variable, I'd be sure to point that out in my description of the problem. The fact that it was not pointed out implies (or, at least, strongly suggests) that such considerations are irrelevant to the problem.

I'm not going to say "If I flip a coin, what is the probability of getting heads?" and then go "Hah! You lose! It was a loaded coin!" when you answer "1/2". The simplest, most reasonable assumption is that the coin is fair. If the coin were unbalanced, or had two heads, or if I would be using different kinds of unfair coins on repetitions of the problem -- if there were any features of my problem which would affect your computation of the answer -- there wouldn't be any sense in my not telling you, because you'd be working on a different problem from the one which I intended to pose.

Yes, you could complain that my wording was ambiguous and allows for all sorts of dirty tricks I could do with my coins, but if you assume that I'd actually like for you to be able to solve the problem I posed to you, you wouldn't bother.

[CurtC]

Again, if you want to assume that the host has decision-making processes which aren't specified, you're adding information to the specification of the problem.

Au contraire, I'm not adding anything, it's you who is doing so. You're adding the assumption that Monty is playing by some unspecified rules where he is forced to always show you a non-prize door. I'm simply pointing out that your assumed rules are not part of the puzzle statement.

I used to watch the TV show when I was younger, and one thing for certain, Monty was not bound by your assumption in real life. I point this out because you were appealing to the reasonableness of the assumption, when it's not reasonable at all.

[AntiTelharsic]

I used to watch the TV show when I was younger, and one thing for certain, Monty was not bound by your assumption in real life.

The question is not about the TV show. It is a self-contained problem involving characters and situations from a TV show.

When you're given the statement of the problem, and that's all the information you're given, then it is reasonable to assume that you've been given enough information to solve the problem. If you assume that part of the problem is that Monty's behavior may deviate from that stated, then you don't have enough information to solve the problem. The problem would not be posed without giving you enough information to solve it.

[AntiTelharsic]

Au contraire, I'm not adding anything, it's you who is doing so. You're adding the requirement that we understand some unspecified decision-making process for Monty.

You're adding the assumption that Monty is playing by some unspecified rules where he is forced to always show you a non-prize door. I'm simply pointing out that your assumed rules are not part of the puzzle statement.

The puzzle's statement says he reveals a goat. It said it the first time I read it, the second time I read it, and indeed on all subsequent readings. Every time I go to solve the puzzle, there it is: he reveals a goat. Lacking any statement that he may one day not reveal a goat, I can only assume that he will reveal a goat the next time I try to solve the puzzle.

If you're supposed to be able to solve the puzzle with only the information given, then you assume that he will always reveal the goat. If you don't assume that he will always reveal the goat, then you need more information and you can't solve the puzzle. I think it was the OP's intent that the puzzle as stated be solvable. You're free, of course, to think it was a trick.

[jsfisher]

The question is not about the TV show. It is a self-contained problem involving characters and situations from a TV show.

When you're given the statement of the problem, and that's all the information you're given, then it is reasonable to assume that you've been given enough information to solve the problem. If you assume that part of the problem is that Monty's behavior may deviate from that stated, then you don't have enough information to solve the problem. The problem would not be posed without giving you enough information to solve it.


For the coin flip question you posed earlier, you said to assume it to be a fair coin (and a fair flip, too) would be reasonable. Your general point was (I think) that given an imprecise problem statement, it is entirely appropriate to make minimal, reasonable assumptions to make the problem unambiguous and solvable.

Ok, fair enough. I see the point. However, in the Monty Hall Problem, I think you have over-reached minimal and reasonable to make your assumptions.

Why is it more appropriate to assume Monty will open a door and behind the door will be a goat than to assume Monty will open a door which in this case has a goat behind it?

Consider this slightly changed coin flip question: What is the probability a flipped coin will come up heads immediately after it being flipped and came up heads?

I'd say your original fair coin / fair flip assumptions were still appropriate, so the answer should be 1/2. The coin coming up heads for the first flip is just the goat that happened to be behind the door. So, wouldn't Monty picks a door at random be at least as valid an assumption as the one you proposed?

[AntiTelharsic]

Your general point was (I think) that given an imprecise problem statement, it is entirely appropriate to make minimal, reasonable assumptions to make the problem unambiguous and solvable.

You put it better than I've been able to :)

Why is it more appropriate to assume Monty will open a door and behind the door will be a goat than to assume Monty will open a door which in this case has a goat behind it?

Because you're supposed to be able to solve the problem. If you don't make that assumption, then you need more information before you can solve it; specifically, you need to know how Monty decides which door to open.


I'd say your original fair coin / fair flip assumptions were still appropriate, so the answer should be 1/2. The coin coming up heads for the first flip is just the goat that happened to be behind the door. So, wouldn't Monty picks a door at random be at least as valid an assumption as the one you proposed?

With coin tosses it's generally understood that there is a random process with a well-known distribution. I think any sort of variable process for Monty, however, random or otherwise, is a more complex situation than him simply acting as described. If it's variable, how does it vary? If it varies randomly, what's the distribution? Too many questions to be asking when there's a way to solve the problem by making the least complex assumption: that he will just act as described.

If you were writing it as an exam question for students, and you intended for Monty's behavior to vary, then you would include information about how it would vary. It's the only fair way to do it. If you received the OP's statement in an exam, how would you solve it?

[CapelDodger]

Look, people, this isn't so hard to grasp. The Monty Hall problem is not some horror movie about a game show host out to cheap unsuspecting players. It's a hypothetical scenario, specifically designed to show how unintuitive probability theory can be by presenting what seems to be a simple problem and showing that the 'gut' solution is incorrect.


My single contribution to this thread is :

What Mobyseven said.

I'm only contributing because it came up across the card-table this evening, and the guy that didn't initally get it did get it the moment it was explained. Serious card-players understand artifical systems with a finite number of discrete elements and a few simple rules about how they relate to each other.

[ddt]

No, that doesn't imply that this is how he always behaves. Consider these two problems:

1. I throw a coin. It comes up heads. Can I deduce that there is anything unusual about the coin?

2. I throw a coin. It always comes up heads. Can I deduce that there is anything unusual about the coin?
I don't think this is a fair comparison. A coin flip, in such a puzzle, is a chance event. Anyone reading the puzzle assumes that. However, the description of Monty's behaviour is a deterministic description.


In any case, I am happy that the OP left the possibility of interpreting the problem in different ways, since it led to the discussion on whether the version where Monty always knows the location of the prize is different from the one where he chooses a door at random. The two versions are indeed different and have different solutions.

I agree that the OP - and Marilyn vos Savant's description in Parade in 1990 which generated so much publicity for it, which was nearly identical - leave indeed both interpretations open. However, I think my interpretation is the more usual one. I cite again the NYT article:
"The problem is not well-formed," Mr. Gardner said, "unless it makes clear that the host must always open an empty door and offer the switch. Otherwise, if the host is malevolent, he may open another door only when it's to his advantage to let the player switch, and the probability of being right by switching could be as low as zero." Mr. Gardner said the ambiguity could be eliminated if the host promised ahead of time to open another door and then offer a switch.

Ms. vos Savant acknowledged that the ambiguity did exist in her original statement. She said it was a minor assumption that should have been made obvious by her subsequent analyses, and that did not excuse her professorial critics. "I wouldn't have minded if they had raised that objection," she said Friday, "because it would mean they really understood the problem. But they never got beyond their first mistaken impression. That's what dismayed me."

In other words: all those angry letter writers who objected to her solution, did indeed follow my interpretation.


Why is it more appropriate to assume Monty will open a door and behind the door will be a goat than to assume Monty will open a door which in this case has a goat behind it?


Because it is thus stated. From the OP:
But before Monty opens that door, he reveals behind another door that there is a goat.
He'd be hard-pressed to reveal a goat behind the door which hides the car.

Or in from the wiki article:
and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat.


So whenever Monty opens a door, it is one with a goat behind it.

[Herzblut]

Set 1 - picks the car on the first guess, Monty then reveals a goat and offers the choice of switching. This group has 100 people.

Why? The group might only have 10 people. Or 250. Probability is not certainty.

[jsfisher]

Because it is thus stated. From the OP:

He'd be hard-pressed to reveal a goat behind the door which hides the car.

Here we disagree. It may be because of a subtlety of the language, or maybe just an ambiguity. My interpretation of the "he reveals behind another door that there is a goat" clause is that it presents two facts: Monty opened another door and there was a goat behind the other door.

Let's agree for the sake of discussion that Monty always opening a door is a minimal, reasonable assumption needed to make the problem solvable. If that is the only assumption made then switching doors is a 50% proposition. You'd have to make a second assumption to connect the two facts and require Monty open a goat door. (And then, it becomes a 67% proposition favoring switching.)

Are two assumptions (Monty always opens a door and it's always a goat door) more minimal and reasonable than just one assumption (Monty always opens a door)?

[Herzblut]

The puzzle's statement says he reveals a goat. It said it the first time I read it, the second time I read it, and indeed on all subsequent readings. Every time I go to solve the puzzle, there it is: he reveals a goat. Lacking any statement that he may one day not reveal a goat, I can only assume that he will reveal a goat the next time I try to solve the puzzle.

If you're supposed to be able to solve the puzzle with only the information given, then you assume that he will always reveal the goat. If you don't assume that he will always reveal the goat, then you need more information and you can't solve the puzzle. I think it was the OP's intent that the puzzle as stated be solvable.
That's my understanding of the problem. Monty reveals a goat after the player has made his choice. Whether the game is played once, twice or 1000 times, in each trial the player makes his choice and then a goat is revealed. Any other case (car, chimpanzee, Britney Spears..) is undefined and must be speculated about, because the description does not consider those options. I don't either.

[Herzblut]

Let's agree for the sake of discussion that Monty always opening a door is a minimal, reasonable assumption needed to make the problem solvable. If that is the only assumption made then switching doors is a 50% proposition.

The minimum problem solvable is the one I presented before, I think:

Monty opens a door and offers the player to stay or switch to another door of his choice.

Always switching then wins in 2/3 of the cases. Like the original game, it corresponds to the offer to either stick to the picked door or to switch to the two other doors. But it's of course a much less ingenious version.

[Mobyseven]

Whatever.

You call it function, he calls it motives, you're talking about the same thing.

No, we're bloody well not. Do me a favour - go to a dictionary. Check the meaning of 'function'. Now check the meaning of 'motive'. Are they the same?

Frickin' no.

Monty's function in the scenario is known and important - he is there to open a door with a goat behind it (not randomly, mind you) and offer you a choice to switch.

Monty's motive in the scenario is unknown and irrelevant - why he does what he does has nothing to do with the solution to the problem, unless you propose a motive that changes his function (as people have been doing), in which case you aren't talking about the Monty Hall problem anymore.

The proper Monty Hall puzzle specifies Monty's behavior as always being forced to show a non-prize door, and offer the switch. However, the puzzle as described in the OP of this thread does not specify that. At best, it's ambiguous. As I mentioned before, I have never seen the puzzle presented in the wild (as opposed to its Wikipedia page), where this constraint was explicit. And without this constraint, you have to make some assumptions about how Monty behaves (his function/motive) in order to come to a solution.

:hb: :hb: :hb: :hb: :hb: :hb: :hb:

[Herzblut]

(Again, I acknowledge that if the host knows where the car is and will show every contestant a goat every time, then he can win 2/3 of the time by switching. But both scenarios are consistent with the OP, and without making an assumption about how Monty operates, you can't say that the 2/3 answer is always correct.)
Yes, as I stated before:

if you make the result of the opened door conditional, then you change winning chances to be based on a conditional probability. The switching-wins probability under the condition that Monty has opened a goat-door is 1/2, if the probability of a goat-door opened by Monty is 2/3. This is the case for instance if Monty has randomly opened the door and

- might as well have been opening a car-door with 1/3 probability,
- in which case the player would have lost the game
- and not permitted to switch to the opened door.

This is crystal clear. But, again, the latter case and its speculations are simply excluded by construction, according to my understanding of the problem.

[jsfisher]

The minimum problem solvable is the one I presented before, I think:

Monty opens a door and offers the player to stay or switch to another door of his choice.


Well, I think we have all now circled to some common ground where we all agree things hinge on how each of us might interpret the wording of the problem. So we are discussing semantics now and not probability.

This may be as much agreement as we can get, and I am willing to leave it there.

That aside, though, the Monty Hall problem is still a beautiful problem (preferably worded with less ambiguity to our mutual satisfaction) because the result is so counter-intuitive.

[Herzblut]

Well, I think we have all now circled to some common ground where we all agree things hinge on how each of us might interpret the wording of the problem. So we are discussing semantics now and not probability.

I'm not a native speaker, so what is unclear about

The player picks a door, then Monty opens one of the two other doors and reveals what is behind and offers the player to switch to another door of his choice.

[jsfisher]

I'm not a native speaker, so what is unclear about

The player picks a door, then Monty opens one of the two other doors and reveals what is behind and offers the player to switch to another door of his choice.

I am a native English speaker, so you have me at a disadvantage. ;)

As written, it makes uses the present tense in a clumsy way. Is it present tense with future meaning (these are things that must be) or present tense with past meaning (these are things that happen to be)?

I favor the latter interpretation, but I can understand why others might prefer the former.

At the risk of confusing things further, consider the following:

The player will pick a door, then Monty will open one of the two other doors that will reveal a goat, and then Monty will offer the player the option to switch to another door of his choice.

Here, I think with everything in future tense, the ambiguity is gone. Now, let's try it all in past tense:

The player picked a door, then Monty opened one of the two other doors that revealed a goat, and then Monty offered the player the option to switch to another door of his choice.

With everything in the past tense, is it clearer that Monty's motives may have a role in the analysis? I am hoping so.


Unfortunately, the original problem statement wasn't at all clear. The use of the present tense gives in characteristics of both the future and past tense versions. And, as I said, I favor the past tense interpretation, but your actual mileage may vary.

[Michael C]

Herzblut, you just wrote this:

Yes, as I stated before:

if you make the result of the opened door conditional, then you change winning chances to be based on a conditional probability. The switching-wins probability under the condition that Monty has opened a goat-door is 1/2, if the probability of a goat-door opened by Monty is 2/3. This is the case for instance if Monty has randomly opened the door and

- might as well have been opening a car-door with 1/3 probability,
- in which case the player would have lost the game
- and not permitted to switch to the opened door.

This is crystal clear. But, again, the latter case and its speculations are simply excluded by construction, according to my understanding of the problem.

But yesterday you wrote this:

OK, now if you change the rules and assume, Monty is randomly choosing a door (but not yours) with two possible outcomes, then changing to another door (of your choice) still gives you 2/3 winning chance.

So which answer do you consider to be correct?

[Herzblut]

So which answer do you consider to be correct?
Both, the two versions describe different scenarios. They differ in what happens when Monty reveals a car (lost/won).

[Herzblut]

As written, it makes uses the present tense in a clumsy way. Is it present tense with future meaning (these are things that must be) or present tense with past meaning (these are things that happen to be)?

It is present tense of a technical description. The description of a random experiment. http://cnx.org/content/m13470/latest/

And I didn't say that Monty revealed a goat.

[Michael C]

Both, the two versions describe different scenarios. They differ in what happens when Monty reveals a car (lost/won).

Fine, let's look at this version

OK, now if you change the rules and assume, Monty is randomly choosing a door (but not yours) with two possible outcomes, then changing to another door (of your choice) still gives you 2/3 winning chance.

In this version, you don't say what happens if Monty reveals a car. What does happen if he reveals a car in this version? Does it mean win or lose?

[Herzblut]

In this version, you don't say what happens if Monty reveals a car.
Of course I do. The player switches to another door of his choice and wins.

[Michael C]

Of course I do. The player switches to another door of his choice and wins.

Ah: a new version. The player may also choose the open door. A strange game, but never mind:

If the car is revealed, the player can either:
- switch to the door with a car (duuh), so his chances of winning are 1
- switch to the other closed door, so his chances of winning are 0
- keep his door, so his chances of winning are 0
If a goat is revealed, the player may either
- switch to the door with a goat, so his chances of winning are 0
- switch to the other closed door, so his chances of winning are 1/2
- keep his door, so his chances of winning are 1/2

In no case does switching increase his chances to 2/3.

[articulett]

actually it does-- at the outset his odds were one in 3-- once he picked the goat door-- it still is one in 3... that means the remaining door still has 2/3 chance of being the car. He doesn't change his original odds that his own door has the car-- ever. He just gets information by opening the other door-- the information he gets is either: He definitely has a goat (as would be the case if the car was revealed) or he his original 1/3 chance of having the other goat if a goat was revealed. Really.

You are incorrect MichaelC. Humans equate 2 choices with 50/50 odds... but that is a logical fallacy. Surely you'd understand this if there were 100 doors, right? and one car-- We'd expect him to reveal a goat... it is more likely than not that a goat will be revealed. In your scenario, If he has the car behind his door--there is 100% chance he'll be opening a door with a goat... if he has a goat behind the door-- he still has a 50/50 chance of turning over a goat. But he doesn't know whether he has a goat or a car. Getting a goat doesn't make it more likely that he has a car. His odds don't suddenly get better; he just gets information. Getting a car makes it 100% likely that he has a goat. The same is true whether there are 3 choices or 10 choices or a 100 choices. The odds are always in favor of the one who switches.

This is counter intuitive but very worth understanding.

His original odds of his door having a goat can never get better; --it just shows itself to be worse if he reveals a car behind another door.

[Herzblut]

Ah: a new version.

Can't you read? What's your comprehension problem with OK, now if you change the rules and assume..? You also forget, that it was your idea to randomize Monty's output, not mine. You changed the rules.


The player may also choose the open door.

What's your comprehension problem with switching to a door of his choice?


If a goat is revealed, the player may either
- switch to the door with a goat, so his chances of winning are 0
- switch to the other closed door, so his chances of winning are 1/2
- keep his door, so his chances of winning are 1/2

In no case does switching increase his chances to 2/3.

Yeah, right. Because

If a goat is revealed, and the player switches to the other closed door there may be
- a goat behind that door, so his chances of winning are 0
- a car behind the door, so his chances of winning are 1.

In no case does switching give any chance other than 0 or 1. Zero if you lose and one if you win. Same as in any other game you play. :D

I see you have fully incorporated the concept of probabilities.

[articulett]

A funny thing about humans is that they tend to go with their first choice-- as soon as they place a bet they tend to see their odds as increasing-- they feel more confident with their choice... but your odds don't suddenly change when you place your bet or pick your door... your feeling that you are more likely to win, does not make you more likely to win. This is true even with lottery winners. Before playing, they may correctly assess their odds-- but once they've committed, they "feel" as if their odds go up.

My whole state (Nevada) is built on gambling odds and the bizarre psychology and irrationalities people have in regards to the subject-- their belief in "winning streaks" and so forth.

[CFLarsen]

but your odds don't suddenly change when you place your bet or pick your door...

That's precisely what the "Monty Hall problem" shows: That your odds do change when you pick a certain door.

[Herzblut]

actually it does-- at the outset his odds were one in 3-- once he picked the goat door-- it still is one in 3... that means the remaining door still has 2/3 chance of being the car. He doesn't change his original odds that his own door has the car-- ever. He just gets information by opening the other door-- the information he gets is either: He definitely has a goat (as would be the case if the car was revealed) or he his original 1/3 chance of having the other goat if a goat was revealed. Really.

Not really. If Monty does not consistently (necessarily), but conditionally, reveal a goat the player's winning-by-switching chances in case a goat has been revealed become dependent on how probable it was the goat showed up.

For instance, assume Monty consistently opens the car-door whenever possible. Monty now opens a goat-door and that means that the player has picked the winning door with 100% probability and will always lose by switching.


But he doesn't know whether he has a goat or a car. Getting a goat doesn't make it more likely that he has a car. His odds don't suddenly get better;

Yes, they might do so. In the above scenario, it's even certain that he has the car.

However, the strategy of consistently switching gives the player an average winning chance of 2/3, no matter what. Actually, that's my point. #sigh#

[lionking]

Not really. If Monty does not consistently (necessarily), but conditionally, reveal a goat the player's winning-by-switching chances in case a goat has been revealed become dependent on how probable it was the goat showed up.

For instance, assume Monty consistently opens the car-door whenever possible. Monty now opens a goat-door and that means that the player has picked the winning door with 100% probability and will always lose by switching.


Yes, they might do so. In the above scenario, it's even certain that he has the car.

However, the strategy of consistently switching gives the player an average winning chance of 2/3, no matter what. Actually, that's my point. #sigh#
Jeez how many times does it have to be repeated?

[articulett]

That's precisely what the "Monty Hall problem" shows: That your odds do change when you pick a certain door.



Wrong. The odds of the car being behind the door you originally chose is never more than 1/3. It doesn't change. The car always has a 2/3 chance of being behind one of the other doors.

[CFLarsen]

Wrong. The odds of the car being behind the door you originally chose is never more than 1/3. It doesn't change. The car always has a 2/3 chance of being behind one of the other doors.

Educate yourself. (http://en.wikipedia.org/wiki/Monty_Hall_problem#Solution)

[Herzblut]

Wrong. The odds of the car being behind the door you originally chose is never more than 1/3. It doesn't change. The car always has a 2/3 chance of being behind one of the other doors.
No. Let's define:

A = car is behind one of the other doors
B = Monty reveals a goat

The (conditional) probability of A under the condition that B has taken place is given by

P(A|B) = P(AnB)/P(B)

where P(AnB) shall be the probability of "A and B". In case P(B)=1 this reduces to

P(A|B) = P(A) = 2/3

which is the Monty Hall case.

However, if P(B) is not 100% this 2/3 is no longer necessarily correct. For instance, if Monty selects a door without knowing where the car is:

P(B) = 2/3
P(AnB) = 2/3 * 1/2 = 1/3 ==> P(A|B) = 1/2

I presented this to avoid any useless debate about a given mathematical truth.

[articulett]

Claus, it says what I said. Your odds of the car being behind the first door you chose remain at 1/3. Always.

Herzblut-- I stopped reading you long ago... I cannot make sense of your tangents.

Claus you are wrong... and you are too stubborn and daft to admit it. Are you actually claiming that at some point the odds of you having the car behind the door you chose is more than 1/3? If you are saying something other than this, you've communicated it poorly. If this is what you really think... go check with someone you find intelligent... find one intelligent source that will agree that your odds of your original door being the correct door is ever more than 1/3. (You are just so embarrassingly wrong so often that it amazes me that you imagine yourself a skeptical expert. You aren't making sense to anyone but yourself, I suspect.) No matter what you believe or how much you believe it-- in the Monty Hall scenario the odds of the car being behind the first door you chose is NEVER more than 1/3. Never.

Imagine 10 cars. Your odds of having chosen the correct car is 1 in 10-- those odds never get better. If they reveal a goat, your odds go up by switching to 1 in 9-- if you keep your same door-- you still have 1 in 10 chance of being right.

Really. Ask someone whom you know to be smart.

And I accept your apology. Your bizarre attempts to derail threads with trying to prove me wrong at every turn only makes you look increasingly foolish.

[ddt]

Here we disagree. It may be because of a subtlety of the language, or maybe just an ambiguity. My interpretation of the "he reveals behind another door that there is a goat" clause is that it presents two facts: Monty opened another door and there was a goat behind the other door.
As I interpret it, the "he reveals ... that" contains the intent to reveal a door-with-goat.


Are two assumptions (Monty always opens a door and it's always a goat door) more minimal and reasonable than just one assumption (Monty always opens a door)?
Is that a rhetorical question? :)


I am a native English speaker, so you have me at a disadvantage. ;)
As Herzblut, I'm not a native speaker, so I have the same advantage of being able to claim things about English and afterwards plead ignorance :p


As written, it makes uses the present tense in a clumsy way. Is it present tense with future meaning (these are things that must be) or present tense with past meaning (these are things that happen to be)?

I favor the latter interpretation, but I can understand why others might prefer the former.
That's a nice categorization. Note that, for instance, kitchen recipes aren't put in future tense either. And the actions in a recipe aren't chance occurrences either - they have to be done, if you want to arrive at the goal. In the same vein, I've never seen puzzles stated in the future tense either: if you want to arrive at the solution, you have to take the actions of (possibly) deterministic agents as rules.


The player will pick a door, then Monty will open one of the two other doors that will reveal a goat, and then Monty will offer the player the option to switch to another door of his choice.

Here, I think with everything in future tense, the ambiguity is gone.
I actually had begun to write a post with all kinds of possible formulations of the problem, without changing tenses. Let's strip this one of the future tense:

The player picks a door. Then Monty opens another door that has a goat behind it. Then Monty offers the player the option to switch to another door.

To me, the intent of revealing a door with a goat - opposed to the door with the car - is signified by the use of a restrictive relative clause. It may not be 100% clear that he will in all cases open a door, but if he opens one, it's one with a goat. Compare with:

The player picks a door. Then Monty opens another door, which has a goat behind it. Then Monty offers the player the option to switch to another door.

which has a non-restrictive relative clause. Now it's a chance occurrence that the door he opens has a goat behind it. A Brit might use "which" in the first case too, so then it boils down to a comma what's meant. :rolleyes:


Now, let's try it all in past tense:

The player picked a door, then Monty opened one of the two other doors that revealed a goat, and then Monty offered the player the option to switch to another door of his choice.

With everything in the past tense, is it clearer that Monty's motives may have a role in the analysis? I am hoping so.
I absolutely concur with that.


Unfortunately, the original problem statement wasn't at all clear. The use of the present tense gives in characteristics of both the future and past tense versions. And, as I said, I favor the past tense interpretation, but your actual mileage may vary.
That begs another question. If you were to formulate the puzzle, would you favour using the future tense for everything, or would you favour keeping it in the present tense and add some words to signify that these are really rules that have to be followed by the game host?

Just focusing on the part which door has to be opened by the host (and thus not the part whether he opens a door at all), I think the original formulation by vos Savant:

and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat.


by adding the "knows what's behind the doors" signifies the intent of Monty to open a door with a goat behind it.

[Herzblut]

Claus you are wrong... and you are too stubborn and daft to admit it. Are you actually claiming that at some point the odds of you having the car behind the door you chose is more than 1/3?

Well. What about the odds of you having the car behind the door at the point when you've opened the door to find that you are having the car behind the door?

Gosh! This post shows the most stupid ignorance I've ever seen.

[sol invictus]

Claus, it says what I said. Your odds of the car being behind the first door you chose remain at 1/3. Always.

Maybe you're thinking of the odds before Monty opens a door? Everyone agrees that's 1/3, but once a door has been opened the odds (that the car is behind your door) can certainly change.

Imagine Monty opens both of the other doors - clearly then the odds have changed, right?

[ddt]

A funny thing about humans is that they tend to go with their first choice--

On this and other psychological aspects of the problem, see this paper (http://www.usd.edu/~xtwang/Papers/MontyHallPaper.pdf) that is referenced from the wiki page.

[articulett]

Herzblut... your posts show the most pompous ignorance I've ever seen. You keep changing the original scenario to make others wrong.

I think all the smart people understand it is a question about odds. After you win the lottery you have a 100% chance of your numbers being drawn too. That doesn't change your odds of you having picked the right numbers in the first place. The model is not about post hoc reasoning... it's about odds. Yes, if I flip a coin and it ends up heads... then the 50/50 probability that it will end up heads becomes 100%. But that's a silly game for the silly to play... as are all your rewritten scenarios where Monty reveals the car and so forth. They are tangential and off topic.

The problem is often used to show how people think irrationally. Most people think like Claus... that your odds suddenly change when there's 2 doors left. However, that is never the case. The odds of your having chosen the correct door is 1 in 3 in the scenario as described. The question is should you switch? Your bizarre mental games are probably why few people engage you in dialogue.

[articulett]

Maybe you're thinking of the odds before Monty opens a door? Everyone agrees that's 1/3, but once a door has been opened the odds (that the car is behind your door) can certainly change.

Imagine Monty opens both of the other doors - clearly then the odds have changed, right?

No. Your odds don't change. If he opens both doors you suddenly know with 100% certainty where the car is. But the game as described do not up your odds of having chosen the correct door in any way. It's counter intuitive, but worth understanding.

Think of a situation with 9 goats and one car. Certainly you odds of choosing the right door the first time are 1 in 10... this doesn't get any better when he shows you a goat... He just eliminates a losing entry... that doesn't make you more likely to be a winner. However switching ups your odds because now there only 8 goats and one car... it's 90% likely that the car is amongst those 8 goats. So instead of the 10 percent chance you had, you now have slightly more than an 11% chance by switching... not the best odds... but better than 10%.... If he eliminates another goat your odds go up to almost 13% by switching.

There's all sorts of programs on line-- applets etc. where you can prove this to yourself...

But your odds of having picked the right door the first time, do not change-- the elimination of goats just increases the chance that it's in your favor to switch. But statistically it's always in your favor to switch.

I'm glad to explain this to people who want to understand. It is true whether you believe it or not. But if no amount of evidence will change your mind, then the loss is yours. Like an optical illusion, it's important to understand how your perceptions can be wrong. It's an important part of critical thinking, in fact.

[sol invictus]

No. Your odds don't change. If he opens both doors you suddenly know with 100% certainty where the car is. But the game as described do not up your odds of having chosen the correct door in any way. It's counter intuitive, but worth understanding.

articulett, I've never been even slightly confused by this problem. I understood it immediately the first time I read about in the paper years ago (and was shocked by the resulting public debate, part of which I attributed to sexism - but that's another story). Probability and statistics, specifically conditional probabilities, are a big part of the work I do every day.

Think of a situation with 9 goats and one car. Certainly you odds of choosing the right door the first time are 1 in 10... this doesn't get any better when he shows you a goat... He just eliminates a losing entry... that doesn't make you more likely to be a winner. However switching ups your odds because now there only 8 goats and one car... it's 90% likely that the car is amongst those 8 goats. So instead of the 10 percent chance you had, you now have slightly more than an 11% chance by switching... not the best odds... but better than 10%.... If he eliminates another goat your odds go up to almost 13% by switching.

That depends on the rules governing Monty's behavior.

Please take a moment and consider a simpler situation. You pick a door (out of 3) at random. At that point, we all agree the odds that the car is behind it are 1/3. Now suppose Monty opens both of the other doors, revealing goats. What are the odds now?

The odds the car is behind the door you picked can certainly change when Monty opens more doors. In the classic Monty hall problem they do not - because of the particular rules Monty follows (he knows where the car is and he will never open your door or the door to the car). But this is by no means guaranteed, as the last several pages of this thread - or my example above - show.

[articulett]

There are all sorts of variations on this...
Here's one-- you give two people two coins and ask them to flip them...
You then ask person A whether he has any heads, "she says, yes--in fact the first coin I flipped was a head". The second person says. "Yes". What are the odds that the first person has 2 heads? What are the odds that the second person does? The odds are not equal, I'll tell you that. Which person do you think is more likely to have 2 heads?

[CFLarsen]

The problem is often used to show how people think irrationally. Most people think like Claus... that your odds suddenly change when there's 2 doors left. However, that is never the case. The odds of your having chosen the correct door is 1 in 3 in the scenario as described. The question is should you switch? Your bizarre mental games are probably why few people engage you in dialogue.

That's precisely what happens: The odds do change, because the situation has changed: You suddenly know something you didn't know before.

Did you read the article I linked to?

[articulett]

No sol-- what's changing is whether you are going to switch doors or not. If he opened both doors, then the riddle is over-- everyone knows exactly where the car is and whether the person should switch. But that never made the original choice better than 1 in 3-- if he shows 2 goats-- it just so happens that the 1 in 3 chance turned out to be the winner. If he shows the car-- that shows the 2/3 chance that switching was right. 1/3 of the time in this new scenario, he'll end up showing you 2 goats. 2/3 of the time, he'll be showing you a goat and a car.

It's not a sexist thing. It's basic math that anyone could do. It's just that some men feel so certain that they are right about this that they cannot imagine that they've fooled themselves. When you change the scenario, you aren't changing the odds of your first door being right!

This has been demonstrated countless times but there are still some people -- always men-- who insist it cannot be so... but it is so. Your odds of having chosen the correct door the first time do not change-- just the information...

I won't argue this. Feel free to ask any expert... read the wikipedia entry... design your own damn program... Facts are facts

And in the coin scenario... the first person is more likely to have 2 heads (50/50)
The second person has a 1 in 3 chance of having 2 heads. Their tosses were either: HT, TH, or HH. See? It's very similar with the Monty hall problem. You are just thinking about it incorrectly-- but it's a very common mental mistake. There are tons of books on this. Predictably Irrational by Dan Ariely is one I'm reading currently. Don't let your ego get in the way of understanding this. It will be your loss.

Thomas Kida's book, Don't Believe Everything You Think is also good. Skepdic's dictionary has little lessons on these things. The truth doesn't change just because you don't understand it.

[articulett]

That's precisely what happens: The odds do change, because the situation has changed: You suddenly know something you didn't know before.

Did you read the article I linked to?

Yes Claus, I read it. I do this demonstration in my class every year, and I'm used to 15 year old boys sounding just like you. Your information changes... the odds of your original choice being the correct one-- do not. Your odds of having picked the correct door the first time remain 1 in 3-- even after he shows you a goat.

There was always a 2/3 chance that you did not have the car-- when he shows you the goat-- as he always can-- that means the 2/3 chance is on the remaining door.

Did YOU read the article??

I actually love teaching this stuff, because there's always egotistical young men in class who cannot believe that their brains can fool them. And they are the easiest to fool and their egos get in the way of their understanding this. But it's a valuable lesson to learn the ways your brain fools you. And it's fun, because as each student has an "aha" moment they begin to try and teach the blowhards... we can recreate with cups and keep track of the odds--

I'm right. You're wrong. Again. Deal with it.

[Herzblut]

you give two people two coins and ask them to flip them...
You then ask person A whether he has any heads, "she says, yes--in fact the first coin I flipped was a head". The second person says. "Yes". What are the odds that the first person has 2 heads?

50%?


What are the odds that the second person does?

1/3?

This is well known and has also been discussed in this thread.

[sol invictus]

No sol-- what's changing is whether you are going to switch doors or not. If he opened both doors, then the riddle is over-- everyone knows exactly where the car is and whether the person should switch.

In other words, the odds have changed.

But that never made the original choice better than 1 in 3-- if he shows 2 goats-- it just so happens that the 1 in 3 chance turned out to be the winner.

No one is disputing that the odds are 1/3 before Monty opens any doors! What is at issue here is your claim that the odds do not change after he opens the door or doors. That's just not true (in general).

It's not a sexist thing. It's basic math that anyone could do. It's just that some men feel so certain that they are right about this that they cannot imagine that they've fooled themselves.

There certainly was some degree of sexism involved in the response to vos Savant's column - see here (http://www.marilynvossavant.com/articles/gameshow.html), especially the last comment. That's what I was taking about. However you are wrong that only men claim the odds can change. In fact, vos Savant herself says the odds change when Monty (the gameshow host) doesn't know which door the car is behind.

When you change the scenario, you aren't changing the odds of your first door being right!

For the third time, no one is arguing about what the odds were originally, before Monty did anything.

This has been demonstrated countless times but there are still some people -- always men-- who insist it cannot be so... but it is so. Your odds of having chosen the correct the first time do not change-- just the information...

Always men? Really?

If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch

If the host is clueless, the odds the car is behind the original door have changed from 1/3 to 1/2.

I won't argue this. Feel free to ask any expert... read the wikipedia entry... design your own damn program... Facts are facts

I am an expert, articulett.

And in the coin scenario... the first person is more likely to have 2 heads (50/50)
The second person has a 1 in 3 chance of having 2 heads. Their tosses were either: HT, TH, or HH.

Correct.

It's very similar with the Monty hall problem. You are just thinking about it incorrectly-- but it's a very common mental mistake.

Please take a step back and think again, slowly. No one is disputing that the odds are initially 1/3. No one is disputing that they stay 1/3 in the standard problem. No one is disputing that switching gives you a 2/3 chance of winning, again in the standard setup. However, your claim that the odds never change - no matter what Monty's rules are - is simply false.

[Michael C]

Articulett: the odds can change when Monty opens a door, depending on what new information can be inferred from what is revealed behind the door. CurtC summed up the different possibilities here: http://forums.randi.org/showpost.php?p=3906664&postcount=418

Marilyn vos Savant (who is known to be very smart indeed and an expert on this particular problem) also explained the difference between the version where Monty chooses a door at random and the original problem, where he knows where the car is. The odds really do change to 50/50 when Monty opens a door at random, happening to reveal a goat. In the original version, where Monty always reveals a goat because he knows where the car is, the chance of the prize being behind your chosen door stays at 1/3, while that of it being behind the remaining door is 2/3.

You can read this in the Wikipedia article, but for convenience I'll quote vos Savant again here. The question is: is it in the contestant's interest to switch doors after the host has revealed the goat? The answer is: it depends on what information the host had when choosing the door. As Marilyn vos Savant puts it:

If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch

[articulett]

Wrong, Sol. Ask any expert you know... run a program. Your odds of your first choice being the winner are always 1 in 3 in the Monty Hall problem.

Even if he reveals both doors every time-- You will still only have chosen the car on your first choice 1/3 of the time. You still will switch to get the car 2/3 of the time. Of course, the more he does this the more greater you see these probabilities.

The odds that you chose the right door on your first pick are always 1 in 3-- through the whole game. Sometimes that 1 in 3 will pay off-- in fact 1/3 of the time it will.

In your new scenario the odds never change. 1/3 of the time, he'll be revealing 2 goats and you'll have chosen the correct door. 2/3 of the time he'll show you a goat and a car and you will change.

The greater the numbers of tests in this scenario the easier you'll see this. In the original scenario you have a 2/3 chance of winning by switching. You have a 1/3 chance of winning by staying with your first choice.

You can try to convince yourself otherwise, but these are well understood odds.

YOUR ODDS OF HAVING CHOSEN THE RIGHT DOOR THE FIRST TIME DO NOT CHANGE THOUGHOUT THE GAME. It does not suddenly go up to 50% because he's eliminated one of the doors.

[Skeptical Greg]

Knowing there are 4 aces in a deck of cards, does not change the odds of one of them being dealt. Stacking the deck does..


The host ' knowing ' where the car is does not change the odds.. The steps taken because of his knowledge does..

[hgc]

Can't believe this thread lives on after all these years. I thought the question was settled on about page 3.

Go ahead and keep flailing, articulett. You'll come to understand eventually. It's inevitable.

eta: Huh? I don't even understand what anyone's aguing about anymore. What are you claiming, articulett?

[sol invictus]

Wrong, Sol. Ask any expert you know... run a program. Your odds of your first choice being the winner are always 1 in 3 in the Monty Hall problem.

Please be aware you're not only disagreeing with every poster in this thread, but with Marilyn vos Savant herself.

Even if he reveals both doors every time-- You will still only have chosen the car on your first choice 1/3 of the time. You still will switch to get the car 2/3 of the time.

You're simply not understanding the situation. Again, suppose Monty opens both other doors, and both have goats behind them. What are the odds that the car is behind your door?

Note that I am not asking you how often this situation will occur if the game is played many times - I'm asking about a particular class of instances, where both other doors are revealed to have goats behind them.

In your new scenario the odds never change. 1/3 of the time, he'll be revealing 2 goats and you'll have chosen the correct door. 2/3 of the time he'll show you a goat and a car and you will change.

That was not my scenario. You're arguing with yourself here.


YOUR ODDS OF HAVING CHOSEN THE RIGHT DOOR THE FIRST TIME DO NOT CHANGE THOUGHOUT THE GAME. It does not suddenly go up to 50% because he's eliminated one of the doors.

Then you should write to vos Savant and tell her she is wrong about the "clueless" case.

[CFLarsen]

Yes Claus, I read it. I do this demonstration in my class every year

Seriously?

If you tell them what you have argued here, then you are demonstrably miseducating your students.

Did YOU read the article??

I actually love teaching this stuff, because there's always egotistical young men in class who cannot believe that their brains can fool them. And they are the easiest to fool and their egos get in the way of their understanding this. But it's a valuable lesson to learn the ways your brain fools you. And it's fun, because as each student has an "aha" moment they begin to try and teach the blowhards... we can recreate with cups and keep track of the odds--

I'm right. You're wrong. Again. Deal with it.

Sorry, but you are the one who is wrong here. Not because you are of a certain gender, but simply because you are wrong - period.

I sincerely hope that you don't in reality single out a gender in your classroom, deriding them not just for being wrong (although you are in fact the one being wrong), but being wrong because they are of the opposite sex than you.

What you are describing here is not in any way an acceptable educational method. You are not just teaching them false logic, you are teaching them that they are wrong because they are not women.

[articulett]

Articulett: the odds can change when Monty opens a door, depending on what new information can be inferred from what is revealed behind the door. CurtC summed up the different possibilities here: http://forums.randi.org/showpost.php?p=3906664&postcount=418

Marilyn vos Savant (who is known to be very smart indeed and an expert on this particular problem) also explained the difference between the version where Monty chooses a door at random and the original problem, where he knows where the car is. The odds really do change to 50/50 when Monty opens a door at random, happening to reveal a goat. In the original version, where Monty always reveals a goat because he knows where the car is, the chance of the prize being behind your chosen door stays at 1/3, (while that of it being behind the remaining door is 2/3.

You can read this in the Wikipedia article, but for convenience I'll quote vos Savant again here. The question is: is it in the contestant's interest to switch doors after the host has revealed the goat? The answer is: it depends on what information the host had when choosing the door. As Marilyn vos Savant puts it:

Yes... I agree... with the alternate scenarios instead of the basics (where the host always reveals the goat and gives you an option to switch) that whether it's better to switch or not IS altered. However, your odds of having picked the correct door the first time don't change... your information as to whether to switch or not is changed by alternating host scenarios... but your odds of having picked correctly on the first choice remain 1 in 3... in the random scenario the host has the odds of picking a goat 2/3 of the time... and if he does... then the remaining door has an equal chance of being the car as the one you have. But even still-- in this case if you played it again and again-- you'd still have only picked the correct door the first time in 1/3 of the cases... and in 1/3 of those cases Monty will reveal a car in his blind guess... whether he offers you a chance to switch or not changes the game.

But my claim is that no matter how you fix the game or what the host does... your odds of having chosen the correct car on the first choice is 1 in 3-- whether you are allowed to switch or not or whether he reveals a car or a goat blindly or on purpose or only offers you a chance to switch if you are wrong doesn't change the fact that your original choice has a 1 in 3 chance of being right. That doesn't change... what the host does or doesn't offer you and his motives for doing so certainly could change whether you switch or not-- but your odds of having chosen correctly on the first choice don't change.

Yes, if he only offers to switch when you are a winner (that would be in 1/3 of the cases), then you should never switch. If it's blind, then it's just as good to stay as switch if he reveals a goat-- if he reveals the car (1/3 of the time) and he lets you switch-- you switch 100% of the time. If what he offers a person is based on whether they have a car or a goat--then the offering will be different in 1/3 of the cases. But these are all side stories. None of these change the fact that you first choice has a 1/3 chance of having been correct. Whether it's correct or not may change what the host offers you and whether you should switch... but it doesn't change the fact that your first choice has a 1 in 3 chance of being correct. It's never more than that. You can increase your odds by switching. But the odds that you have chosen correctly the first time remain at 1 in 3.

I agree that it's not always best to switch given the alternating scenarios. However I still maintain that your original odds for having chosen right the first time are not more than 1 in 3. Claus contended that your odds of having chosen correctly changed. In fact, that's what this derail is about. That isn't the case. Your information regarding whether to switch or not IS altered by these differing scenarios. Having more information doesn't change the fact that your first choice has a 1 in 3 chance of being correct.

(And because of this, if given the opportunity to switch-- without knowing anything else-- the odds are better on average if you do switch... the whole "lesson" in this little scenario.)

[sol invictus]

Whether it's correct or not may change what the host offers you and whether you should switch... but it doesn't change the fact that your first choice has a 1 in 3 chance of being correct. It's never more than that.

I agree that it's not always best to switch given the alternating scenarios. However I still maintain that your original odds for having chosen right the first time are not more than 1 in 3. Claus contended that your odds of having chosen correctly changed. That isn't the case. Your information regarding whether to switch or not is altered by these differing scenarios.

So let me get this straight.

You now agree that (in the alternate scenario where Monty is "clueless" and has opened a goat door) your odds of winning are not improved by switching; i.e., that the odds the car is behind your door are equal to the odds that it's behind the other unopened door. But at the same time you continue to insist that the odds the car is behind your door are 1/3 and never change. Those two claims imply that the probability that the car is behind any of the three doors is 2/3.

That's what physicists call a violation of unitarity... it's possible only if someone might have driven the car away while you were dithering over whether to switch and replaced it with a goat :).

[CFLarsen]

But my claim is that no matter how you fix the game or what the host does... your odds of having chosen the correct car on the first choice is 1 in 3-- whether you are allowed to switch or not or whether he reveals a car or a goat blindly or on purpose or only offers you a chance to switch if you are wrong doesn't change the fact that your original choice has a 1 in 3 chance of being right.

Why do you keep on with this? Who has disputed it?

[Vorpal]

I agree with Sol. Earlier, I calculated (post #400) that a clueless Monty that opens a random door that is not the player's pick gives equal chances of winning the game regardless of keep or switch. I thought this was fairly obvious, but since it appears that some people won't be convinced without a program and I've a few minutes to spare, here's an ugly MATLAB script:
trials = 2e5; cars = 0; valid_trials = 0; for n=1:trials
car = 1+floor(3*rand); player = 1+floor(3*rand); %
monty = player; % Monty guesses until his pick
while ( monty == player ) % is different from the player's
monty = 1+floor(3*rand); % (equivalent to picking a random
end % door that is not the player's)
if ( monty ~= car ) % Car not revealed
valid_trials = valid_trials+1;
% Player switches to the other door
for k=1:3 if (k~=player & k~=monty) guess2 = k; end; end
if ( guess2 == car ) cars = cars+1; end
end
end; valid_trials, cars/valid_trials
The result was 133172 trials in which Monty did not accidentally reveal the car, 49.82% of the player won by switching.

[Herzblut]

That's what physicists call a violation of unitarity... it's possible only if someone might have driven the car away while you were dithering over whether to switch and replaced it with a goat :).
Men do those nasty things!! "Violation of unitrinity", pshaw! Physicists like yourself are the devote, stupid apologists of the evil male! Pigs! Right, articulett?

[boooeee]

Monty Hall is a lucky man. Marilyn Vos Savant has granted him cultural immortality. In three hundred years, people will will still be arguing about this problem. Just extrapolate the progression of this zombie thread.

[AntiTelharsic]

I think I found the source of some of the confusion here.

From Vorpal's program, it looks like the trial isn't counted when Monty, choosing randomly, reveals the car. That's really important.

Consider the following cases:

1. If Monty chooses randomly and you always switch to the remaining, unopened door (even if he revealed a car), you will win 1/3 of the time.

2. If Monty chooses randomly and you always switch to his door if he revealed a car and to the unopened door otherwise, you will win 2/3 of the time.

3. If Monty chooses randomly and you always switch to the remaining, unopened door, but it doesn't count when he reveals a car, you will win 1/2 of the times that count.

In the first case, half of Monty's would-have-been goat revelations in situations where you'd chosen a goat are now car revelations, so half of your winning switches have been turned to losses. The second case is identical to the standard problem. The third case is just like the first, but since some of the trials have been invalidated your winning choices make up a larger proportion of the total number of trials.

Hope that helps.

[Scott1972]

3. If Monty chooses randomly and you always switch to the remaining, unopened door, but it doesn't count when he reveals a car, you will win 1/2 of the times that count.


It might be more clear to say if Monty reveals the car the game ends and if Monty reveals a goat a new game, with new probabilities, begins.

[AntiTelharsic]

It doesn't have to end, though.

I pick a door; Monty opens a random other door; if Monty's door hid a goat, I switch to the third door; otherwise this iteration isn't valid.

As the number N of valid iterations of the above goes to infinity, the number of times I switched and got the car approaches 1/2 N. I will win 1/3 of the total number of attempts, but the number of valid attempts will be 2/3 of the total number.

[Scott1972]

It doesn't have to end, though.

I pick a door; Monty opens a random other door; if Monty's door hid a goat, I switch to the third door; otherwise this iteration isn't valid.

As the number N of valid iterations of the above goes to infinity, the number of times I switched and got the car approaches 1/2 N.

I understand what you are saying. I just think it's more clear to say the probability changed because a new, different game began rather than the probability changed because we choose to ignore the instances in which Monty reveals the car.

[sol invictus]

I think I found the source of some of the confusion here.

From Vorpal's program, it looks like the trial isn't counted when Monty, choosing randomly, reveals the car. That's really important.

Yes, I said that several times already in trying to explain this to articulett.

The statement of this alternative version is that Monty does not know where the car is, and that in the instance in question he has opened a door with a goat. We want to know what the odds are given that data. To compute odds for that situation we should only consider games in which he opens a door and reveals a goat - not ones where something else happens - and then ask in what fraction of those games the car is behind the player's door.

[Skeptical Greg]

How many times to we have to reiterate the OP ?

He opens a door and REVEALS A GOAT ..

Really folks, this is not rocket science ...

[CFLarsen]

How many times to we have to reiterate the OP ?

He opens a door and REVEALS A GOAT ..

...because he has to:

The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it.
Source (http://en.wikipedia.org/wiki/Monty_hall_problem)

That's the whole point of The Monty Hall Problem.

This is not about the 1/3rd chance of what the first choice is - or bigoted misandry, for that matter.

This is about the counter-intuitive choice that one has to make, when the additional, inside information is revealed.

[Scott1972]

Yes, I said that several times already in trying to explain this to articulett.

The statement of this alternative version is that Monty does not know where the car is, and that in the instance in question he has opened a door with a goat. We want to know what the odds are given that data. To compute odds for that situation we should only consider games in which he opens a door and reveals a goat - not ones where something else happens - and then ask in what fraction of those games the car is behind the player's door.

I agree that everything you wrote is clear and correct.

I think articulett's argument is that the 1/3 probability for each door at the beginning doesn't change, the game changes. At least that is what I got from her posts. By only considering games in which Monty randomly reveals a goat you have essentially created a new game with two doors, 1 goat, 1 car, and no knowledge about what is behind each door. Unless I'm way off base, I don't think anyone is claiming the probability of finding the car behind each door is not 50% for this scenario.

[sol invictus]

Unless I'm way off base, I don't think anyone is claiming the probability of finding the car behind each door is not 50% for this scenario.

Articulett claimed that the odds the car is behind the door you pick were always 1/3 and never change, even in this scenario. That is false. The odds are perfectly well-defined, and if the door is opened and a goat is revealed they change to 1/2. If the door is opened and a car is revealed they change to 0.

There is no need to discard the games where the car is revealed, but including them has no effect on the odds in cases where a goat is revealed.

[articulett]

Why do you keep on with this? Who has disputed it?

You did, Claus-- in post 535--in your everlasting quest to prove me wrong by knocking down a straw man version of what I actually said.


but your odds don't suddenly change when you place your bet or pick your door...
That's precisely what the "Monty Hall problem" shows: That your odds do change when you pick a certain door.

I think anyone can see that if you always stick with your current choice--you will have 1 in 3 chance of winning-- the only way to increase those odds is to change your choice after a goat is shown. The host can always show a goat. The host does not always need to offer you the switch. However, in the scenario--with no added information-- your odds increase by switching. Moreover, even if you add information-- you always have 1 in 3 chance of winning by not switching--that doesn't change with the added information. In your made up scenarios you are only switching whether the host is going to be offering you a switch or not.

Because the only information we have as the scenario is laid out is that he's shown a goat and offered a switch. That is the only information in the scenario. But no matter how much you add information, you are never increasing the odds over 1 in 3 by staying. If the host only offers a switch if you have the car... then the information you get when if he asks you to switch, is that you have the car. But that means that in 2/3 of the cases, he'll never ask you to switch. And because you don't know this is the case, it's more likely that he's offering you to switch because he always does (the original scenario) or because he blindly chose a goat... (meaning that if he blindly chose the car... he'd not be offering you the chance to switch.)

The salient part of this exercise is that your odds if you stay are not more than 1 in 3-- ever. The only way to increase those odds is to switch if given the opportunity. (Of course 1 in 3 isn't bad odds)... and even with the 2/3 increased odds with a switch-- you still lose 1/3 of the time.

[articulett]

Articulett claimed that the odds the car is behind the door you pick were always 1/3 and never change, even in this scenario. That is false. The odds are perfectly well-defined, and if the door is opened and a goat is revealed they change to 1/2. If the door is opened and a car is revealed they change to 0.

There is no need to discard the games where the car is revealed, but including them has no effect on the odds in cases where a goat is revealed.

Wrong.

Your odds of having picked the car on the first choice are always 1 in 3. The host can always reveal a goat. Revealing a goat doesn't change the fact that you have a 1 in 3 chance of being a winner if you stay with your first choice. The only way to increase these odds is to switch if offered the choice to switch and you have no further information.

[RecoveringYuppy]

Can I clarify something articulett?

If you accept that there is only one car behind three doors and one door is open revealing a car, what are the odds that there is a car behind either of the two closed doors?

[articulett]

So let me get this straight.

You now agree that (in the alternate scenario where Monty is "clueless" and has opened a goat door) your odds of winning are not improved by switching; i.e., that the odds the car is behind your door are equal to the odds that it's behind the other unopened door. But at the same time you continue to insist that the odds the car is behind your door are 1/3 and never change. Those two claims imply that the probability that the car is behind any of the three doors is 2/3.

That's what physicists call a violation of unitarity... it's possible only if someone might have driven the car away while you were dithering over whether to switch and replaced it with a goat :).

That is only true if you KNOW he's chosen blindly... this presumes however that if he'd accidentally revealed the car, he'd change whether you could switch or not. What would change is whether you are offered a choice to switch. Your odds don't go up of having chosen correctly the first time... It's just that the odds of rechoosing a car go down. And/or whether you are offered the choice to switch or not go down. You cannot change the odds that staying with your first choice every time gives you a 1 in 3 chance of winning the car. You can however increase your odds of getting a car by changing your choices depending on how many extra conditions you want to add to the equation. Does Monty ALWAYS give a choice to switch? Does he always reveal a goat? Do you know of ulterior motives or statistics which show the percentage of times he offers to switch and why?

The added information can only tell you whether you are increasing your odds by switching-- it cannot make "staying with your original choice" better than 1 in 3. What your changing is whether you are offered that choice to stay or not.

[Scott1972]

Just to be clear articulett, if Monty randomly opens a door and it reveals a goat and he ALWAYS gives you the option to switch do agree that the probability of the car being behind your originally chosen door is now 1 in 2?

Giving Monty the choice of offering or refusing a switch obviously changes the problem.

Edit: The last sentence is incorrect. The odds are 1 in 2 regardless. I confused myself by thinking about Monty having the choice to either open a door or not.

[articulett]

I agree with Sol. Earlier, I calculated (post #400) that a clueless Monty that opens a random door that is not the player's pick gives equal chances of winning the game regardless of keep or switch. I thought this was fairly obvious, but since it appears that some people won't be convinced without a program and I've a few minutes to spare, here's an ugly MATLAB script:
trials = 2e5; cars = 0; valid_trials = 0; for n=1:trials
car = 1+floor(3*rand); player = 1+floor(3*rand); %
monty = player; % Monty guesses until his pick
while ( monty == player ) % is different from the player's
monty = 1+floor(3*rand); % (equivalent to picking a random
end % door that is not the player's)
if ( monty ~= car ) % Car not revealed
valid_trials = valid_trials+1;
% Player switches to the other door
for k=1:3 if (k~=player & k~=monty) guess2 = k; end; end
if ( guess2 == car ) cars = cars+1; end
end
end; valid_trials, cars/valid_trials
The result was 133172 trials in which Monty did not accidentally reveal the car, 49.82% of the player won by switching.

Yes... but what would happen if he DID reveal the car as he would do 1 out of 3 times when choosing blindly? Do you still get to choose or is the choice suddenly taken away?

If you don't know whether Monty chose blindly or not, you still know that you only have a 1 in 3 chance of having chosen correctly the first time.

[AntiTelharsic]

If Monty always opens a door, always chooses the door randomly, and has revealed a goat, then you have a 50% chance of already having chosen the door with the car.

Sure, in 1/3 of all possible situations you will have already chosen the car, but you will have chosen it in fully half of this restricted set of situations.

It could be that this is not what anyone's arguing (I think several people are arguing several different things) but I thought I'd offer it just in case this is where a misunderstanding is happening.

[AntiTelharsic]

Yes... but what would happen if he DID reveal the car as he would do 1 out of 3 times when choosing blindly? Do you still get to choose or is the choice suddenly taken away?

It's taken away; that iteration doesn't count because it's stipulated that Monty reveals a goat.

[articulett]

Can I clarify something articulett?

If you accept that there is only one car behind three doors and one door is open revealing a car, what are the odds that there is a car behind either of the two closed doors?

0%-- but then the game is over. If you are told to pick a number between 1 and 3 you have a 1 in 3 chance of picking the right number. Once the correct number is revealed-- you have a 0% chance or a 100% chance of having picked the right number. If you do a multitude of these trials... you will be in the 100% category 1/3 of the time and in the 0% category 2/3 of the time. What you are really doing in the alternating scenarios is changing whether the hosts offers you a chance to switch or not. If he always offers you the switch... and he reveals the car-- you switch 100% of the time. If he doesn't (2/3 of the time) it doesn't matter what you do. This still means, that overall-- if always given the chance to switch-- you increase your odds by doing so.

The real question is-- if the host reveals the car-- do you still get to switch? Staying with your first choice in every case never increases your odds more than 1 in 3 for winning the car. If you don't know the hosts reasons or you have no additional information, and you want to improve your odds of winning the car, then the only way to increase you odds is to switch if given the opportunity.

Hindsight is always 20/20-- all possibilities have collapsed to 0 or 1.

[Vorpal]

Yes... but what would happen if he DID reveal the car as he would do 1 out of 3 times when choosing blindly? Do you still get to choose or is the choice suddenly taken away?
I've covered two cases of this back in post #400. If the choice is taken away completely, i.e., that Monty revealing the car counts as an automatic loss, then the chances of winning are 1/3 regardless of whether one resolves to keep or switch on a goat. If the choice is not taken away and the player plays it competently (i.e., whenever Monty reveals the car, go for that door), then this counts as an automatic win and the chances of winning are 2/3 regardless of whether one resolves to keep or switch on a goat.
Conclusion: resolving to keep or switch on a goat give the same probability of winning if Monty picks a random door that is not the player's.

I suppose there is a third case that I neglected, that of an incompetent player who is allowed to pick again even if the car is revealed but who resolves to pick not to pick a revealed car. However, I don't think that's what Sol was talking about, and I didn't think that's what you were talking about either.

If you don't know whether Monty chose blindly or not, you still know that you only have a 1 in 3 chance of having chosen correctly the first time.
Edit: Hmm... I think I see where you're going with this. You're really are talking about a player who keeps the same door even if Monty reveals a car. Yes, you're right--probability of winning the game for such a player is 1/3, although I don't believe that's what Sol disputed (but he can clarify his own position better himself).

[Scott1972]

Yes... but what would happen if he DID reveal the car as he would do 1 out of 3 times when choosing blindly? Do you still get to choose or is the choice suddenly taken away?

If you don't know whether Monty chose blindly or not, you still know that you only have a 1 in 3 chance of having chosen correctly the first time.

If you don't know whether Monty is choosing randomly AND he opens a door to reveals a goat results in two scenarios:

1. Monty chose randomly: The possibly for for each remaining door is 1 in 2. Changing doesn't matter.

2. Monty knowingly chose a goat door: Switching increases your win chance to 2 in 3.

I think your objection is with removing the cases in which Monty randomly opens the door with the car from the possible ways you could lose in the original game and you consider that changing the game.

[Herzblut]

That is only true if you KNOW he's chosen blindly...
Of course. The factual truth is only true, if you know it's true. Otherwise the truth is false.

[Scott1972]

If Monty always opens a door, always chooses the door randomly, and has revealed a goat, then you have a 50% chance of already having chosen the door with the car.

Sure, in 1/3 of all possible situations you will have already chosen the car, but you will have chosen it in fully half of this restricted set of situations.

It could be that this is not what anyone's arguing (I think several people are arguing several different things) but I thought I'd offer it just in case this is where a misunderstanding is happening.

This is what I've been trying to say, but nowhere near as clearly as you have.

[RecoveringYuppy]

0%-- but then the game is over. If you are told to pick a number between 1 and 3 you have a 1 in 3 chance of picking the right number. [...snip]
OK. Never mind. I think there was some miscommunication going on when I questioned you.

[articulett]

It's taken away; that iteration doesn't count because it's stipulated that Monty reveals a goat.

Correct. So what this scenario is actually doing is changing the odds of Monty asking you if you want to switch or not... it's not changing the fact that staying with your choice gives you a 1 in 3 chance of winning a car. It just means that you have a 1 in 3 chance of not having been offered a choice and a 1 in 3 chance of having the car behind the door monty didn't reveal... yes, your odds increase to 50% by switching... but your odds of having chosen the correct door the first time are still 1 in 3. What you are really getting in this scenario is information that you are not in the situation where the Host has revealed a car. You are in the situation you'd expect to be in 2/3 of the time... half of the time you are in such a situation (which will be 2/3 of the time), you will have the car (you'll be the 1 in 3)-- half of the time you won't. You've eliminated the possibility that the host will be taking away your choice by revealing the car in the blind guess.

Information only tells you whether it's smarter to switch or not. It doesn't change the fact that staying with your first choice gives you a 1 in 3 chance of winning... varying host motivations just give you more information as to whether you might be that 1 in 3-- If you know the host only asks people to switch if they have the car-- then-- if you are asked to switch you don't-- you KNOW you are the 1 in 3. But, this means that you won't be given such an opportunity 2 of 3 times... and so you still never increase you odds of winning to more than 1 out of 3. You've just collapsed the possibilities so that you've become the 1 in 3.

[AntiTelharsic]

Correct. So what this scenario is actually doing is changing the odds of Monty asking you if you want to switch or not... it's not changing the fact that staying with your choice gives you a 1 in 3 chance of winning a car. It just means that you have a 1 in 3 chance of not having been offered a choice and a 1 in 3 chance of having the car behind the door monty didn't reveal... yes, your odds increase to 50% by switching... but your odds of having chosen the correct door the first time are still 1 in 3.


We're assuming that Monty always asks.

Your odds of having chosen the correct door are 1 in 3 over the entire range of possibilities, yes, but if you consider only the subset of those possibilities where Monty always gives a choice, chooses randomly, and has revealed a goat, then you'll have chosen the correct door in half of those.

[Michael C]

I think anyone can see that if you always stick with your current choice--you will have 1 in 3 chance of winning-- the only way to increase those odds is to change your choice after a goat is shown. The host can always show a goat. The host does not always need to offer you the switch. [...] The salient part of this exercise is that your odds if you stay are not more than 1 in 3-- ever.

No. The host could be using the strategy of only offering a switch if the car is behind your currently chosen door. In that case, the odds if you stay go up to 1 and the odds if you switch go down to 0.

[Vorpal]

Maybe I'm misunderstanding someone's position here, but this is the way this dispute appears to me:
Articulett: chance of winning via sticking with the first choice no matter what happens is just P(first pick is car).
Others: incorporating the added information once an event has happened gives the conditional probability P(first pick is car|info), which is not equal to P(first pick is car)
I don't understand why there is such a case of miscommunication. Those two claims are perfectly consistent.

[articulett]

Yes... I was derailed by Claus' claim that your odds change when you pick a certain door.

They don't. The thing that actually is changing in the alternate scenarios is whether the host offers you a choice in changing. You always have a 1/3 chance of winning by staying with the original door-- when offered the option of switching after being shown a goat, you are just getting more information as to whether your odds increase by switching. Unless you have knowledge that the host only offers prizes if he knows you have the car, you can increase your odds of getting the car to 1 in 2 or 2 in 3 by switching. Therefore, if you have no additional knowledge or constraints on the problem-- you increase your odds of getting the car by switching. In fact, the only way you CAN increase your odds is by switching if given the choice. Claus just has a need to prove me wrong whenever he can... even if it means arguing a straw man version or tangent of what I actually said. He inferred that I was mistaken or stupid regarding my claim. I was, however, correct.

[articulett]

No. The host could be using the strategy of only offering a switch if the car is behind your currently chosen door. In that case, the odds if you stay go up to 1 and the odds if you switch go down to 0.

Yes... but then the host would only be offering you that choice 1/3 of the time. If the host was using that strategy-- then 2/3 of the time you would never even be offered the choice. If you have no knowledge that the host does this... then the only knowledge you have is that he's shown you a goat (which he can always do) and he's offered you a choice (which he can always do.) Unless you have reason to suspect otherwise (a host only giving you the choice because you picked the car)... you can only increase you odds by changing. Sure, in 100% of the cases where you know why the host gave you the choice-- you will lose by changing. But you will also lose 2/3 of the time because he will never offer you the choice-- he'll just reveal that you didn't pick it the first time.

Yes, you are correct... but if you don't know that the host is using that strategy... then you can only hypothesize that it's the 1 in 9 possibility that the host IS using that strategy and you have managed to choose the right door which is why he's offered you the choice. That's the only info. that you are getting without having prior knowledge or reason to suspect such a thing. If you think this is the strategy and you are right-- you will win every time he offers you the choice-- he just won't offer you the choice if you are wrong. If you think this is the strategy and you are wrong you will only win 1/3 of the time and lose 2/3 of the time.

[articulett]

Articulett claimed that the odds the car is behind the door you pick were always 1/3 and never change, even in this scenario. That is false. The odds are perfectly well-defined, and if the door is opened and a goat is revealed they change to 1/2. If the door is opened and a car is revealed they change to 0.

There is no need to discard the games where the car is revealed, but including them has no effect on the odds in cases where a goat is revealed.

No. If you include the games where the cars are revealed and you still allow the choice-- then the person wins by switching 100% of the time. If a goat is revealed, you win by switching 50% of the time... If you don't know which if these scenarios you are dealing with... then you have a 1/3 chance of winning by keeping your choice and a 2/3 chance by switching. If you know that the host always offers a chance to change your choice-- you have a 50% chance of winning by changing your choice if you see a goat and a 100% chance of winning by seeing a car. Your odds of your first choice being the car do not change from 1 in 3. You are only getting info. as to whether you can improve those odds by switching... that's it. If you know the host's choice is random and that he will offer you a choice either way... you still are better off switching... because 1/2 of the time you will DEFINITELY have the car. And 1/2 the time you will have 50% of a chance of getting the car by switching. IF he doesn't offer the choice when he reveals the car-- you still have the 50% odds left if he reveals the goat-- and 0% chance if he reveals the car. What you are really getting is information about whether and how your odds increase if you are given the opportunity to switch. You don't increase your odds of winning the car more than 1/3 by keeping that choice-- you just don't get offered the chance to switch as often in the alternate scenarios.

[Michael C]

Yes, you are correct... but if you don't know that the host is using that strategy... then you can only hypothesize that it's the 1 in 9 possibility that the host IS using that strategy and you have managed to choose the right door which is why he's offered you the choice. That's the only info. that you are getting without having prior knowledge or reason to suspect such a thing. If you think this is the strategy and you are right-- you will win every time he offers you the choice-- he just won't offer you the choice if you are wrong. If you think this is the strategy and you are wrong you will only win 1/3 of the time and lose 2/3 of the time.

Now I see what was worrying me in what you said earlier. You said The only way to increase those odds is to switch if given the opportunity.

It would be clearer to say
"The only way to change those odds is to switch if given the opportunity".

Obviously if your strategy is to choose a door and never to switch, you will, in the long run, win 1/3 of the games. But if your strategy is always to switch when given the chance, you may increase or decrease your odds, depending on what strategy Monty is using.

For instance, if Monty (unknown to you) only gives the opportunity of switching when there is a car behind your chosen door, the strategy of switching if given the choice will ensure that you never win: you have decreased your odds from 1/3 to 0.

A real-life game show host will probably use a mixture of strategies, to keep up the suspense and make sure that nobody can work out how to cheat him. You can only work out if it's better for you to switch or to keep your door if you know precisely what mixture of strategies he uses.

[articulett]

Just to be clear articulett, if Monty randomly opens a door and it reveals a goat and he ALWAYS gives you the option to switch do agree that the probability of the car being behind your originally chosen door is now 1 in 2?

Giving Monty the choice of offering or refusing a switch obviously changes the problem.

Edit: The last sentence is incorrect. The odds are 1 in 2 regardless. I confused myself by thinking about Monty having the choice to either open a door or not.


No, if Monty always reveals a goat and always gives you a chance to switch
then you double your odds by switching from the 1/3 you had originally to 2/3. The odds are not 50/50 despite the fact that there are two choices. All he's done is increased the odds in favor of switching. Really. Do the math. If he chooses blindly then he's going to be taking away your choice when he chooses the car (forcing you to lose) and/or revealing the correct answer --and letting you switch guaranteeing you'll win. This will happen 1/3 of the time. If he chooses the goat (2/3 of the time)-- then you have information that tells you that 1 of the two options left is the car... that is the 50-50 thing. It would be expected to happen 2/3 of the time. See?

Knowing Monty's motives only tells you 2 things-- what are your odds of being offered a choice... and what are your odds that you increase your odds by switching. There's 2 factors... not one. They are co-dependent. In the scenario it's stipulated that he always shows a goat and you always get a choice. In that case, switching doubles your odds.

I think everyone can understand that if you keep your first choice every time no matter what you know about Monty's motives or whether you are offered a chance to change or not-- you will have 1 in 3 chance of winning a car (good odds). The question then becomes is there any way to increase your odds of winning the car. The only way to increase your odds is to change if given the choice. If you are always given a choice (even when you picked a goat or the host reveals a car) then you win 2/3 of the time by changing. No matter what the motives of the host.

If you aren't always given a choice, then it's a different game. But your odds don't increase from the 1 in 3 regarding the original choice. The only way you can increase those odds is by changing if offered the choice. Changing doesn't always increase your odds-- but if the host always shows a goat and always offers a switch, you will always increase your odds by changing. Increasing your odds does not mean that you are guaranteed to win... it just means that doing this consistently or amongst many trials or huge numbers of people-- you will win twice as much as if you don't switch.

I think that should suffice. This is true--basic probabilities. Ask any expert on the subject or run the figures yourself.

[AntiTelharsic]

No, if Monty always reveals a goat and always gives you a chance to switch then your double your odds by switching from the 1/3 you had originally to 2/3.

That is the standard version of the problem.

The odds are not 50/50 despite the fact that there are two choices.

They are 50/50 if Monty chooses randomly AND has revealed a goat, because his randomly choosing a goat happens 2/3 of the time, and your 1/3 is half of that.

[Michael C]

Maybe I'm misunderstanding someone's position here, but this is the way this dispute appears to me:
Articulett: chance of winning via sticking with the first choice no matter what happens is just P(first pick is car).
Others: incorporating the added information once an event has happened gives the conditional probability P(first pick is car|info), which is not equal to P(first pick is car)
I don't understand why there is such a case of miscommunication. Those two claims are perfectly consistent.

You're absolutely right. For me, the Monty Hall problem is clearly asking a question about the chances of winning at the point where Monty offers the switch. What I find interesting is that the distribution of probabilities between the two remaining doors does not depend solely on the condition that Monty revealed a goat, it also depends on how he came to the decision of which door to open.

[CFLarsen]

You did, Claus-- in post 535--in your everlasting quest to prove me wrong by knocking down a straw man version of what I actually said.

Nonsense.

I think anyone can see that if you always stick with your current choice--you will have 1 in 3 chance of winning-- the only way to increase those odds is to change your choice after a goat is shown. The host can always show a goat. The host does not always need to offer you the switch.

Yes, he does. That's the whole idea of the Monty Hall Problem:

The host knows what is behind the doors.

The host will offer to switch.

If the host doesn't know, and doesn't offer to switch, it isn't the Monty Hall Problem.

You are miseducating your students if you claim otherwise.

However, in the scenario--with no added information-- your odds increase by switching. Moreover, even if you add information-- you always have 1 in 3 chance of winning by not switching--that doesn't change with the added information. In your made up scenarios you are only switching whether the host is going to be offering you a switch or not.

That is not a "made up scenario". That is the Monty Hall Problem.

Because the only information we have as the scenario is laid out is that he's shown a goat and offered a switch. That is the only information in the scenario. But no matter how much you add information, you are never increasing the odds over 1 in 3 by staying. If the host only offers a switch if you have the car... then the information you get when if he asks you to switch, is that you have the car. But that means that in 2/3 of the cases, he'll never ask you to switch. And because you don't know this is the case, it's more likely that he's offering you to switch because he always does (the original scenario) or because he blindly chose a goat... (meaning that if he blindly chose the car... he'd not be offering you the chance to switch.)

The salient part of this exercise is that your odds if you stay are not more than 1 in 3-- ever. The only way to increase those odds is to switch if given the opportunity. (Of course 1 in 3 isn't bad odds)... and even with the 2/3 increased odds with a switch-- you still lose 1/3 of the time.

Please provide evidence that the Monty Hall Problem (http://en.wikipedia.org/wiki/Monty_Hall_problem) does not include these two:

The host knows what is behind the doors.

The host will offer to switch.

Evidence. Not your own explanations.

[articulett]

Now I see what was worrying me in what you said earlier. You said

It would be clearer to say
"The only way to change those odds is to switch if given the opportunity".

Obviously if your strategy is to choose a door and never to switch, you will, in the long run, win 1/3 of the games. But if your strategy is always to switch when given the chance, you may increase or decrease your odds, depending on what strategy Monty is using.

For instance, if Monty (unknown to you) only gives the opportunity of switching when there is a car behind your chosen door, the strategy of switching if given the choice will ensure that you never win: you have decreased your odds from 1/3 to 0.

A real-life game show host will probably use a mixture of strategies, to keep up the suspense and make sure that nobody can work out how to cheat him. You can only work out if it's better for you to switch or to keep your door if you know precisely what mixture of strategies he uses.

I was correct in saying that the only way you can increase your 1 in 3 chance of winning a car is by switching. This does not mean that you always increase your odds by switching-- it means that it's the ONLY way you can increase your odds from the 1 in 3 chance you have by staying. If Monty is only offering you the choice because you chose the car-- then your odds do not increase by switching.-- but he'd only be giving you that option in 1 out of 3 cases! In the other cases you'd just be revealed to be a loser without being offered a choice. Your odds of winning by keeping your first choice are never more than 1 in 3-- you are just changing whether you are being offered a chance to switch. Yes, Monty could be using such a strategy... and if you believe that, you will keep your first choice and win the car 1/3 of the time. If it's true, he won't offer you a chance to switch 2/3 of the time-- if it's not true, he will offer you a chance to switch and you won't do it, making you a winner 1/3 of the time. Your odds don't increase by keeping the door. You CAN increase your odds... but only by switching if given the opportunity.

[articulett]

Claus... the evidence IS saying what I"m saying. Go ask any expert. Your odds are never more than 1 in 3 by staying with your first choice.

If the host must give you a choice and must show a goat then you double your chance of winning the car by switching. Period. Your odds remain 1 in 3 if you keep your first choice. It never becomes 50-50 even though there are 2 choices.

You are wrong. Embarrassingly so. Again. As usual. No matter what the motives of the host-- you never have more than 1 in 3 chance of winning the car if you stick with your first choice given that the host must offer a choice and must show a goat. Yes, you can still win-- and you will-- 1/3 of the time. 2/3 of the time you'll be the sorry loser saying "I thought my odds were 50-50"

[CFLarsen]

Claus... the evidence IS saying what I"m saying.

OK - in the absence of evidence...

You mentioned that you regularly teach this to 15-year olds. How do you present this problem to your students?

I'm not talking about the explanation that follows.

I'm talking about how you describe the scenario and the rules, before you let the solve the problem.

Please write that down and post it here.

[articulett]

Exactly as you did Claus... that I will always show the goat and always offer a choice-- I use cups and little plastic cars and goats. There are also computer applets that they can use.

Like you, they think their odds are 50-50 when they have two choices left. But when they run the statistics they always learn that they win 1/3 of the time by keeping their original choice and 2/3 of the time by switching-- that is, they learn that despite their convicetion that their odds are 50-50 because there are two choices-- their odds of winning actually double from 1/3 to 2/3 by switching... the whole point of the lesson. The more trials they run the more clear these statistics become.

Even you could learn this.

[AntiTelharsic]

Articulett, are you addressing the case where Monty chooses randomly? I think that's what CFLarsen is addressing, not the standard problem. Apologies if I'm mistaken here, but I think that's where the disconnect is (I fear there are several different versions being simultaneously discussed).

[CFLarsen]

Exactly as you did Claus... that I will always show the goat and always offer a choice-- I use cups and little plastic cars and goats. There are also computer applets that they can use.

Like you, they think their odds are 50-50 when they have two choices left. But when they run the statistics they always learn that they win 1/3 of the time by keeping their original choice and 2/3 of the time by switching-- that is, they learn that despite their convicetion that their odds are 50-50 because there are two choices-- their odds of winning actually double from 1/3 to 2/3 by switching... the whole point of the lesson. The more trials they run the more clear these statistics become.

Even you could learn this.

Then, why are you going on and on and on about something that is not the Monty Hall Problem?

[articulett]

Then, why are you going on and on and on about something that is not the Monty Hall Problem?

because other people proposed other scenarios.

My first response to you was because you claimed that your odds change when you are shown the goat-- they don't-- the only thing that changes is that you increase your odds by switching. That's it. Then you denied saying that and kept with your quest that I was saying something false or wrong-- when I was not.

Then other people proposed various scenarios but no matter what the scenario, you never increase your odds above the 1 in 3 chance of having picked the car the first time no matter what is or isn't revealed or what choices you are and aren't offered. You are only gaining information as to whether it's better to switch if offered.

You can't make your offer more likely to be a winner by choosing it again. But it sure feels like you can. Choosing an option gives a person a conviction that wasn't there before the choice causing them to choose irrationally. It's a known psychological quirk of humans. Moreover, we see 2 choices and think 50-50. God or "no God "seems like a 50-50.

But is Xenu vs "no Xenu "a 50/50? Unicorns vs "no unicorns"? I have to teach my students how they get things wrong before I can teach them the methods for correcting it.

[CFLarsen]

because other people proposed other scenarios.

Then, stop telling people they are wrong about the Monty Hall Problem when they're not.

My first response to you was because you claimed that your odds change when you are shown the goat-- they don't-- the only thing that changes is that you increase your odds by switching. That's it.

Whatever gave you the idea that I was talking about other scenarios than the Monty Hall Problem?

Then other people proposed various scenarios but no matter what the scenario, you never increase your odds above the 1 in 3 chance of having picked the car the first time no matter what is or isn't revealed or what choices you are and aren't offered. You are only gaining information as to whether it's better to switch if offered.

Nobody has argued otherwise.

[articulett]

Don't give me advice Claus... I would never take advice from someone I would never want to be like. You are the one who told me I was wrong when I was not. I never told someone they were wrong, when they were not.

My words are there for everyone to see. If I said something wrong, it's easy to highlight it... it should also be easy to show that I am wrong with an opposing version on the page you linked.

I didn't think YOU were talking about anything different-- I was only answering your allegations that I was wrong in saying that your odds don't change from 1 in 3 just because a goat was revealed. You contended that they do. That makes you wrong.

I quoted your words in post #535. You accused me of being wrong-- you were wrong. And now you are trying to weasel your way out of your wrongness by tsk-tsking me. I was never wrong in any of my statements. You were wrong... as you often are-- and yet, I've never seen you admit it. Instead you attack the people who reveal you gaffes and change the subject and pretend you have nothing to do with the derail.

[CFLarsen]

That's precisely what the "Monty Hall problem" shows: That your odds do change when you pick a certain door.

Wrong. The odds of the car being behind the door you originally chose is never more than 1/3. It doesn't change. The car always has a 2/3 chance of being behind one of the other doors.

I was talking about the Monty Hall Problem.

If you focus on the first part and leave out the parts where the host opens the door to the goat and asks the person if he wants to switch, then you are not talking about the Monty Hall Problem.

The Monty Hall Problem isn't the Monty Hall Problem without these two things. Without them, it's just a trite probability question.

[articulett]

Yes... and you are still wrong. Your odds don't change when you pick the door. Your odds of being correct are 1 in 3 if you keep the door--the same as if you never made the switch... and 2 in 3 if you switch.

Thanks for finally admitting your error and for highlighting it so everyone else can see.

[CFLarsen]

Yes... and you are still wrong. Your odds don't change when you pick the door. Your odds of being correct are 1 in 3 if you keep the door--

That isn't the Monty Hall Problem. That's just a trite probability problem.

the same as if you never made the switch... and 2 in 3 if you switch.

That is the Monty Hall Problem.

You can't talk about the first part, leave out the other, and claim you are talking about the Monty Hall Problem. You are not.

[articulett]

I'm talking about the Monty Hall problem the whole time. You never increase your odds from 1 in 3 by keeping the same door. You can only increase your odds by changing. If you don't agree with this-- you are wrong.... if you don't understand this... your English is the problem. If you doubt it-- ask someone you trust that is smarter than you on one or both topics.

[AntiTelharsic]

Thing is, there's more than just the classic Monty Hall Problem being discussed here. Since the original statement has been interpreted by some to be ambiguous -- with Monty not necessarily revealing a goat each time, but having simply revealed a goat that time -- they have put forth for consideration the situation where Monty chooses randomly. But to keep it consistent with the original statement they have been considering only those situations where Monty chose randomly but revealed a goat.

Certainly, in the classic Monty Hall problem, switching gives you a 2/3 chance of winning.

But in the variant which is also being discussed, switching gives you a 1/2 chance of winning (for rationale, see my and others people's earlier posts; for experimental proof see Vorpal's program or I could post mine or someone else could write another).

Since the argument seems to be between 2/3 and 1/2, I wonder if people aren't arguing about those two different scenarios.

To help clarify:

Is there anyone here who believes that in the classic Monty Hall Problem (You pick a door, Monty always opens a goat door, you're always allowed to choose the remaining door) you DO NOT get a 2/3 chance of winning if you switch?

I'm not convinced anyone here believes that. I think many people here are failing to clearly specify which version they are talking about.

Personally, I don't find the Monty-choosing-randomly variant very interesting, but it does appear to have been contentious here, so it's worth being careful to specify which variant one is addressing.

[Herzblut]

To help clarify:

Is there anyone here who believes that in the classic Monty Hall Problem (You pick a door, Monty opens a goat door always, you're allowed to choose the remaining door) you DO NOT get a 2/3 chance of winning if you switch?
No. There is no need to clarify the obvious.

It was in Egypt, I think, where they recently found a 5.000 years old papyrus bobbin. Guess what, the hieroglyphics revealed exactly this explanation!

[articulett]

Yes, A-T, I agree... my explanation at first was geared towards Claus because he said you odds change in the middle of the game... they don't.

And then other people brought up other scenarios... but they either involved the possibility that a car could be shown (rather than a goat) and/or the possibililty that you wouldn't be allowed to make a choice... so that changes the game... it doesn't change the fact that you have a 1/3 chance of winning the car by staying with your first choice.

More here:

http://www.shodor.org/interactivate/activities/
http://www.shodor.org/interactivate/activities/SimpleMontyHall/
http://www.cut-the-knot.org/probability.shtml

(interactive on probabilities.... if you are really curious I'll be tossing punnett squares up there.)

[CFLarsen]

Yes, A-T, I agree... my explanation at first was geared towards Claus because he said you odds change in the middle of the game... they don't.

That's precisely what they do: When you are given the choice of changing doors.

[articulett]

What odds change, Claus? Your odds are the same - 1 in 3 if you don't change and your odds are alway 2 in 3 if you do. Period. That doesn't change in the basic Monty Hall problem. Your odds of winning the car do not suddenly become 50-50 just because there are two choices left.

Saying it, doesn't make it so.

[CFLarsen]

What odds change, Claus? Your odds are the same - 1 in 3 if you don't change and your odds are alway 2 in 3 if you do. Period. That doesn't change in the basic Monty Hall problem.

The odds change when you are given the option of switching doors.

You simply don't understand the Monty Hall Problem, if you don't understand this. This is the key element in the problem.

Your odds of winning the car do not suddenly become 50-50 just because there are two choices left.

Saying it, doesn't make it so.

Huh? Where did I say that the odds suddenly became 50-50?

[sol invictus]

Maybe I'm misunderstanding someone's position here, but this is the way this dispute appears to me:
Articulett: chance of winning via sticking with the first choice no matter what happens is just P(first pick is car).
Others: incorporating the added information once an event has happened gives the conditional probability P(first pick is car|info), which is not equal to P(first pick is car)
I don't understand why there is such a case of miscommunication. Those two claims are perfectly consistent.

Those two statements are correct (and therefore consistent). I repeated over and over to articulett that no one was disputing that the odds are 1/3 before taking into account the information from Monty opening the door, but that after taking it into account the odds change. She said I was wrong, and that the odds never change.

It's a bit like arguing with someone that says "the weather is clear" because it was clear on April 4th 1963, and when you point out its raining outside responds "the weather never changes" because, indeed, the weather on Aril 4th 1963 never changes.

[articulett]

Can you please get a smart person to confirm or can you cut and paste from the article that says your odds change when you are given the option of switching doors.

If you can't do this... then clarify what you mean. Your odds change from what to what? And you say this happens when he asks you "do you want to switch doors", correct? What does this mean? Does this mean anything to anyone besides Claus? What is it your odds change to? From what? Given what reasoning?

And you think you are an expert on this problem? Claus, has anyone ever given you a reason to feel like you have expertise on this problem? Has anyone other than you actually said your odds change when you are given the option of switching?

If not 50-50-- what odds are changing to what exactly... give me numbers please.

My claim is that your odds never switch. That you always have 1/3 a chance of winning when you don't switch and 2/3 a chance of winning if you do. Always-- when he gives the option and before he gives you the option.

You agree that you have a 1/3 chance of winning if you were never given a choice, right? How does his showing you the goat, change the odds that the car is behind your door-- that is, how does it change the odds that you should keep your first choice-- What do you think the odds change to and what choice would you make with this new information?

I say, it doesn't change the odds of your first choice being a car- it just lets you know that the remaining door is less likely to be a goat. Do you agree? Do you agree that the Monty Hall problem shows that, contrary to intuition, you double your odds by switching (from 1/3 to 2/3)? If not, you are wrong. If so, then it's a communication problem.

In either case, I have not been wrong-- but you derailed this thread to claim that I was --and alluded to your imagined expertise on the topic claiming that it was me who didn't understand something. This is something you do quite frequently. I don't think anyone responds well to such behavior.

[sol invictus]

And then other people brought up other scenarios... but they either involved the possibility that a car could be shown (rather than a goat) and/or the possibililty that you wouldn't be allowed to make a choice... so that changes the game... it doesn't change the fact that you have a 1/3 chance of winning the car by staying with your first choice.

Yes, articulett, it does. Look - in any version of the game, if it's played over and over again and you always stick with your first choice, the number of times you win will be approximately 1/3.

No one is disputing that.

However, in the version where Monty does not know where the car is and opens a goat door, your odds of winning by staying CHANGE to 1/2. That statement is not in conflict with the one above - the information provided by Monty allows you to narrow down which type of game you're likely to be in (it tells you you're not in a game where the car is behind door 2, if that's the one Monty opened) and taking that into account CHANGES THE ODDS.

[sol invictus]

Can you please get a smart person to confirm or can you cut and paste from the article that says your odds change when you are given the option of switching doors.

I think what CFLarsen is saying - and he is free to correct me if not - is that in the standard problem the odds the car is behind door #3 (if, say, you picked #1 and Monty opened #2) change from 1/3 to 2/3 when Monty opens door #2, and hence your odds of winning by switching to #3 also change from 1/3 to 2/3.

In any case that is a correct statement.

[CFLarsen]

Those two statements are correct (and therefore consistent). I repeated over and over to articulett that no one was disputing that the odds are 1/3 before taking into account the information from Monty opening the door, but that after taking it into account the odds change. She said I was wrong, and that the odds never change.

I have no idea why articulett is stuck in the first part of the scenario. Nobody has argued that the chances aren't 1 in 3 of choosing the car at the beginning. Nobody.

What articulett doesn't understand is that the changing of the odds is the whole point of the Monty Hall Problem.

It is easy to determine what the odds are at the beginning. You have three doors, one of them have a car, what are the chances of you choosing the door with the car? 1 in 3. Any kid can do that. But that is just a trite probability problem.

The point of the Monty Hall Problem is that the conditions change, from the time where you have to choose between 3 doors, to the time where you have chosen one, are told what is behind one of the remaining doors - and given the option to switch.

I can understand why her students have such a hard time with this. Not because it is a tough nut to crack, and certainly not because some of them are boys. She is just really, really bad at explaining even a simple setup. Her explanations are way too long and unnecessarily complex. Instead, they are downright confusing.

A lengthy stream-of-consciousness is not a substitute for a clear explanation. Complexity is never preferable to clarity - especially when it comes to logical problems like this one.

[CFLarsen]

I think what CFLarsen is saying - and he is free to correct me if not - is that in the standard problem the odds the car is behind door #3 (if, say, you picked #1 and Monty opened #2) change from 1/3 to 2/3 when Monty opens door #2, and hence your odds of winning by switching to #3 also change from 1/3 to 2/3.

In any case that is a correct statement.

Yes, yes, yes. The odds change, because the conditions change.

When teaching logic, it is bad paedagogy to complexify. Instead of teaching students the issue, you confuse them, thereby making them think that they are stupid and you are clever. That's not teaching, that's just trampling on your students.

[articulett]

That's what YOU Are doing CLAUS! Recall-- I said this:

A funny thing about humans is that they tend to go with their first choice-- as soon as they place a bet they tend to see their odds as increasing-- they feel more confident with their choice... but your odds don't suddenly change when you place your bet or pick your door... your feeling that you are more likely to win, does not make you more likely to win. This is true even with lottery winners. Before playing, they may correctly assess their odds-- but once they've committed, they "feel" as if their odds go up.


You responded quoting the bold with this challenge: That's precisely what the "Monty Hall problem" shows: That your odds do change when you pick a certain door.

I was clearly talking about your first choice... Your odds are never better than 1 in 3 that you have chosen the car when he shows you the goat--so your odds can never be better than 1 in 3 by staying with your first choice. You replied that your odds DO change. You either changed the situation when you found out you were wrong and decided to add more pedantry to cover for your anger at my showing you to be wrong-- or you didn't pay attention in the first place in your eagerness to prove me wrong. MY statement above IS correct. Your response that is specifically in regard to that statement is NOT correct... it does not imply the sol invictus scenario that you latched onto to save your butt--but I can see why you would want to claim it does... because then it looks like a language problem rather than the fact that you are being exactly the kind of person you are lecturing others (specifically me) not to be.

If you had an ounce of character you'd admit it and apologize.

[CFLarsen]

Where did I say that the odds suddenly became 50-50?

[sol invictus]

I was clearly talking about your first choice... Your odds are never better than 1 in 3 that you have chosen the car... not in any case. You replied that your odds DO change.

articulett - he's talking about the standard problem, and the odds you will win if you switch to the door Monty doesn't open. Not the non-standard problem, and not the odds you will win if you stay.

He didn't make that very clear, in part because he is a bit of a troll (judging by his posting history). But he is correct in this case.

[articulett]

My students don't have a hard time. I have about 4 boys out of a hundred every year that sound like you however... but the other students and the trials... eventually shut them up. Some still stay mad because they don't like to know that they can be wrong. But most are glad to learn the way their brain fools them.

BTW, your response to Sol makes no sense... so NOW the conditions change, eh?-- Before it was the odds... Neither make sense. That's why you can't find such a quote.

The only thing that changes is that you now know that the door with the goat that he revealed has 100% chance of not being a car-- leaving the remaining door with a 2/3 chance of being a car. That's it.... that's the only thing that changes. Not the odds of your first choice being the winner. Not the odds that you should stay or switch... not the "conditions" as you now claim... just the fact that you now know the car is NOT behind that door.

[articulett]

articulett - he's talking about the standard problem, and the odds you will win if you switch to the door Monty doesn't open. Not the non-standard problem, and not the odds you will win if you stay.

He didn't make that very clear, in part because he is a bit of a troll (judging by his posting history). But he is correct in this case.


Correct in what way... that your odds change? What odds change to what exactly. I maintain that your odds are alway 1/3 by keeping it (and I'm talking the standard scenerio) and always 2/3 by changing it. Are you saying something different? If so-- you are wrong.

[articulett]

Where did I say that the odds suddenly became 50-50?

Why don't you clarify Claus...

I said that you never have more than a 1/3 chance of being correct by switiching... You claimed the odds change when he offers the choice.... what odds change to what, Claus?

I claim that even before he shows you a goat your odds of winning the car are doubled (in the classic scenario) if you switch-- this doesn't change when he shows you the goat... the only thing that changes is you now are left with the option you must switch to in order to get those 2/3 odds.

So what odds are changing when he reveals the goat-- You claimed that the odds change... and then you claimed that the condition changes... I claim that the only thing that changes-- is that you can no longer choose the door the host revealed-- that doesn't change any odds or any conditions as far as I can see. If that is what you meant-- then you've expressed it so poorly that I doubt you could convey the problem to anyone. I can and have conveyed it to many. I am used to your kind of pettiness and denial-- but usually it's coming from 15 year old boys not grown men.

I have all Freshman this fall (14 year olds). I suspect the majority of them will be able to understand the Monty Hall problem better than you AND communicate it better to others.

[CFLarsen]

My students don't have a hard time. I have about 4 boys out of a hundred every year that sound like you however... but the other students and the trials... eventually shut them up. Some still stay mad because they don't like to know that they can be wrong. But most are glad to learn the way their brain fools them.

Ah, OK. Now, it is only a few "boys". Take a look at your earlier blanket statement:

Yes Claus, I read it. I do this demonstration in my class every year, and I'm used to 15 year old boys sounding just like you.
...
I actually love teaching this stuff, because there's always egotistical young men in class who cannot believe that their brains can fool them. And they are the easiest to fool and their egos get in the way of their understanding this. But it's a valuable lesson to learn the ways your brain fools you. And it's fun, because as each student has an "aha" moment they begin to try and teach the blowhards... we can recreate with cups and keep track of the odds--

I'm right. You're wrong. Again. Deal with it.

Why do you single out boys? Do you have any evidence that girls are better at logic than boys?

BTW, your response to Sol makes no sense... so NOW the conditions change, eh?-- Before it was the odds... Neither make sense. That's why you can't find such a quote.

The only thing that changes is that you now know that the door with the goat that he revealed has 100% chance of not being a car-- leaving the remaining door with a 2/3 chance of being a car. That's it.... that's the only thing that changes. Not the odds of your first choice being the winner. Not the odds that you should stay or switch... not the "conditions" as you now claim... just the fact that you now know the car is NOT behind that door.

Where did I say that the odds suddenly became 50-50?

[articulett]

Yes, articulett, it does. Look - in any version of the game, if it's played over and over again and you always stick with your first choice, the number of times you win will be approximately 1/3.

No one is disputing that.

However, in the version where Monty does not know where the car is and opens a goat door, your odds of winning by staying CHANGE to 1/2. That statement is not in conflict with the one above - the information provided by Monty allows you to narrow down which type of game you're likely to be in (it tells you you're not in a game where the car is behind door 2, if that's the one Monty opened) and taking that into account CHANGES THE ODDS.


Yes, but that's not the classic story... in Claus' scenario-- Monty always reveals a goat and always gives you a choice.

In the blind scenario-- the host reveals a CAR 1/3 of the time-- so he either takes away your choice or gives you a 100% of a chance of winning by allowing you to choose... so either way... you have a 1/3 of a chance of winning the car by remaining with your first choice (that's average of 0% of 1/3 of the time and 50% of 2/3 of the time).... see? (conversely you have a 2/3 chance of winning by switching... 100% of the 1/3 of times he shows you the car and 50% of the 2/3 of the time he reveals a goat.) You can't really throw out the fact that 1/3 of the time-- he will be revealing a car if he's choosing blindly.

But Claus gets mad if we are talking about different scenarios like that. We are only talking about the scenario where the host always shows a goat and always gives a choice... Claus is very clear that we are ONLY talking about that scenario.

I maintain it's also the same odds when you don't know if this is the case or not... but that's another chapter...

http://www.shodor.org/interactivate/activities/SimpleMontyHall/
that's the simple applet... the door you choose first has the pointer on it...

You can test the more advanced scenarios here:
http://www.shodor.org/interactivate/activities/AdvancedMontyHall/

[articulett]

I'm not singling out boys... in my years of teaching... it's the boys who are always angry that their conviction has been proven wrong... not all the boys.... but only the boys. The girls are embarrassed, but they never had the conviction in their rightness the boys have. That's just my experience.... Every year I have 3 or 4 students that are "like you"-- very resistant and angry-- and they are always boys... But when we run the tests.... they can't deny the numbers. But they will be angry at me for exposing the fact that they can be fooled... it's just like the guy in Randi's class when he exposes the astrology trick (in his Nova program)... instead of being humbled by the knowledge, they are mad at the messenger. It's rare... but it's less rare in boys then in girls.

I never made a blanket statement. I've always maintained that it's just a few boys....I'm not responsible for what you add to my words in your head.

You never said 50-50... you said your odds changed... and in my experience most people who believe the odds change, believe that the odds become 50-50 because there are 2 choices.

YOU SAID THE ODDS CHANGE-- SO TELL US-- WHAT ODDS CHANGE TO WHAT? Moreover you said the odds change WHEN he offers you a choice. On top of that you said this in RESPONSE to my statement that your odds are never better than 1/3 for the first choice... it's always 1 in 3 if you keep the first choice.

If you don't answer what you meant when you said the odds change then I will take it as an admission of your error in response to my words. I was right-- you claimed I didn't understand the problem.

I recognize your straw man as an attempt to throw everyone off of the fact that YOU WERE WRONG and you can't admit it.

You make allegations about me all the time because you are angry that I revealed your error. But I'm not claiming that I am the only person you do this to.

[sol invictus]

Correct in what way... that your odds change? What odds change to what exactly.

I told you that very clearly in the post you quoted. If you can't be bothered to even skim the posts you respond to, there's no point in continuing this.

[CFLarsen]

Yes, but that's not the classic story... in Claus' scenario-- Monty always reveals a goat and always gives you a choice.

It's not my scenario. It's the Monty Hall scenario.

Don't make it look as if I am inventing special conditions, or describing special scenarios. I am describing the Monty Hall Problem.

I'm not singling out boys... in my years of teaching... it's the boys who are always angry that their conviction has been proven wrong... not all the boys.... but only the boys. The girls are embarrassed, but they never had the conviction in their rightness the boys have. That's just my experience.... Every year I have 3 or 4 students that are "like you"-- very resistant and angry-- and they are always boys... But when we run the tests.... they can't deny the numbers.

But "not singling out boys"... :rolleyes:

Where did I say that the odds suddenly became 50-50? Are you ever going to provide evidence of that claim?

[sol invictus]

I claim that even before he shows you a goat your odds of winning the car are doubled (in the classic scenario) if you switch-- this doesn't change when he shows you the goat...

That doesn't make any sense. Switch to what? If you switch to a random door (your only option, since Monty hasn't yet shown you anything) your odds remain at 1/3.

[articulett]

I agree... but the only scenario you want to discuss is the one where he always reveals a goat and he always gives a choice... That IS the classic problem... and it is the only one you want to discuss... You showed anger at me when I discussed others because other people brought up such scenarios. I maintain that my statements are true rather it's the classic scenario or you don't know what scenario you are in. They are true for every case that you would agree with-- plus others.


And I already said that you never claimed 50 -50. It was my presumptions because you said the odds change and that is what everybody else has meant when they say the odds change... they think your odds suddenly become 50-50. You never explained what you meant by "the odds change"-- though you insisted that "the odds changing" is what the Monty Hall problem is about.

So tell us what you meant when you claimed that the the precise thing the Monty Hall problem shows is that the odds change when the host gives you a choice. What the hell did you mean? You said it in response to my bolded statement above. If you didn't mean that the odds suddenly become 50-50-- What DID you mean?? I can think of no way to interpret that statement in response to my statement. You have never clarified. And I think I know why. It's because your odds DON"T change-- my statement was correct and your pedantry was revealed for the sophistry it was.

No one but you thinks I singled out boys... I made it quite clear that in my experiences most students aren't belligerent during this lesson... but the ones that are-- have always been boys. I can't change those facts... and revealing those facts is not singling out boys even if you (singular) imagine it is. Even with the little smiley it doesn't.

[articulett]

That doesn't make any sense. Switch to what? If you switch to a random door (your only option, since Monty hasn't yet shown you anything) your odds remain at 1/3.

Yes but he must show a door and the door will be a goat... and so your odds are always doubled from 1/3 (by staying) to 2/3 by switching. You don't have to know what door to switch to... just that you will switch to the door he doesn't open--because you can't choose the one he does (nor would you want to... since, in the situation stipulated, it's always a goat.) The only way it can always be a goat is if the host knows which door to choose (or which doors to choose from). The host cannot always reveal a goat if he's choosing blindly.

Try the links... you'll see...

[sol invictus]

Yes but he must show a door and the door will be a goat... and so your odds are always doubled from 1/3 (by staying) to 2/3 by switching. You don't have to know what door to switch to... just that you will switch to the door he doesn't open--because you can't choose the one he does (nor would you want to... since in the situation stipulated, it's always a goat.)

No offense, but I'm going to bow out of this exchange now. As far as I can tell there is zero information going in either direction.

[articulett]

That's fine. I'm done too. I provided links... you can see the classic odds in the first link...

(you can play as many versions of the game as you want) Or you can try the alternating scenarios (blind host, etc. in the second link.

I realize that Claus will never explain what he means by saying your odds change when the host reveals a goat. The only thing that changes.... is that you can't or won't pick that door (the one the host revealed)... that means the remaining door has 2/3 of a chance of having the car behind it. Your original choice still has 1/3

[CFLarsen]

That's fine. I'm done too. I provided links... you can see the classic odds in the first link...

(you can play as many versions of the game as you want) Or you can try the alternating scenarios (blind host, etc. in the second link.

I realize that Claus will never explain what he means by saying your odds change when the host reveals a goat. The only thing that changes.... is that you can't or won't pick that door (the one the host revealed)... that means the remaining door has 2/3 of a chance of having the car behind it. Your original choice still has 1/3

Where did I say that the odds suddenly became 50-50? Are you ever going to provide evidence of that claim?

[RecoveringYuppy]

Try the links... you'll see...

FWIW I'm sure he sees. I'm fairly sure Claus gets it too, but I haven't read every detail in his posts. It appears to me you (articulett) are being obtuse in your wording in order to inject "Claus is wrong" phrases.

It's obtuse to object to someone saying that conditions change, or that probabilities changes as conditions (known information) changes.

It's also obtuse to object to someone pointing out what is and isn't the Monty Hall problem. As with this thread, confusion over a nonintuitive idea isn't helped by a misstatement of the problem.

Take a break.

[GreyICE]

Articulett vs. Claus round... god. What are we on now?

Claus, you are one of the least likable human beings its ever been my distinct displeasure to have discourse with on the internets. I don't know what it is. We have many opinions in common, you seem reasonably well educated, you don't troll, flame, or otherwise behave blatantly poorly, and therefore I am at a loss to explain why, three posts into any thread you're posting in, I want to strangle you with a garden hose. Maybe it's the fact that you back down even less frequently than I do, and have a bulldozer-like obstinacy combined with an unerring knack for annoying even people who agree with you.

Articulett, you're ornery, have no capacity to admit you're wrong, frequently post long rambling posts that go nowhere, are one of the most passive-aggressive people I've ever interacted with, have the logical capacity of a rhubarb, and generally have all the sense of proportion of your average 16 year old.

Therefore, I've got to ask, since we can basically assume that the two of you will disagree just to be disagreeable, can we just manually insert 'three pages of articulett and claus arguing over some nonsense point that adds nothing to the discussion' at the bottom of pretty much every OP and then we don't have to, I don't know, do it?

[Michael C]

Things to get straight:

1. The odds of having picked the correct door the first time never change.

2. The odds of winning can change at some point in the game.

In the case of the "classic" Monty Hall problem, the odds of winning do not change at the point where Monty opens the door with a goat: they are still 1/3. They can change at the point where Monty offers the choice of switching doors, if the contestant chooses to switch.

In the case of the variant of the problem where Monty chooses a door at random and there happens to be a goat behind it, the odds of winning do change when Monty opens this door: they change from 1/3 to 1/2.

[sol invictus]

Things to get straight:

1. The odds of having picked the correct door the first time never change.
<snip>
In the case of the variant of the problem where Monty chooses a door at random and there happens to be a goat behind it, the odds of winning do change when Monty opens this door: they change from 1/3 to 1/2.

These two statements appear to contradict each other. At best, they are sufficiently imprecise that one can't tell for sure.

Statement 1 is correct only if:

a) it's intended to apply only to the standard variant, or

b) you mean the odds with no conditions or extra information from opening doors applied - in which case it's a tautology, since the odds trivially cannot change.

It's extremely confusing to say "the odds never change" when you're talking about something which by definition cannot change. It's like saying "the weather at noon on April 12 1963 never changes".

[Michael C]

These two statements appear to contradict each other. At best, they are sufficiently imprecise that one can't tell for sure.

Statement 1 is correct only if:

a) it's intended to apply only to the standard variant, or

b) you mean the odds with no conditions or extra information from opening doors applied - in which case it's a tautology, since the odds trivially cannot change.

It's extremely confusing to say "the odds never change" when you're talking about something which by definition cannot change. It's like saying "the weather at noon on April 12 1963 never changes".

I mean "b": the opening odds. Yes, I agree that it's confusing to say "the odds never change" if we are talking about something which by definition cannot change. So in statement 1, I agree that I am stating the obvious.

This, however, seems to be the main point of Articulett. It confused me until I realised that she really was talking about the beginning odds of 1/3, which by definition never change, and I was talking about the odds of winning calculated at a given point in time, which change throughout the game. The odds of winning start at 1/3, and at the very end they are either 1 or 0; in between they may change to 2/3, or 1/2, or something else, depending on what version of the game Monty is playing and what choices you make.

[Herzblut]

These two statements appear to contradict each other. At best, they are sufficiently imprecise that one can't tell for sure.

Right. Articulett doesn't know statistics or statistical lingo, hence she's talking in mysterious terms like "the odds I have chosen noodles for dinner yesterday". This might help her to understand a problem, but it doesn't allow for a meaningful discussion.

What she arguably means, in proper terms, is the marginal probability or prior probability. Which is, of course, not what the problem is all about. It's about optimizing winning chances in a concrete game situation. This is what playing a game is about anyway.


Conditional probability is the probability of some event A, given the occurrence of some other event B. Conditional probability is written P(A|B), and is read "the probability of A, given B".

Marginal probability is then the unconditional probability P(A) of the event A; that is, the probability of A, regardless of whether an event B did or did not occur.

Joint probability the probability of two events in conjunction. That is, it is the probability of both events together. The joint probability of A and B is written http://upload.wikimedia.org/math/b/5/d/b5d8f79c12183ae00f4ace6c94d71795.png or http://upload.wikimedia.org/math/0/f/a/0fa667edc39216b95b3e7f3719a73a11.png
http://en.wikipedia.org/wiki/Conditional_probability


A prior probability is a marginal probability, interpreted as a description of what is known about a variable in the absence of some evidence. The posterior probability is then the conditional probability of the variable taking the evidence into account.
http://en.wikipedia.org/wiki/Prior_probability_distribution

The problem in question, of course, tasks us with determining a posterior/conditional probability, the probability of winning by switching or not switching, given Monty has opened a goat door. Obviously, the prior probability of winning by not switching is 1/3. I've shown how to calculate the posterior probability by simple, clear mathematics. I encourage the reader to have the heart to take a look at the mathematical concept, which is very simple and elegant, two lines of calculation replace 1000 lines of needless discussion.

For further discussion, if any, we might use the correct terms prior probability (before opening a door) and posterior probability (after a door has been opened), because they are clearly defined and nicely descriptive.

[Rolfe]

Articulett vs. Claus round... god. What are we on now?

Claus, you are one of the least likable human beings its ever been my distinct displeasure to have discourse with on the internets. I don't know what it is. We have many opinions in common, you seem reasonably well educated, you don't troll, flame, or otherwise behave blatantly poorly, and therefore I am at a loss to explain why, three posts into any thread you're posting in, I want to strangle you with a garden hose. Maybe it's the fact that you back down even less frequently than I do, and have a bulldozer-like obstinacy combined with an unerring knack for annoying even people who agree with you.

Articulett, you're ornery, have no capacity to admit you're wrong, frequently post long rambling posts that go nowhere, are one of the most passive-aggressive people I've ever interacted with, have the logical capacity of a rhubarb, and generally have all the sense of proportion of your average 16 year old.

Therefore, I've got to ask, since we can basically assume that the two of you will disagree just to be disagreeable, can we just manually insert 'three pages of articulett and claus arguing over some nonsense point that adds nothing to the discussion' at the bottom of pretty much every OP and then we don't have to, I don't know, do it?


Rolfe nods enthusiastically, entirely in agreement with GreyICE on every point.

[Rolfe]

At the risk of repeating myself.... (well, the whole thread is on a repeat cycle I suppose)

According to Wikipedia, the first publication of the puzzle dates back to 1975. (When did the Monty Hall TV show start, does anyone know?) The exact wording of that publication isn't stated, but going from CurtC's and Claus's insistence that the "Monty will always open one of the two unchosen doors and will always reveal a goat" version is The Only True Authentic Monty Hall Puzzle, I'm assuming that perhaps it did make this clear.

However, as far as I can see, the perennial fascination owes a great deal to the fact that subsequent wordings of the scenario have become more ambiguous. I can't see how a puzzle clearly and precisely defined as above could be anything more than a nine-day wonder.

Certainly, when I first encountered the puzzle in 1994 or thereabouts, the wording was similar to the OP, and arguably ambiguous. And then, as now, the bulk of the argumentation seemed to revolve around exactly what rule Monty is to be supposed to be working to. And to be honest, I think this raises the debate to a whole new level.

If we just consider what Claus and CurtC and others call the "classic Monty Hall" scenario, yes, it's intriguing, and counterintuitive, but the answer is perfectly clear once you've got your brain around it. I think it took me a day or two, just puzzling at it on my own, to get there. Really, really, even though there might be some dimwits who will never get it, there isn't a debate. Switching doubles your chances of getting the car, full stop. End of.

All everybody has been arguing about for most of this thread is interpretation. Exactly what game are we playing anyway?

I know how I got on to that aspect, and I think it's how many people have got there. I imagined the 100-doors version (99 goats and a car), while I was thinking about the basic puzzle. This illustrates how the odds change during the process, but it also clarifies the need to know what the exact rules are.

Contestant chooses a door. Monty opens another, a goat. Switch? Wouldn't make much difference, you've still got 98 other doors. But as the process is repeated and repeated, the situation becomes much clearer. The longer the car remains unrevealed, the clearer it becomes that Monty is deliberately avoiding it. By the time there are only two doors left, two things are obvious. The probability that Monty is deliberately avoiding the car is 98/100. And it should be completely intuitive that you should switch. Because the probability that the car is behind the door you originally chose is and always was 1/100, therefore the probability that it is behind the other door is 99/100.

For me, it was this exercise that clarified the need to stipulate the exact rules. Because in the 100-doors version, if Monty was opening doors at random, with no more idea than you have of where the car is, then in the overwhelming majority of games, he'd have revealed the car behind one of the other doors way before you're left with only two closed doors.

And once you think about it, that is also a reasonable interpretation of the wording of the puzzle as usually presented. And, indeed, a reasonable way of running a game show.

Contestant picks a door. Monty, who has no idea where the car is, opens one of the other two doors at random. A third of the time he reveals the car. Oh too bad sir, thanks for playing, good game, hope you and your goat will be very happy together. Two thirds of the time he reveals a goat. Well sir, so far so good, but would you like to change your choice? In that situation, switching makes bugger-all difference.

Cue argument that lasts for (to date) 33 years.

But then it gets worse. Now that the matter of the exact rules has been opened to scrutiny, people start getting even more creative. What if Monty is a complete bastard, and will only open the door if he knows you've already got the car? Well, unlike the random-opening scenario, that is neither a sensible game show, or a reasonable brain-teaser. As one of a range of possibilities incorporated in my next scenario (capricious Monty) it has a place. However, as a consistent stratagem, it's a nonsense. It's just a hypothetical scenario that is dragged in for the sake of argument.

But what if Monty is totally capricious? Maybe he'll help some contestants and be a bastard to others. Maybe he has a different scenario in his head every time the game is played! From what I've heard from those who have watched the original show, this may well have been the actual case. And presumably it was a perfectly viable game show.

However, it's not a perfectly viable brain teaser. To have any validity as an abstract puzzle, there has to be consistency. To put the question to anyone and then say, well, there's no right answer because Monty is entirely capricious, isn't a lot of fun.

For these reasons I maintain that it is reasonable to exclude both the "Monty-is-a-bastard" scenario and the "Monty-is-capricious" scenario. No game show could possibly persist if the former scenario was being operated, and no rational brain-teaser could employ the latter.

Nevertheless, what I will call the "B" scenario, the one where Monty doesn't know where the car is but simply opens one of the two doors the contestant hasn't chosen, is valid as a brain teaser and as a game show, and is in accordance with the usual, ambiguous, wording of the puzzle. (I'm designating the "real" Monty Hall puzzle, where he knows where the car is and will avoid opening that door at this stage, as the "A" scenario.)

This is the conundrum I find completely and utterly fascinating.

The puzzle is as presented. It is not clear whether the A or the B scenario is intended. (That is, you deduce that Monty opening one of the doors you didn't choose is a prerequisite, but you don't know whether or not he knows where the car is.) Where does that leave you?

It leaves you in the peculiar position where the correct answer depends entirely on what is going on in the mind of the host. And in my experience it is that very surreal situation that causes the most outcry from those who have difficulty with the puzzle.

I can only really get my brain round it by imagining that I'm playing one of two computer-game versions. One is programmed with the "A" version, and the other with the "B". I've got to the point where one of the doors I didn't choose has been opened, and there is a goat there, but I still don't know which one I'm playing. If it's the "A" version, I know that switching will double my chances of winning, but if it's the "B" version, I know that switching won't make a blind bit of difference.

Seems sensible enough when looked at that way.

However, I'm not playing with computers. There is an actual person there. Monty just opened a door to reveal a goat. Will switching improve my chances of winning? Well, the answer depends entirely on whether or not Monty himself knows where the car is.

Surreal, what?

Rolfe.

[articulett]

I want to point out something that might clarify it for those who haven't got it.

If the host is "blind" (doesn't know what he's choosing)-- then you are not doing the Classic Monty Hall problem (which stipulates that a goat is always shown.)

The same is true in the case where Monty only offers the choice when you have the car (because the classic situation says he always offer the choice.)

However, whether you are involved in the classic Monty Hall situation or have no knowledge as to which of the 3 options you are dealing with-- on average, your odds double from 1/3 to 2/3 if you are offered to switch and do so. If the host is in one of the latter 2 categories-- you will either have 1/3 of a chance of having your choice taken away because the host accidentally reveals a car (blind Monty) OR you will have a 2/3 chance of having your choice taking away, because the choice is only offered if you have car hidden behind the door you chose.

The odds aren't changing... just your odds of being offered a choice.

[articulett]

Where did I say that the odds suddenly became 50-50? Are you ever going to provide evidence of that claim?

I'll show when you explain exactly what you meant when you insisted that "the odds change when the host offers you the option to switch." What odds change to what, Claus? You were adamant about this... and said I was wrong to say they don't.

[articulett]

Articulett vs. Claus round... god. What are we on now?

Claus, you are one of the least likable human beings its ever been my distinct displeasure to have discourse with on the internets. I don't know what it is. We have many opinions in common, you seem reasonably well educated, you don't troll, flame, or otherwise behave blatantly poorly, and therefore I am at a loss to explain why, three posts into any thread you're posting in, I want to strangle you with a garden hose. Maybe it's the fact that you back down even less frequently than I do, and have a bulldozer-like obstinacy combined with an unerring knack for annoying even people who agree with you.

Articulett, you're ornery, have no capacity to admit you're wrong, frequently post long rambling posts that go nowhere, are one of the most passive-aggressive people I've ever interacted with, have the logical capacity of a rhubarb, and generally have all the sense of proportion of your average 16 year old.

Therefore, I've got to ask, since we can basically assume that the two of you will disagree just to be disagreeable, can we just manually insert 'three pages of articulett and claus arguing over some nonsense point that adds nothing to the discussion' at the bottom of pretty much every OP and then we don't have to, I don't know, do it?

Just for the record, my opinion of you is on par with your opinion of me. I find you self-important, rambling, and strident; moreover, I find it hypocritical that you claim I don't admit I'm wrong. I do and have. I've never witnessed you doing so, however. Nor Claus.

However, if I've made a wrong statement here-- no-one has highlighted it nor shown a link or reasoning as to why I am wrong. Feel free to do so.

I was trying to explain the problem as I do to my students, when Claus derailed insisting I was wrong about something and that I didn't know anything about the problem. I think any perusal of my explanations by anyone who really understands the problem will affirm that my answers were correct--and that I've always maintained the following:

In the classic Monty Hall Problem (or in situations where you don't have knowledge as to whether it's the classic Monty Hall Problem but it could be-- because you are shown a goat and offered a choice)-- you have a 1/3 chance of winning by keeping the first choice and 2/3 of a chance of winning by switching to the other door.

If you have a problem with that statement, highlight the problem and prove me wrong. Otherwise, I'll take it you were wrong, like Claus, and are mad at me for revealing your pompous buffoonery.

The reason this thread is so rambling is because Claus derailed to make allegations about me-- just as you have done. I can let your dishonest allegations stand, or I can defend myself revealing you guys for what you are. This time I chose the latter, because some people actually wanted to understand the problem. But there won't be a next time, because I won't waste my time reading your posts.

[articulett]

FWIW I'm sure he sees. I'm fairly sure Claus gets it too, but I haven't read every detail in his posts. It appears to me you (articulett) are being obtuse in your wording in order to inject "Claus is wrong" phrases.

It's obtuse to object to someone saying that conditions change, or that probabilities changes as conditions (known information) changes.

It's also obtuse to object to someone pointing out what is and isn't the Monty Hall problem. As with this thread, confusion over a nonintuitive idea isn't helped by a misstatement of the problem.

Take a break.

He was the one who claimed I was wrong and didn't know anything about the problem. What is "obtuse" about my wording? Highlight it and tell me how you would have explained it differently. And what odds change to what --as per Claus' claim that "your odds change when the host offers you an option to choose"? No odds NOR conditions change-- it is stipulated that you are always shown a goat and always offered a switch. He made this claim to assert I was wrong. So what are these odds that change exactly?

[TrueSceptic]

At the risk of repeating myself.... (well, the whole thread is on a repeat cycle I suppose)

According to Wikipedia, the first publication of the puzzle dates back to 1975. (When did the Monty Hall TV show start, does anyone know?) The exact wording of that publication isn't stated, but going from CurtC's and Claus's insistence that the "Monty will always open one of the two unchosen doors and will always reveal a goat" version is The Only True Authentic Monty Hall Puzzle, I'm assuming that perhaps it did make this clear.

However, as far as I can see, the perennial fascination owes a great deal to the fact that subsequent wordings of the scenario have become more ambiguous. I can't see how a puzzle clearly and precisely defined as above could be anything more than a nine-day wonder.

Certainly, when I first encountered the puzzle in 1994 or thereabouts, the wording was similar to the OP, and arguably ambiguous. And then, as now, the bulk of the argumentation seemed to revolve around exactly what rule Monty is to be supposed to be working to. And to be honest, I think this raises the debate to a whole new level.

If we just consider what Claus and CurtC and others call the "classic Monty Hall" scenario, yes, it's intriguing, and counterintuitive, but the answer is perfectly clear once you've got your brain around it. I think it took me a day or two, just puzzling at it on my own, to get there. Really, really, even though there might be some dimwits who will never get it, there isn't a debate. Switching doubles your chances of getting the car, full stop. End of.

All everybody has been arguing about for most of this thread is interpretation. Exactly what game are we playing anyway?

I know how I got on to that aspect, and I think it's how many people have got there. I imagined the 100-doors version (99 goats and a car), while I was thinking about the basic puzzle. This illustrates how the odds change during the process, but it also clarifies the need to know what the exact rules are.

Contestant chooses a door. Monty opens another, a goat. Switch? Wouldn't make much difference, you've still got 98 other doors. But as the process is repeated and repeated, the situation becomes much clearer. The longer the car remains unrevealed, the clearer it becomes that Monty is deliberately avoiding it. By the time there are only two doors left, two things are obvious. The probability that Monty is deliberately avoiding the car is 98/100. And it should be completely intuitive that you should switch. Because the probability that the car is behind the door you originally chose is and always was 1/100, therefore the probability that it is behind the other door is 99/100.

For me, it was this exercise that clarified the need to stipulate the exact rules. Because in the 100-doors version, if Monty was opening doors at random, with no more idea than you have of where the car is, then in the overwhelming majority of games, he'd have revealed the car behind one of the other doors way before you're left with only two closed doors.

And once you think about it, that is also a reasonable interpretation of the wording of the puzzle as usually presented. And, indeed, a reasonable way of running a game show.

Contestant picks a door. Monty, who has no idea where the car is, opens one of the other two doors at random. A third of the time he reveals the car. Oh too bad sir, thanks for playing, good game, hope you and your goat will be very happy together. Two thirds of the time he reveals a goat. Well sir, so far so good, but would you like to change your choice? In that situation, switching makes bugger-all difference.

Cue argument that lasts for (to date) 33 years.

But then it gets worse. Now that the matter of the exact rules has been opened to scrutiny, people start getting even more creative. What if Monty is a complete bastard, and will only open the door if he knows you've already got the car? Well, unlike the random-opening scenario, that is neither a sensible game show, or a reasonable brain-teaser. As one of a range of possibilities incorporated in my next scenario (capricious Monty) it has a place. However, as a consistent stratagem, it's a nonsense. It's just a hypothetical scenario that is dragged in for the sake of argument.

But what if Monty is totally capricious? Maybe he'll help some contestants and be a bastard to others. Maybe he has a different scenario in his head every time the game is played! From what I've heard from those who have watched the original show, this may well have been the actual case. And presumably it was a perfectly viable game show.

However, it's not a perfectly viable brain teaser. To have any validity as an abstract puzzle, there has to be consistency. To put the question to anyone and then say, well, there's no right answer because Monty is entirely capricious, isn't a lot of fun.

For these reasons I maintain that it is reasonable to exclude both the "Monty-is-a-bastard" scenario and the "Monty-is-capricious" scenario. No game show could possibly persist if the former scenario was being operated, and no rational brain-teaser could employ the latter.

Nevertheless, what I will call the "B" scenario, the one where Monty doesn't know where the car is but simply opens one of the two doors the contestant hasn't chosen, is valid as a brain teaser and as a game show, and is in accordance with the usual, ambiguous, wording of the puzzle. (I'm designating the "real" Monty Hall puzzle, where he knows where the car is and will avoid opening that door at this stage, as the "A" scenario.)

This is the conundrum I find completely and utterly fascinating.

The puzzle is as presented. It is not clear whether the A or the B scenario is intended. (That is, you deduce that Monty opening one of the doors you didn't choose is a prerequisite, but you don't know whether or not he knows where the car is.) Where does that leave you?

It leaves you in the peculiar position where the correct answer depends entirely on what is going on in the mind of the host. And in my experience it is that very surreal situation that causes the most outcry from those who have difficulty with the puzzle.

I can only really get my brain round it by imagining that I'm playing one of two computer-game versions. One is programmed with the "A" version, and the other with the "B". I've got to the point where one of the doors I didn't choose has been opened, and there is a goat there, but I still don't know which one I'm playing. If it's the "A" version, I know that switching will double my chances of winning, but if it's the "B" version, I know that switching won't make a blind bit of difference.

Seems sensible enough when looked at that way.

However, I'm not playing with computers. There is an actual person there. Monty just opened a door to reveal a goat. Will switching improve my chances of winning? Well, the answer depends entirely on whether or not Monty himself knows where the car is.

Surreal, what?

Rolfe.
Just saw this thread, which has a life beyond all reason or probability. ;)

Just to follow on from your post, I heard about this only a few years ago. The rules were not explained, only what happened in the show, so the accepted answer of switching increasing the probability from 1/3 to 2/3 was not at all obvious. To make this the case, one has to assume that Monty not only knows where the car is but is also obliged to open one door. As you say, the puzzle changes if either of those do not apply.

To take this further, if we assume that either your A or B apply but we don't know which, surely we should still switch because not switching leaves us with a 1/3 or 1/2 chance, whereas switching gives us 1/2 or 2/3, i.e., improving our odds?

[Rolfe]

Just for the record, my opinion of you is on par with your opinion of me. I find you self-important, rambling, and strident; moreover, U find it hypocritical that you claim I don't admit I'm wrong. I do and have. I've never witnessed you doing so, however.


Point of information.

GreyICE admitted a mistake several pages back on this thread, with perfectly good grace.

I'm going away now, because you're only quibbling about semantics.

And I don't think anyone is reading my posts anyway.

Rolfe.

[Rolfe]

To take this further, if we assume that either your A or B apply but we don't know which, surely we should still switch because not switching leaves us with a 1/3 or 1/2 chance, whereas switching gives us 1/2 or 2/3, i.e., improving our odds?


Yes, absolutely!

This now gets me back to my original point, which is that excluding a rigged game, you should switch. Because even if you're on the latter programme, switching won't decrease your chances. But there is a less-than zero possibility that you are on the former programme, when switching is beneficial. Therefore you should rationally decide to switch, to allow for the possibility that the former is the game in town.

I thiink this is what I like about it. The surreal conclusion that Monty's very intent affects the odds (made a bit less surreal if we substitute Monty with a pre-programmed computer game). And then, the extra leap that says, stop arguing. If you don't know which scenario you're dealing with, and in one the switch is beneficial while in the other it's neutral, then just switch anyway - you can't lose by it, and by switching you are in a position to exploit the possibility of the beneficial scenario.


That's why I specifically said "will switching improve my chances of winning?" in the post you quoted, rather than simply "should I switch?"

Because when you boil it right down like that, then the answer is a simple "yes".

I said this about six times already, but as far as I can see, you're the only person who has read any of my posts recently.

Rolfe.

[articulett]

Just saw this thread, which has a life beyond all reason or probability. ;)

Just to follow on from your post, I heard about this only a few years ago. The rules were not explained, only what happened in the show, so the accepted answer of switching increasing the probability from 1/3 to 2/3 was not at all obvious. To make this the case, one has to assume that Monty not only knows where the car is but is also obliged to open one door. As you say, the puzzle changes if either of those do not apply.

To take this further, if we assume that either your A or B apply but we don't know which, surely we should still switch because not switching leaves us with a 1/3 or 1/2 chance, whereas switching gives us 1/2 or 2/3, i.e., improving our odds?

Yes... and there is the 3rd option... Monty is only offering the choice because you have the car. But if you don't know which of these options are in play your odds are the same as in the classic situation. See post #252.

In the classic situation the host cannot be blind, because you are always shown a goat and the host cannot be in the 3rd category, because it's stipulated that you are always offered a choice. So you are either in the classic situation or you don't know which of the 3 situations you are in. This means that IF you are offered the choice (after being shown a goat)-- you increase your odds by taking it. You won't always win-- .)

[TrueSceptic]

Yes, absolutely!




That's why I specifically said "will switching improve my chances of winning?" in the post you quoted, rather than simply "should I switch?"

Because when you boil it right down like that, then the answer is a simple "yes".

I said this about six times already, but as far as I can see, you're the only person who has read any of my posts recently.

Rolfe.
As I said, this thread has only just got my attention. I read only the first few and last few posts and the post I replied to seemed to include all the options that whizz around my brain whenever I think about the problem. I confess to focussing on the tail of your post because it is the nub of the puzzle IMO. Anything else is not the Monty Hall Problem as I see it, and Articulett seems of the same view. :)

[TrueSceptic]

Yes... and there is the 3rd option... Monty is only offering the choice because you have the car. But if you don't know which of these options are in play your odds are the same as in the classic situation. See post #252.

I assumed that that was ruled out, but as you say, if you don't know if it has, then you are right.

Of course, a capricious Monty might want to you to win because he likes you, or he's just, well, capricious. ;)


In the classic situation the host cannot be blind, because you are always shown a goat and the host cannot be in the 3rd category, because it's stipulated that you are always offered a choice. So you are either in the classic situation or you don't know which of the 3 situations you are in. This means that IF you are offered the choice (after being shown a goat)-- you increase your odds by taking it. You won't always win-- but the odds are double what they will be if you never take the choice. (Your actual odds in the "unknown" situation are "probability that the car is behind the door you pick" multiplied by the "probability that you will be given a choice"-- but it ends up being the same odds as in the classic problem.)
You've obviously played this through many times (with your class?). I'm glad to see this discussion because this problem has a habit of taking over my mind in unguarded moments and it's sort of nice to see those quandaries expressed so well here ;)

I take it that you know the similar Three Prisoners Problem? :D

[GreyICE]

Just for the record, my opinion of you is on par with your opinion of me. I find you self-important, rambling, and strident; moreover, I find it hypocritical that you claim I don't admit I'm wrong. I do and have. I've never witnessed you doing so, however. Nor Claus.
Well given the fact that you have me on ignore 4/5ths of the time, at least according to you, I'm really not surprised. I'm shocked you saw that post.

Lets see, what's the most passive-aggressive response you could make to this? Back to the ignore list with you?

However, if I've made a wrong statement here-- no-one has highlighted it nor shown a link or reasoning as to why I am wrong. Feel free to do so. Nobody cares anymore. Really.

I was trying to explain the problem as I do to my students, when Claus derailed insisting I was wrong about something and that I didn't know anything about the problem. I think any perusal of my explanations by anyone who really understands the problem will affirm that my answers were correct--and that I've always maintained the following:

In the classic Monty Hall Problem (or in situations where you don't have knowledge as to whether it's the classic Monty Hall Problem but it could be-- because you are shown a goat and offered a choice)-- you have a 1/3 chance of winning by keeping the first choice and 2/3 of a chance of winning by switching to the other door.

If you have a problem with that statement, highlight the problem and prove me wrong. Otherwise, I'll take it you were wrong, like Claus, and are mad at me for revealing your pompous buffoonery.
My pompous buffoonery. Well, now I know how you think I don't back down. You fail to read threads.

If you're shown 3 doors, pick one, the host opens one, and its a goat, you don't have enough information to determine whether its a good choice to switch.

The reason this thread is so rambling is because Claus derailed to make allegations about me-- just as you have done. I can let your dishonest allegations stand, or I can defend myself revealing you guys for what you are. This time I chose the latter, because some people actually wanted to understand the problem. But there won't be a next time, because I won't waste my time reading your posts. Oh wait! I didn't even have to wait for the next post for the passive-aggressive back to the ignore list.

Do we have a facepalm smiley?

I did reply to this so I didn't pull an 'oracle of the ages' and write some pithy response and zinger out of here, but you continue to exemplify exactly what I was talking about. Look, you and Claus disagree. We know you disagree, we know you'll fight about it for 10k pages. Can you just, I dunno, put him on ignore or something?

[ddt]

Articulett vs. Claus round... god. What are we on now?

Claus, [...] Maybe it's the fact that you back down even less frequently than I do, and have a bulldozer-like obstinacy combined with an unerring knack for annoying even people who agree with you.

Articulett, you're ornery, have no capacity to admit you're wrong, frequently post long rambling posts that go nowhere, [...]

Therefore, I've got to ask, since we can basically assume that the two of you will disagree just to be disagreeable, can we just manually insert 'three pages of articulett and claus arguing over some nonsense point that adds nothing to the discussion' at the bottom of pretty much every OP and then we don't have to, I don't know, do it?
I'm not long enough around to judge on all you wrote here GreyICE, but I can heartily agree with the above. Especially, articulett, where do you get the time to write all that? I'd need two clones to just read the stuff...

BTW, with all this talking of goats: where is the Marquis? :D

[Herzblut]

Yes... and there is the 3rd option... Monty is only offering the choice because you have the car. But if you don't know which of these options are in play your odds are the same as in the classic situation. See post #252.

This is incomprehensible. In this version, the winning chances of a player changing the door when possible are identically zero.

[TrueSceptic]

This is incomprehensible. In this option, the winning chances of a player changing the door when possible are identically zero.
Zero means you know the outcome. We don't even know the rules. ;)

[TrueSceptic]

Damn you, Monty Hall!

You trapped me in your endless mind-games again!

[Myriad]

Yes, absolutely!




That's why I specifically said "will switching improve my chances of winning?" in the post you quoted, rather than simply "should I switch?"

Because when you boil it right down like that, then the answer is a simple "yes".


That's an excellent point. The correct answer to "should I switch?" depends entirely on whether or not I picked the car the first time!

Respectfully,
Myriad

[articulett]

Point of information.

GreyICE admitted a mistake several pages back on this thread, with perfectly good grace.

I'm going away now, because you're only quibbling about semantics.

And I don't think anyone is reading my posts anyway.

Rolfe.

Point of information: GrayICE made a false allegation about my capacity to admit error which you and others have agreed with. Claus has made this same allegation and multiple people and postings were shown proving it wrong. Did you have actual evidence of a single case where I was shown to be wrong and did not admit it?

Moreover, my statement was in response to GreyICE's allegation--and it was stated as an opinion and it is TRUE. I claimed that I had never seen GreyICE admit error. I will change that if you want to cut and paste the event. I will then be able to claim that I have witnessed him admitting error. However, I won't hold my breath that those who've made the accusation against me will apologize for making a false allegation. Nor will they show any interest in the evidence which proves they made a false allegation.

You've accused me of things that you are more guilty of... as has GreyICE and others who agreed with his statement.

I agreed with everything you wrote in your little piece-- but I did not respond because you agreed with a false allegation made about me AND went on to accuse me of making a false allegation when I stated that I had not witnessed (the truth) GreyICE apologizing for an error. All you needed to do was cut and paste so I could "witness" it. I might add that I've never witnessed an admission of error from any of the people who allege that I have no capacity to admit I am wrong or jumped on that bandwagon. To say I have no capacity to admit error is a LIE that is betrayed by multiple posts and witnesses on this forum. Are you going to be apologizing for this error? Why don't you try modeling the kind of person you would prefer I be... because as far as I can tell, the allegations GreyICE made (and you agreed with) are more fitting of you than of me. Why don't you show us all how good you are at admitting this error?

If you think I made an error and didn't admit it, feel free to highlight it. If it is an error-- I guarantee, I'll admit it.

If you can show me that GreyICE has admitted error... then I will be able to change my claim that I have never witnessed it.

As for the Monty Hall Problem... I think we can all see that in the classic case, you always have a 1/3 chance of winning by not switching and 2/3 chance by switching. Those odds don't change so long as you are staying in the classic problem despite Claus' claim (and this derail where allegations have been made about me) that supposedly "the Monty Hall problem is all about the odds changing when the host gives you a choice."

[Mobyseven]

Point of information.

GreyICE admitted a mistake several pages back on this thread, with perfectly good grace.

I'm going away now, because you're only quibbling about semantics.

And I don't think anyone is reading my posts anyway.

Rolfe.

Nah, I'm reading. Just nothing really to say other than a hearty, "I agree." :)

[Lisa Simpson]

Please - let's not let this thread slip into personal bickering and the magical land of AAH and infractions. Discuss the topic and not each other.

[jsfisher]

Lisa,

Monty says thanks.

:)

[Michael C]

At the risk of repeating myself.... (well, the whole thread is on a repeat cycle I suppose)...
Thanks, Rolfe, for that clear and eloquent post. I agree with you that much of the fascination comes from looking at the different possible scenarios that the ambiguous wording of the problem permits. It seems that some people, having twigged to the fact that in the so-called "classic" version you double your chances of winning by switching, just stop there and refuse to consider other interpretations. I was particularly bugged by the posters on this thread who flatly refused to admit that the answer could change, depending on whether or not Monty knows where the prize is.

[...]But what if Monty is totally capricious? Maybe he'll help some contestants and be a bastard to others. Maybe he has a different scenario in his head every time the game is played! From what I've heard from those who have watched the original show, this may well have been the actual case. And presumably it was a perfectly viable game show.

With my limited knowledge of game shows, I tend to think that the "Monty-is-capricious" strategy is the one most likely to be used in real life. I don't think that the strategy where Monty chooses the door at random is often used. If the host doesn't know what is behind the doors, it's more likely that he lets the contestant choose a door to be opened, with the understanding that whatever is found behind the door is lost to the contestant.

It's also clear that the "Monty-is-a-bastard" technique will be found out if it is used consistently. The "classic" scenario (the host always opens a door himself, there's always a goat behind it and he always gives the contestant the chance of switching), is also too boring if it's used every time. In order to keep up the suspense and make sure that no contestant has a chance of creating a consistent winning strategy, a good game show host has a whole repertoire of variations: not only does he sometimes open a door himself, sometimes let the contestant do it, sometimes let the contestant switch and sometimes not: he has other tricks such as offering the contestant a cash sum for whatever might be behind the chosen door.

Finally: there's no doubt for me that the OP does present the problem in an ambiguous way, leaving open the possibility of the "Monty-is-capricious" strategy. For that reason one could consider that the only answer is "there is not enough information to decide", but I think the much better answer is to expose the differing odds that result from Monty's different possible strategies. Which is what you've done.

[sphenisc]

..But what if Monty is totally capricious?...

Yee gads! Could you have chosen a more ambiguous turn-of-phrase??! :p

[Mashuna]

Ok, I think I've followed this so far. But what if you open a door to find it completely empty, then turn around to see a car heading for the exit driven by a goat?

[Rolfe]

Yee gads! Could you have chosen a more ambiguous turn-of-phrase??! :p


Oh wow! My puns are so good I don't even spot them myself until someone else points them out!

By the way, can anyone tell me what year this Monty Hall show started?

Rolfe.

[TrueSceptic]

Oh wow! My puns are so good I don't even spot them myself until someone else points them out!

By the way, can anyone tell me what year this Monty Hall show started?

Rolfe.
From Wiki (http://en.wikipedia.org/wiki/Monty_Hall),

Hall was the host of the long-running game show Let's Make a Deal, which he developed and produced with partner Stefan Hatos. Let's Make a Deal aired on NBC daytime from December 30, 1963 to December 27, 1968 and on ABC daytime from December 30, 1968 to July 9, 1976, along with two primetime runs. It also aired in syndication from 1971 to 1977, from 1980 to 1981, from 1984 to 1986, and again on NBC briefly from 1990 to 1991.


Now to read the rest of this improbable thread. :eek:

[Ivor the Engineer]

<snip>

However, I'm not playing with computers. There is an actual person there. Monty just opened a door to reveal a goat. Will switching improve my chances of winning? Well, the answer depends entirely on whether or not Monty himself knows where the car is.

Surreal, what?

Rolfe.

The answer depends on what you know or assume about the host's behaviour and the result of one of the not-chosen doors being opened.

I agree with your argument that the only reasonable behaviour of the host for a gameshow would be to always open a not-chosen door. Under this condition, if the host is guessing or knowledgeable but malevolent, the show might finish early with the host revealing the car.

[Rolfe]

The answer depends on what you know or assume about the host's behaviour and the result of one of the not-chosen doors being opened.


Actually, since the behaviour hasn't happened yet, it depends on what we assume his intentions are.

Rolfe.

[Rolfe]

I agree with your argument that the only reasonable behaviour of the host for a gameshow would be to always open a not-chosen door. Under this condition, if the host is guessing or knowledgeable but malevolent, the show might finish early with the host revealing the car.


That wasn't exactly what I said. I think the most likely behaviour for a game show host in real life is to be capricious (pun now intentional!).

Consistent "bastard" behaviour would be very poor entertainment as it would soon become obvious that nobody who switched ever won the car, indeed that the very offer of the switch was a guarantee that you'd chosen right the first time.

The other two consistent behaviour patterns (always opening an unchosen door to reveal a goat, and always opening an unchosen door at random) would be perfectly viable game show scenarios, but because they would present less variety than capriciousness, they might well be less good entertainment value.

However, when the puzzle is presented as a brain-teaser, I think it's unreasonable for the basic premise to be capricious behaviour, because in that case there is simply no answer, making the question pointless. Capricious behaviour means that all three of the possible scenarios may be included, including the "bastard" scenario, in which switching is fatal. Thus, Monty may be running a scenario where switching is beneficial, or one where switching is neutral, or one where switching is fatal. If you can't say which one, then the brain-teaser is just silly.

We may therefore assume consistent behaviour as the premise for a brain-teaser. And we may exclude consistent "bastard" behaviour on the basis that the brain-teaser is about a game show, and consistent "bastard" behaviour is incompatible with a realistic game show.

Thus we're left with the two possible consistent behaviours, either the "classic" one where Monty knows where the car is and will always reveal a goat at step 2, and the alternative which is introduced by the ambiguous wording of the question, in which Monty has no idea where the car is and has opened a door at random.

The interesting one to ponder is the situation where one of these two scenarios is in operation, but you don't know which.

Rolfe.

[Ivor the Engineer]

Rolfe,

The only behaviours Monty could be employing are:

1) "Bastard": Reveal the car if the contestant had not chosen the door with it behind, else reveal a goat.

2) "Random": Monty doesn't know which door has the car behind it and opens a door at random, which could be a goat or the car.

3) "Kind": Monty knows which door has the car behind it and if it is one of the unchosen doors, always opens the door with the goat behind it.


Behaviour 1 results in contestants either never getting the opportunity to switch doors, since Monty reveals the car, or switching doors results in them loosing the car.

Behaviour 2 results in contestants either never getting the opportunity to switch doors, since Monty reveals the car, or switching giving contestants a 2/3rd chance of winning the car.

Behaviour 3 results in contestants winning the car 2/3rds the time if they switch doors.

Now, if it is made clear that contestants always gets the opportunity to switch doors (and it always has been in the versions of the problem I've seen), then behaviours 1 and 2 are not possible.

Alternatively, if sometimes Monty is allowed to reveal the car after the contestant's first choice, either because he doesn't know which door it is behind (behaviour 2), or he doesn't want the contestant to win (behaviour 1), then capriciousness is possible and it becomes impossible to answer the question, as you have already said.

Here's another similar puzzle:

A random two-child family whose older child is a boy is chosen. What is the probability that the younger child is a girl?

A random two-child family with at least one boy is chosen. What is the probability that it has a girl?

The answers and explanation are on wiki, under 'Boy or Girl paradox'.

[Michael C]

The answer depends on what you know or assume about the host's behaviour and the result of one of the not-chosen doors being opened.

I agree with your argument that the only reasonable behaviour of the host for a gameshow would be to always open a not-chosen door. Under this condition, if the host is guessing or knowledgeable but malevolent, the show might finish early with the host revealing the car.

I doubt that the host would knowingly open a door with a car, nor run the risk of opening a door with a car by choosing at random. It wouldn't make for a good game show. Imagine the situation if the host says "OK, you've chosen door A: let's see what's behind door B"; he opens door B and there's the car. End of game. People will immediately suspect him of knowing where the car was, even if he didn't.

It's a completely different thing if he lets the contestant choose which door to open: the contestant dithers between doors B and C, people in the audience are shouting out which door they think should be chosen, the contestant finally makes a decision and the door is opened. If the car is found there it's a disappointment, but it was the contestant who sealed his own fate: it looks as if he's been given a fair shot.

So I think a real-life game show host will be switching between knowingly revealing a goat, or letting the contestant decide. Of course, he can also try to push the contestant to make a certain decision...

[CurtC]

The exact wording of that publication isn't stated, but going from CurtC's and Claus's insistence that the "Monty will always open one of the two unchosen doors and will always reveal a goat" version is The Only True Authentic Monty Hall Puzzle, I'm assuming that perhaps it did make this clear.Actually, in the game show, Monty was nothing if not capricious. The very idea of the show was Monty presenting people with various and sometimes bizarre choices. For example, in a situation like the puzzle, he may offer a switch or not, or may offer a wad of cash instead of the chosen door, etc. etc. That said, the Only True Authentic Monty Hall Puzzle is not that - the proper puzzle explicitly says that not only does Monty know where the prize is located, he must first show you a non-prize door and offer the switch.


And once you think about it, that is also a reasonable interpretation of the wording of the puzzle as usually presented. And, indeed, a reasonable way of running a game show.Thank you - the way the puzzle is usually presented does not properly constrain Monty's rules. In this case, I thought at first it was the same, but I can see some ambiguity where people (especially non-native English speakers) could interpret the wording in the OP to specify that Monty was forced to behave as he did.

However, it's not a perfectly viable brain teaser. To have any validity as an abstract puzzle, there has to be consistency. To put the question to anyone and then say, well, there's no right answer because Monty is entirely capricious, isn't a lot of fun.
Sure, but no one presents it as a brain teaser with the intention of bringing in all the other variations of Monty's behavior. I've seen this come up here and in other forums over the years, and the pattern is this: Someone posts a poorly-worded formulation of the puzzle, answers are offered that the answer is to switch, you'll win 2/3 of the time. Then some curmudgeon like me points out that, according to the puzzle given, you can't say that for sure, and it's actually interesting to consider the different possibilities of the rules Monty may be playing by, and how that affects the probabilities.

Usually, it ends there, but there has been a lot of bickering in this thread revolving around misunderstandings of people's interpretations of wordings, etc.

[ZouPrime]

OMGLOL at anyone trying to redefine a problem that has been as much discussed as the Monty Hall puzzle. Seriously. There are so many web pages dedicated to it, any attempt of redefinition is ridiculous.

http://en.wikipedia.org/wiki/Monty_Hall_problem

It's really not hard.

Anyway.

The odds may be 2/3 if you switch, but 0.999... < 1.0 and there's nothing anyone can say to make me change my mind of that! :-P

[CurtC]

2) "Random": Monty doesn't know which door has the car behind it and opens a door at random, which could be a goat or the car.

...

Behaviour 2 results in contestants either never getting the opportunity to switch doors, since Monty reveals the car, or switching giving contestants a 2/3rd chance of winning the car.

I hope you mis-typed. IF we're considering the case where Monty picks a door at random, and it just so happens that when it's your turn, he revealed the goat, then (in that case), it's a 50/50 proposition - the odds are equal for the two remaining doors.

[Ivor the Engineer]

I hope you mis-typed. IF we're considering the case where Monty picks a door at random, and it just so happens that when it's your turn, he revealed the goat, then (in that case), it's a 50/50 proposition - the odds are equal for the two remaining doors.

I was considering the case where the contestant chooses a door and then Monty opens one of the other two doors at random. If Monty does not open the door with the car behind it (doing so would end the show there and then), the contestant would win 2/3rds of the time by switching to the door Monty did not open.

[CurtC]

I was considering the case where the contestant chooses a door and then Monty opens one of the other two doors at random. If Monty does not open the door with the car behind it (doing so would end the show there and then), the contestant would win 2/3rds of the time by switching to the door Monty did not open.
Imagine that you're on a game show where the hosts always operates by letting you pick, then he, not knowing where the prize is, opens one of the other two doors at random, then if the prize is still hidden, offers you the choice of switching. Now, at this point, what should you do? Which door has the best chance of winning?

This is the situation we're discussing right now, do you agree? I have to clearly state that up front so that some others here won't complain that this isn't the classic version of the puzzle, it's a digression imagining some different rules.

In this case, each door has equal probability. I think that once someone understands the 2/3 answer for the classic puzzle, this variation is just as interesting. Here's how they add up. Imagine 300 people who over the years start playing the game, and let's count their numbers.

Group 1: picked a goat door on their first guess, and then Monty revealed the car. Game over. There are ~100 people in this group.

Group 2: picked a goat door on their first guess, and then Monty revealed the other goat door. There are ~100 people in this group, and they would all win by switching.

Group 3: picked the car door on their first guess, then Monty (of course) reveals a goat door. There are ~100 people in this group, and they would all lose by switching.

Now, only the people in groups 2 and 3 have found themselves in the situation described. Group 1 does not count towards this analysis, because the problem statement has already ruled them out. Out of those remaining 200 people, 100 would win by switching, 100 would win by sticking.


ETA: Actually, I think that this is a much more interesting concept than the classic "2/3" puzzle. If the puzzle is not worded properly, and leaves out the rules Monty must play by, the correct answer depends on what assumptions you make about why Monty offered the switch.

[sol invictus]

I was considering the case where the contestant chooses a door and then Monty opens one of the other two doors at random. If Monty does not open the door with the car behind it (doing so would end the show there and then), the contestant would win 2/3rds of the time by switching to the door Monty did not open.

No! Switching will win 1/2 the time in that situation. That's what the last 5 pages of this thread are about....

[CurtC]

I'm not a native speaker, so what is unclear about

The player picks a door, then Monty opens one of the two other doors and reveals what is behind and offers the player to switch to another door of his choice.
Some people here have really gotten their panties in a wad about this very aspect, so I'm trying to be polite a clear...

That statement reads to me like a description of what has already happened. Even though it doesn't use the past tense, it's a common way of speaking informally. On the other hand, I can see how someone else may interpret the present tense as specifying the rules Monty must play by. I haven't done a poll, but my guess is that most native English speakers would, in similar circumstances, interpret the wording as simply reporting what had already happened.

That's what's unclear about it.

[ZirconBlue]

A funny thing about humans is that they tend to go with their first choice-- as soon as they place a bet they tend to see their odds as increasing-- they feel more confident with their choice... but your odds don't suddenly change when you place your bet or pick your door... your feeling that you are more likely to win, does not make you more likely to win. This is true even with lottery winners. Before playing, they may correctly assess their odds-- but once they've committed, they "feel" as if their odds go up.

My whole state (Nevada) is built on gambling odds and the bizarre psychology and irrationalities people have in regards to the subject-- their belief in "winning streaks" and so forth.
[bolding added]

I thought I'd point out that articulett's original claim was entirely correct. The act of selecting a door (or placing a bet) does not, in and of itself, change the odds of that door (or bet) being the correct (winning) choice, at that point in time. The act of selecting a door does not make that door special or somehow more likely to be the correct door, but people often behave as if it did, favoring their first choice, even when the situation changes. If the solution to the classic Monty Hall problem were obvious to most people, then it wouldn't be very interesting. That it defies the "common sense" understanding is what makes it worth discussion in the first place, IMO.

[Ivor the Engineer]

Sol & CurtC, you are correct (but you already knew that;)). I've just drawn out the tree diagram and convinced myself my previous intuition was wrong.

The difference is the probability Monty opens a particular door in the original problem is conditional in one case on the door the contestant chooses, whereas in the "dumb" Monty version, he (typically) chooses the door to open with equal probability.

Where it gets interesting is if you now say although Monty is ignorant to which door has the car behind, he still alters the probability of opening a particular door based on the door the contestant chooses.

[articulett]

[bolding added]

I thought I'd point out that articulett's original claim was entirely correct. The act of selecting a door (or placing a bet) does not, in and of itself, change the odds of that door (or bet) being the correct (winning) choice, at that point in time. The act of selecting a door does not make that door special or somehow more likely to be the correct door, but people often behave as if it did, favoring their first choice, even when the situation changes. If the solution to the classic Monty Hall problem were obvious to most people, then it wouldn't be very interesting. That it defies the "common sense" understanding is what makes it worth discussion in the first place, IMO.

Thank you.
I am forever in debt to you.
The whole derail was because of that sentence.

*kisses zirconblue all over*

[Rolfe]

Sol & CurtC, you are correct (but you already knew that ;)). I've just drawn out the tree diagram and convinced myself my previous intuition was wrong.

The difference is the probability Monty opens a particular door in the original problem is conditional in one case on the door the contestant chooses, whereas in the "dumb" Monty version, he (typically) chooses the door to open with equal probability.

Where it gets interesting is if you now say although Monty is ignorant to which door has the car behind, he still alters the probability of opening a particular door based on the door the contestant chooses.


Mmmm, yes.

If you've got to the point in the game where you have chosen a door and Monty has then opened another to reveal a goat, the answer to the question "will switching to the other unopened door improve your chances of getting the car" depends entirely on Monty's motive in choosing which door to open.

A. "Classic" scenario, where Monty knows where the car is and has deliberately opened a goat door, switching doubles your odds of winning (from 1/3 to 2/3).

B. Alternative possibility, not excluded by the way the puzzle is usually worded, where Monty does not know where the car is (or if he knows, is capable of entirely disregarding this information) and has opened one of the two unchosen doors at random. Switching does not change the probability of getting the car. (Originally 1/3 chance you chose correctely the first time, 1/3 chance the remaining door has the car, and 1/3 chance that Monty might have revealed the car - we are now at the stage where that last possibility has been excluded, and you're left with two equal-probability coices.)

C. "Monty-is-a-bastard" scenario, where he knows you have already chosen the correct door and is only offering you the switch to entice you away from the winning choice. Switching would lose you the car 100% of the time. However, as this is not a strategy (if employed consistently) that would produce a viable game show, it is arguable that it may be discounted.

The inclusion of the last scenario is mainly of importance when considering the possibility that Monty may be running a different scenario every time (which is probably what actually happened in the real show). While scenario C is non-viable if used consistently, it could form part of the mix in this situation.

Scenario C can only be excluded if we know or assume that Monty always offers a choice. If we can't assume that Monty will always do that, then we can't exclude the possibility that scenario C may be invoked. In that case, the question is unanswerable, because you don't know if you'll improve your chances (scenario A), leave them unchanged (scenario B) or destroy them completely (scenario C).

If, however, we can assume, deduce or stipulate that Monty will always offer a choice, then scenario C can be excluded. In that case, even if we don't know whether scenario A or scenario B is in operation, then the answer is clear. Since switching cannot reduce your chances of winning, and may improve them (if A is the game in town), then switch anyway.

Rolfe.

[GreyICE]

C. "Monty-is-a-bastard" scenario, where he knows you have already chosen the correct door and is only offering you the switch to entice you away from the winning choice. Switching would lose you the car 100% of the time. However, as this is not a strategy (if employed consistently) that would produce a viable game show, it is arguable that it may be discounted.
And yet it is. The show is called "Lets make a deal" and Monty Hall is the host.

http://query.nytimes.com/gst/fullpage.html?res=9D0CEFDD1E3FF932A15754C0A9679582 60

After the 20 trials at the dining room table, the problem also captured Mr. Hall's imagination. He picked up a copy of Ms. vos Savant's original column, read it carefully, saw a loophole and then suggested more trials.

On the first, the contestant picked Door 1.

"That's too bad," Mr. Hall said, opening Door 1. "You've won a goat."

"But you didn't open another door yet or give me a chance to switch."

"Where does it say I have to let you switch every time? I'm the master of the show. Here, try it again."

On the second trial, the contestant again picked Door 1. Mr. Hall opened Door 3, revealing a goat. The contestant was about to switch to Door 2 when Mr. Hall pulled out a roll of bills.

"You're sure you want Door No. 2?" he asked. "Before I show you what's behind that door, I will give you $3,000 in cash not to switch to it."

"I'll switch to it."

"Three thousand dollars," Mr. Hall repeated, shifting into his famous cadence. "Cash. Cash money. It could be a car, but it could be a goat. Four thousand."

"I'll try the door."

"Forty-five hundred. Forty-seven. Forty-eight. My last offer: Five thousand dollars."

"Let's open the door."

"You just ended up with a goat," he said, opening the door. The Problem With the Problem

Mr. Hall continued: "Now do you see what happened there? The higher I got, the more you thought the car was behind Door 2. I wanted to con you into switching there, because I knew the car was behind 1. That's the kind of thing I can do when I'm in control of the game. You may think you have probability going for you when you follow the answer in her column, but there's the pyschological factor to consider."

He proceeded to prove his case by winning the next eight rounds. Whenever the contestant began with the wrong door, Mr. Hall promptly opened it and awarded the goat; whenever the contestant started out with the right door, Mr. Hall allowed him to switch doors and get another goat. The only way to win a car would have been to disregard Ms. vos Savant's advice and stick with the original door.

Monty Hall, it turns out, is also quite smart.

[Michael C]

And yet it is. The show is called "Lets make a deal" and Monty Hall is the host.

http://query.nytimes.com/gst/fullpage.html?res=9D0CEFDD1E3FF932A15754C0A9679582 60

Monty Hall, it turns out, is also quite smart.

Indeed, indeed. From that story, and from experience of other similar game shows, I conclude that Monty always knew where the car was and would not have used the random strategy: he liked to be able to control the outcome as much as possible.

In fact Rolfe missed one strategy in the previous post: the "Monty-is-super-nice" strategy: Monty only offers the chance to switch if there is a goat behind your chosen door. So once Monty has opened a door that reveals a goat, he can choose between the following strategies:

- "Classic" Monty: he always gives the chance to switch.

- "Super-nice" Monty: he only gives the chance to switch if there is a goat behind your door.

- "Bastard" Monty: he only gives the chance to switch if the car is behind your door.

If you play the strategy "always switch if the possibility is given", you will win 2/3 of all games against "classic" Monty, you will always win against "super-nice" Monty and you will never win against "bastard" Monty.

Since Monty will in fact mix the three strategies, there is no consistent way for you to increase your odds of winning.

[Ivor the Engineer]

<snip>

Since Monty will in fact mix the three strategies, there is no consistent way for you to increase your odds of winning.

That is unless you can estimate the probability of Monty using a particular strategy. For example, if it is known he selects between Classic (2/3), Nice (1) and Bastard (0) with equal probability, switching to the other door is still optimal (5/9).

[ddt]

Indeed, indeed. From that story, and from experience of other similar game shows, I conclude that Monty always knew where the car was and would not have used the random strategy: he liked to be able to control the outcome as much as possible.
IRL, I doubt that the presenter would not know where the car is. Even in case he'd forget, the director of the show behind the scenes would whisper it into his ear (piece).


Since Monty will in fact mix the three strategies, there is no consistent way for you to increase your odds of winning.
Not to mention that fact that Monty will try to influence the people into switching or not, after he has shown a second door. It's impossible to quantify that.

That is unless you can estimate the probability of Monty using a particular strategy. For example, if it is known he selects between Classic (2/3), Nice (1) and Bastard (0) with equal probability, switching to the other door is still optimal (5/9).
Which is still quite a way up from the 1/3 if you didn't have the whole switching business at all. Unfortunately, we don't get to see normally post-facto which door it had been (plus I don't care for this kind of shows).

BTW, Happy Birthday, Ivor!

[CurtC]

That is unless you can estimate the probability of Monty using a particular strategy. For example, if it is known he selects between Classic (2/3), Nice (1) and Bastard (0) with equal probability, switching to the other door is still optimal (5/9).

Seems to me that if he sometimes uses Nice and Bastard, then that rules out the Classic Monty, which requires that he always offer the switch. In other words, on those times when he's the Classic Monty, he's really either Nice or a Bastard depending on the contestant's initial choice.

If he selected between Nice and Bastard with equal probability, then the contestant who finds himself with the switch/stick decision still has a 2/3 chance of winning if he switches.

I think Monty's actual strategy was to keep 'em guessing, which would mean that he should offer the switch more often when the first guess was the car door, less often when it was the goat door. If Monty offers the switch twice as often to those who initially pick the car than he does to those who initially pick the goat (for example he will offer the switch 80% of the time to those who chose the car door and 40% of the time to those who chose the goat), then the poor sap contestant who is faced with the decision has a gut-wrenching 50/50 choice. But at least that's better than the 1/3 chance he had if Monty simply revealed the choice every time!

[Rolfe]

And yet it is. The show is called "Lets make a deal" and Monty Hall is the host.

http://query.nytimes.com/gst/fullpage.html?res=9D0CEFDD1E3FF932A15754C0A9679582 60

Monty Hall, it turns out, is also quite smart.


:D :D :D

OK, fair enough, my premise was that this scenario would only be viable as part of a mixed run of scenario-switching, otherwise it would be far too predictable. But if you wanted to smack down a smart-aleck who hadn't taken all the possibilities into account and didn't think Monty would do that, then yes, it makes perfect sense.

Rolfe.

[Rolfe]

Indeed, indeed. From that story, and from experience of other similar game shows, I conclude that Monty always knew where the car was and would not have used the random strategy: he liked to be able to control the outcome as much as possible.

In fact Rolfe missed one strategy in the previous post: the "Monty-is-super-nice" strategy: Monty only offers the chance to switch if there is a goat behind your chosen door. So once Monty has opened a door that reveals a goat, he can choose between the following strategies:

- "Classic" Monty: he always gives the chance to switch.

- "Super-nice" Monty: he only gives the chance to switch if there is a goat behind your door.

- "Bastard" Monty: he only gives the chance to switch if the car is behind your door.

If you play the strategy "always switch if the possibility is given", you will win 2/3 of all games against "classic" Monty, you will always win against "super-nice" Monty and you will never win against "bastard" Monty.

Since Monty will in fact mix the three strategies, there is no consistent way for you to increase your odds of winning.


Yes, I think that in the real show (as opposed to the brain-teaser), it's unlikely that Monty would either not have known where the car was, or would have completely discounted that knowledge when deciding what to do next. It's mainly that that possibility is inherent in the brain-teaser as it is usually presented.

I was considering "super-nice" Monty as a subset of the classic scenario, but you're right, it's technically a different setup. However, given that Monty in the real game was never consistent, you'd never know if he was doing that, so you'd probably never decide to bank on it.

That is unless you can estimate the probability of Monty using a particular strategy. For example, if it is known he selects between Classic (2/3), Nice (1) and Bastard (0) with equal probability, switching to the other door is still optimal (5/9).


But once you're talking about real life rather then the brain-teaser, you can't even bank on Monty's distribution of strategies remaining constant. As in the example GreyICE quoted, if he thought someone had him sussed, he'd just swing to a different approach to confound them. Sort of like Curt C. was suggesting. Sweet.

Rolfe.

[ZirconBlue]

Thank you.
I am forever in debt to you.
The whole derail was because of that sentence.

*kisses zirconblue all over*

:blush:

Don't mention it.

[Ivor the Engineer]

<snip>

BTW, Happy Birthday, Ivor!

Thankyou.

[billydkid]

The premise of the problem is that Monty opens a door with a goat, regardless of his motives.

[billydkid]

actually it does-- at the outset his odds were one in 3-- once he picked the goat door-- it still is one in 3... that means the remaining door still has 2/3 chance of being the car. This horse in quite completely decayed so - this above is what I said, right? The person who make the major point of saying I was EXACTLY WRONG, with oversized and bolded type - that it doesn't matter what Monty knows or doesn't know is wrong, right? This is all we know and all we need to know - you pick one of three doors. Monty opens another door to reveal a booby prize. And the only question is whether you odds improve by choosing the remaining door or they do not. They improve from 1/3 to 2/3. This is not about what Monty knows or doesn't know. The only things stipulated by the puzzle is what I have described above. In exactly the same way if you had a thousand doors and you picked one (your odds being 1/1000. Then Monty went on to reveal goats behind 998 of the other doors (regardless of why he chose those particular doors - it's not germaine to the problem). The odds of the prize being behind the unchosen door are 1000/1. This is correct isn't it?

Where I got really confused was with the Wikipedia page and they started talking about Monty's motives being critical. I just think that is wrong if you adhere to the stipulation of the problem - the he does, regardless of his motive or knowlege, pick a door with a goat. It is about statistics and nothing else.

[GreyICE]

This horse in quite completely decayed so - this above is what I said, right? The person who make the major point of saying I was EXACTLY WRONG, with oversized and bolded type - that it doesn't matter what Monty knows or doesn't know is wrong, right? This is all we know and all we need to know - you pick one of three doors. Monty opens another door to reveal a booby prize. And the only question is whether you odds improve by choosing the remaining door or they do not. They improve from 1/3 to 2/3. This is not about what Monty knows or doesn't know. The only things stipulated by the puzzle is what I have described above. In exactly the same way if you had a thousand doors and you picked one (your odds being 1/1000. Then Monty went on to reveal goats behind 998 of the other doors (regardless of why he chose those particular doors - it's not germaine to the problem). The odds of the prize being behind the unchosen door are 1000/1. This is correct isn't it?

Where I got really confused was with the Wikipedia page and they started talking about Monty's motives being critical. I just think that is wrong if you adhere to the stipulation of the problem - the he does, regardless of his motive or knowlege, pick a door with a goat. It is about statistics and nothing else. No, you're failing to understand the nature of the problem.

Lets say Monty gets very, very luck with his thousand doors, and happens to open 998 of them at random without revealing the car.

The chances you now have are 1/2. Monty doesn't know where the car is, so he essentially won the lottery and happened to miss it. Now there's a 50/50 shot what door its behind. Think about it - there are a 999 outcomes on which door is remaining. 998 of those outcomes were eliminated, because he missed the car.

If he can't open the door with the car behind it, every possibility where he would open a door with a car becomes a possibility where he doesn't pick the car. The 998 outcomes where he picks a car aren't eliminated, in this case. They don't exist in the first place, in this scenario. Since no outcomes have been eliminated, the chances are overwhelming that its behind the final door.

I graphed this out (do it with paper) and its infinitely convincing once you see the paper in front of you.

The conceptual problem is Monty's motivations. Remove them, and the problem remains similar.

Flip two coins. But hide them, so you don't know the outcomes. Ask an impartial friend (A computer, if we want no humans in this) to reveal one if it came up heads (if both came up heads, he reveals just one). What are the chances the other coin is tails?

Now reveal one of the two coins at random. If its heads, whats the chances the other one is tails?

Finally program a computer to sometimes randomly reveal a coin, and sometimes reveal one of the coins if one is heads.

If the computer reveals heads, does it matter if its revealing coins at random, or doing the "reveal one if its heads" in terms of the possibility that the other coin is tails?

It seems odd, but which program the computer is running matters.

[billydkid]

No, you're failing to understand the nature of the problem.

Lets say Monty gets very, very luck with his thousand doors, and happens to open 998 of them at random without revealing the car.

The chances you now have are 1/2. Monty doesn't know where the car is, so he essentially won the lottery and happened to miss it. Now there's a 50/50 shot what door its behind. Think about it - there are a 999 outcomes on which door is remaining. 998 of those outcomes were eliminated, because he missed the car.

If he can't open the door with the car behind it, every possibility where he would open a door with a car becomes a possibility where he doesn't pick the car. The 998 outcomes where he picks a car aren't eliminated, in this case. They don't exist in the first place, in this scenario. Since no outcomes have been eliminated, the chances are overwhelming that its behind the final door.

I graphed this out (do it with paper) and its infinitely convincing once you see the paper in front of you.

The conceptual problem is Monty's motivations. Remove them, and the problem remains similar.

Flip two coins. But hide them, so you don't know the outcomes. Ask an impartial friend (A computer, if we want no humans in this) to reveal one if it came up heads (if both came up heads, he reveals just one). What are the chances the other coin is tails?

Now reveal one of the two coins at random. If its heads, whats the chances the other one is tails?

Finally program a computer to sometimes randomly reveal a coin, and sometimes reveal one of the coins if one is heads.

If the computer reveals heads, does it matter if its revealing coins at random, or doing the "reveal one if its heads" in terms of the possibility that the other coin is tails?

It seems odd, but which program the computer is running matters.No, no, no, no. Articulet has it exactly right and the talk about Monty "always having" to pick a certain door and so on is completely irrelevant. It doesn't matter how many doors you are talking about as long as there are more than two. We can leave Monty out altogether. There are 3 doors. When you pick a door you are essentially dividing these 3 doors into two groups with one door on one side and two on the other. Say, the door you picked on the right and the other two on the left. The odds of the the prize being on the right are 1 in 3. The odds of it being on the left are 2 in 3. A door without prize on the left is revealed (regardless of why). The odds of the prize being on the right remain 33% and the odds of it being on the left remain 67%. Take a billion doors with a single prize randomly places behind one of them. You pick a door. The prize is almost certainly behind one of the other doors. All of the other doors except one are revealed to be prizeless (regardless of why - it's not part of the puzzle). The prize is virtually guaranteed to be behind the remaining unchosen door. So you are saying there is a difference between whether he can't open the door with the car and he simply doesn't open that door? My argument would be that he can't by definition, because he doesn't. The definition of the problem is that he can't. How about he merely won't open the door with the car? Suppose this scenario - you choose one of a million doors and then Monty gives you the option of keeping your one door or switching to all of the remaining doors and you do that and then, for whatever reason, Monty reveals all of the remaining doors except the one you chose initially and just one of 999,998 on the other side. The car was a virtual certainty to be on the side with the umpteen doors before Monty revealed all of the remaining doors. Why would that change by him showing you all of the doors where the car is not?

[Michael C]

No, no, no, no. Articulet has it exactly right and the talk about Monty "always having" to pick a certain door and so on is completely irrelevant[...]

Here we go again!

Billy, what is you answer to this slightly different problem:

After you have chosen a door, Monty lets you designate a door to be opened. The door reveals a goat. Now you are offered the choice to switch from your chosen door to the other remaining one. What are now your chances of winning if you stay? If you switch?

[ThatSoundAgain]

Here we go again!

Billy, what is you answer to this slightly different problem:

After you have chosen a door, Monty lets you designate a door to be opened. The door reveals a goat. Now you are offered the choice to switch from your chosen door to the other remaining one. What are now your chances of winning if you stay? If you switch?

I think that this is easier to explain if you take care to remind people of the "hidden branch" - the possibility that the host unwittingly opens the door on the prize in step 2. In the latest scenario from your post, you'll have to take into account that the player could have revealed the prize with their second choice, but didn't.

[CurtC]

Michael C - I really like that variation.

Billy, I thought this was a settled issue, with everyone finally agreeing, until you showed up again. Please think carefully to realize why everyone else here has agreed that you need to know Monty's methods and not just what has happened on this one trial.

[Mobyseven]

I think that earlier on, when I was saying that Monty's motivations 'don't matter', people may have misunderstood what I meant. When we look at the problem, we as human beings with motives see that there is a person in the problem acting in a particular way. Because of this, we automatically assign him a motive to explain his behaviour. More than this, because the difference between the 'classic' Monty Hall problem and the 'random' Monty Hall alternative is that either a door is opened to always show a goat or a door is opened at random, we have to explain how this gels with what is going on inside the head of the person opening the doors - in this case Monty.

If Monty is always going to reveal a goat, then he has to know what is behind each door - otherwise it is improbable to the point of impossibility that he will never accidentally reveal a car.

If, however, Monty opens a door at random, we need to explain that too. The easiest way to do this is to have Monty not know where the prize is (so that it will not influence his door choice). There are other ways to randomize this, but the most intuitive is simply that he doesn't know where the prizes are.

This is where things start to get a bit back-asswards. In real life, motive precedes behaviour. This leads people to think that Monty's motive is the important part of the setup, as it influences his behaviour (and thus what game you are playing). Were this real life, that'd be the right way to do things...but in the context of the Monty Hall problem, it's getting the cart before the horse. Monty doesn't act in a particular fashion because of his motives, his motives have been assigned to match the setup in the scenario. It is entirely possible to set the problem up without including Monty at all, and to simply describe the rules of the scenario - but such a setup is dry, and not as 'fun' (in a manner) as one where you are on a gameshow with a host.

So while it may be fun to assign motives to Monty, they don't actually matter - the motives are there to serve the scenario, rather than define it.

[Michael C]

I think that earlier on, when I was saying that Monty's motivations 'don't matter', people may have misunderstood what I meant. When we look at the problem, we as human beings with motives see that there is a person in the problem acting in a particular way. Because of this, we automatically assign him a motive to explain his behaviour. More than this, because the difference between the 'classic' Monty Hall problem and the 'random' Monty Hall alternative is that either a door is opened to always show a goat or a door is opened at random, we have to explain how this gels with what is going on inside the head of the person opening the doors - in this case Monty.

You're right: I agree that what makes a difference is Monty's behaviour. We have to be careful when we're saying this, though: what matters is not just his behaviour this time, it's his behaviour pattern. To say that he opens a door revealing a goat is not enough information: we need to know if this always happens, or if it only happens when there is a car behind the contestant's door, or if it happens at random, or if some other conditions apply.

It's true that we don't have to ascribe motives to Monty in order to describe his behaviour pattern: we could just as well say that there is a machine that opens a certain door depending on certain conditions. But if we just say "a door is opened, revealing a goat", there is not enough information to give a unique answer. Equally, stating that the function of Monty is to open a door with a goat, or that the problem is so defined that Monty opens a door with a goat, does not provide enough information for a unique answer. We need to define the function of the "Monty machine" more precisely. Does it always open a door with a goat, and, equally important, does it always give the contestant the option of switching doors?

So much hinges on that word "always". Does the wording of the OP imply everything always happens the same way? I say it doesn't, but many posters here were adamantly insisting that it does. Apparently there is something ambiguous there, otherwise we wouldn't have so much disagreement (though I do wonder if a few people weren't just being pig-headed, refusing to even consider alternative interpretations, either because they were too attached to a mythical one and only "real" Monty Hall problem, because they were too attached to what they had first said or because they just wanted to annoy).

For those who insist that the notion that Monty always reveals a goat is implicit in the OP, I offer the following problem:

Minty Hill opens the front door of her home, revealing a goat on her doorstep. What should she do?

[Rolfe]

But once you're talking about real life rather then the brain-teaser, you can't even bank on Monty's distribution of strategies remaining constant. As in the example GreyICE quoted, if he thought someone had him sussed, he'd just swing to a different approach to confound them. Sort of like Curt C. was suggesting. Sweet.


I'm going on from this, because I'm fed up repeating the same ground.

Which is.

We know that if Monty will always open another door at step 2, and will always reveal a goat, irrespective of your initial choice, your chances of winning are doubled if you switch. End of.

We know that if Monty has opened a door at random at step 2, switching doesn't alter your chances of winning. End of, again.

We've gone round this so often I'm getting dizzy.

And then there are the other scenarios where is decision as to whether or not to open a door at all is dependent on whether or not you already have the right door. Once you start allowing for these (that is, scenarios where Monty may not offer you the choice at all), there really is no way to answer the conundrum, because you can't exclude the possibility that Monty is deliberately trying to trap you into switching away from the correct choice.

So, can I stand the problem on its head? Just for a bit of variety? Let's look at it from Monty's point of view, assuming he knows where the car is (as would certainly have been the case in the real show), and he's an active participant rather than a computer model.

Step 1, you choose a door. There are only two possibilities here - you've either picked the prize door or you haven't.

Monty now has one decision to make - will he give you a chance to switch, or won't he?

So there are really only four possibilities.

You already got the prize, and Monty decides to let you have it. He would do this by opening the door you already chose.
You don't have the prize and Monty decides not to give you a second chance. He could do this either by opening the door you chose, or by opening to door with the prize to show that you don't have it.
You already got the prize, and Monty decides to try to entice you away from it. He opens one of the two goat doors and sweetly invites you to change.
You don't have the prize, and Monty decides to give you a second chance. He opens the other goat door, and sweetly invites you to change.
Obviously scenarios 1 and 2 are out of the loop here, you can't do anything about this. By the stage of the game we're looking at, we're either at scenario 3 or 4. If it's 3, switching will lose you the prize 100% of the time. If it's 4, switching will win you the prize 100% of the time.

Looked at this way, it's not "do I feel lucky", but "do I think Monty likes me?"

Really, your chances of a switch being beneficial depend on whether Monty is more likely to offer someone who has initially chosen wrong a second chance, or whether he's more likely to try to snooker someone who chose right initially.

No doubt some information could be gained on this by observing the actual outcomes of a sufficient number of previous games. However, since Monty probably isn't tossing mental coins here, other factors will come into play such as audience reaction, how stubborn he thinks you are, and indeed whether or not he likes you!

Rolfe.

[TrueSceptic]

I'm going on from this, because I'm fed up repeating the same ground.

Which is.

We know that if Monty will always open another door at step 2, and will always reveal a goat, irrespective of your initial choice, your chances of winning are doubled if you switch. End of.

We know that if Monty has opened a door at random at step 2, switching doesn't alter your chances of winning. End of, again.

We've gone round this so often I'm getting dizzy.

And then there are the other scenarios where is decision as to whether or not to open a door at all is dependent on whether or not you already have the right door. Once you start allowing for these (that is, scenarios where Monty may not offer you the choice at all), there really is no way to answer the conundrum, because you can't exclude the possibility that Monty is deliberately trying to trap you into switching away from the correct choice.

So, can I stand the problem on its head? Just for a bit of variety? Let's look at it from Monty's point of view, assuming he knows where the car is (as would certainly have been the case in the real show), and he's an active participant rather than a computer model.

Step 1, you choose a door. There are only two possibilities here - you've either picked the prize door or you haven't.

Monty now has one decision to make - will he give you a chance to switch, or won't he?

So there are really only four possibilities.

You already got the prize, and Monty decides to let you have it. He would do this by opening the door you already chose.
You don't have the prize and Monty decides not to give you a second chance. He could do this either by opening the door you chose, or by opening to door with the prize to show that you don't have it.
You already got the prize, and Monty decides to try to entice you away from it. He opens one of the two goat doors and sweetly invites you to change.
You don't have the prize, and Monty decides to give you a second chance. He opens the other goat door, and sweetly invites you to change.
Obviously scenarios 1 and 2 are out of the loop here, you can't do anything about this. By the stage of the game we're looking at, we're either at scenario 3 or 4. If it's 3, switching will lose you the prize 100% of the time. If it's 4, switching will win you the prize 100% of the time.

Looked at this way, it's not "do I feel lucky", but "do I think Monty likes me?"

Really, your chances of a switch being beneficial depend on whether Monty is more likely to offer someone who has initially chosen wrong a second chance, or whether he's more likely to try to snooker someone who chose right initially.

No doubt some information could be gained on this by observing the actual outcomes of a sufficient number of previous games. However, since Monty probably isn't tossing mental coins here, other factors will come into play such as audience reaction, how stubborn he thinks you are, and indeed whether or not he likes you!

Rolfe.
Once you start thinking about this and then consider Monty's motives, there's no end to it.

One more possibility, and I apologise if it's been mentioned already: what if Monty needs someone to win the prize every nth time, to maintain interest and viewer levels? On these occasions he steers the game towards you winning, changing from the default strategy of steering the game towards you losing. How often would he do this, i.e. what are your chances of being favoured?

Is there any way this puzzle can be resolved without more information?

[lenny]

hi Rolfe

We know that if Monty will always open another door at step 2, and will always reveal a goat, irrespective of your initial choice, your chances of winning are doubled if you switch. End of.
agreed.
We know that if Monty has opened a door at random at step 2, switching doesn't alter your chances of winning. End of, again.

your terms here appear unclear. please just point me to the relevant post if this was already clarified or if you are interested in a nonstandard definition of the problem (like opening the chosen door).

if monty opens one of the other two doors at random, there are two options:

a) the door he opens is the prize door: your chances of winning are now zero and you know it
b) the door he opens is a goat: you can increase your chances of winning by switching (the standard "two doors versus one door" argument holds regardless of Monty's ignorance).

if he opens the door you have, you know you have won, or you know you should switch; even in this nonstandard verison you gain information and can improve your "chance of winning".

so even opening a door at random introduces more information and thus alters your "chances of winning".
(or: which of the 700 other posts should i have read?)

[Paul C. Anagnostopoulos]

Does it change anything if two doors contain goats and the other door contains free will?

~~ Paul

[Paul C. Anagnostopoulos]

if monty opens one of the other two doors at random, there are two options:

a) the door he opens is the prize door: your chances of winning are now zero and you know it
b) the door he opens is a goat: you can increase your chances of winning by switching (the standard "two doors versus one door" argument holds regardless of Monty's ignorance).
If you're saying what it sounds like you're saying, then no.

Make a table and fill it in.

~~ Paul

[articulett]

If you're saying what it sounds like you're saying, then no.

Make a stable and fill it in.

~~ Paul

Right. If you are shown the goat and you are asked to make a choice and you have no other knowledge on the subject (such that Monty always asks a person if they want to change)-- then it doesn't really matter what you do-- it's roughly 50-50.

But if it's stipulated as it is in the classic Monty Hall problem-- Monty always shows a goat (so Monty can't be choosing blindly) and you are always offered a choice (so Monty can't be offering you the choice for nefarious purposes)-- then you double your odds from 1/3 to 2/3 by switching.

And if there is free will behind on door and a dead horse behind the other two...

[lenny]

But if it's stipulated as it is in the classic Monty Hall problem-- Monty always shows a goat (so Monty can't be choosing blindly) and you are always offered a choice (so Monty can't be offering you the choice for nefarious purposes)-- then you double your odds from 1/3 to 2/3 by switching.
agreed.

[ThatSoundAgain]

a) the door he opens is the prize door: your chances of winning are now zero and you know it
b) the door he opens is a goat: you can increase your chances of winning by switching (the standard "two doors versus one door" argument holds regardless of Monty's ignorance).


Ruling out the possibility of a) in the standard version is exactly what increases the players' odds. You can think of the probablilties of a) and b) as merged in the standard version: "Do you want the single door you first picked or whatever prize is larger of the other two?"

This only occurs if a) is - per problem definition - an impossibility. In the random Monty scenario, you'll have 1/3 chance of winning the prize by staying, 1/3 by switching, and a 1/3 risk of having Monty end the game by picking the prize. This is as viewed from the start of the game - if you get to the point in the game where Monty has opened a non-prize door, each remaining door will have a 1/2 chance of being the winner.

Since I mapped it out anyway:


Doors | Outcome
1)Car 2)Goat 3)Goat | Stay Switch Game over
p m | 1 0 0
p m | 1 0 0
m p | 0 0 1
p m | 0 1 0
m p | 0 0 1
m p | 0 1 0

p: Players' choice
m: Monty's choice


ETA: The above is, of course, a table of the non-standard "random Monty" variation. In the classic version, I agree that you increase your odds by switching.

ETA ETA: Your odds increase in one version and not the other, not by magic but by Monty essentially telling you something you don't know. As much of this thread has been about, only if the rules are strictly as in the classic version (and known to you) can you decipher his message.

[lenny]

If you're saying what it sounds like you're saying, then no.
Make a table and fill it in.


OK: i will make a table tomorrow, but i expect it to look rather like the standard table; given additional information, i can exploit this information within the rules to change "my chances" or i know immediately that the proability of success is zero, that this also changes "my chances".

can i assume you only object to (b)? (if

(a) monty opens the door with the prize, we agree that my chances of winning have changed (they are zero)

(c, nonstandard) monty open my door, i have the option to change if i have not won, and "my chances" of winning are increased from zero (if i do not change).

if i learn one of the doors i did not pick was not the winning door, this information is of value. full stop.

[ThatSoundAgain]

lenny, I hadn't thought of your option c. Goes to show that strict definitions are necessary here - my table above operates under the rule that Monty will always choose another door than the player.

ETA: And now I see you wrote about that above. Sorry.

[lenny]

thanks ThatSoundAgain yours appeared as i was typing...
if you get to the point in the game where Monty has opened a non-prize door, each remaining door will have a 1/2 chance of being the winner.
...so you conclude "my chances of winning" have changed, but that there is no value in switching. i had thought there was, but in fact my argument only supported the claim that the "chances of winning" had changed. in this we agree.

[lenny]

lenny, I hadn't thought of your option c. Goes to show that strict definitions are necessary here - my table above operates under the rule that Monty will always choose another door than the player.

ETA: And now I see you wrote about that above. Sorry.

no worries. i was not sure what the rules were!

agreed: strict definitions are required for clear analysis.

[ThatSoundAgain]

thanks ThatSoundAgain yours appeared as i was typing...

...so you conclude "my chances of winning" have changed, but that there is no value in switching. i had thought there was, but in fact my argument only supported the claim that the "chances of winning" had changed. in this we agree.

Yes, I agree completely. At that point in time, the player will have dodged the "instant lose" bullets of scenarios 3 and 5 in the table.

[Rolfe]

One more possibility, and I apologise if it's been mentioned already: what if Monty needs someone to win the prize every nth time, to maintain interest and viewer levels? On these occasions he steers the game towards you winning, changing from the default strategy of steering the game towards you losing. How often would he do this, i.e. what are your chances of being favoured?

Is there any way this puzzle can be resolved without more information?


That was roughly what I was driving at in my last post. How often is he going to try to steer you to rather than away from the prize? It depends on a lot of things, and audience reaction is one of them. If it's time for someone to win, then he's more likely to steer you in the right direction.

On the other hand you can get too far into second-guessing at that point, and I'm sure the guy was a master of suppressio veri, suggestio falsi.

You can't possibly solve the puzzle for all possible variants, because once Monty is behaving inconsistently it's officially insoluble.

Rolfe.

[Rolfe]

....
your terms here appear unclear. please just point me to the relevant post if this was already clarified or if you are interested in a nonstandard definition of the problem (like opening the chosen door).

if monty opens one of the other two doors at random, there are two options:

a) the door he opens is the prize door: your chances of winning are now zero and you know it
b) the door he opens is a goat: you can increase your chances of winning by switching (the standard "two doors versus one door" argument holds regardless of Monty's ignorance).

if he opens the door you have, you know you have won, or you know you should switch; even in this nonstandard verison you gain information and can improve your "chance of winning".

so even opening a door at random introduces more information and thus alters your "chances of winning".
(or: which of the 700 other posts should i have read?)


I think someone has already dealt with this.

I was referring to the point in the game which is dealt with in the standard presentation of the puzzle. You have chosen a door and Monty has opened one of the other two doors to reveal a goat.

If Monty deliberately opened a goat door at that point, your chances of winning are doubled by switching, because your original door remains at the 1/3rd chance it had at the beginning while the remaining door now carries 2/3rds probability of being the car.

If Monty opened either unchosen door at random, indeed there was a chance of an instant lose, another 1/3rd probability. However, once you know that didn't happen, the chances of the car being behind the remaining door are still only 1/3rd, same as the chance of it being behind your original door, so there's no benefot to switching.

This has been demonstrated approximately a gadzillion times since the start of this thread.

Rolfe.

[Herzblut]

You're right: I agree that what makes a difference is Monty's behaviour. We have to be careful when we're saying this, though: what matters is not just his behaviour this time, it's his behaviour pattern. To say that he opens a door revealing a goat is not enough information: we need to know if this always happens, or if it only happens when there is a car behind the contestant's door, or if it happens at random, or if some other conditions apply.

My viewpoint is still slightly different and I don't like talking about what "always" happens. Actually, in a game situation nothing else counts but the evidence probability P(E) in this one game. Formally, if

A= car is behind player's door ("Assumption")
E= another door with goat is opened ("Evidence")

Then, in any case, the probability of A, given E simply is

http://www.randi.org/latexrender/latex.php?$$%20P%28A%7CE%29=%20%5Cfrac%7B%5Cfrac%7 B1%7D%7B3%7D%7D%7BP%28E%29%7D%20$$

depending only on P(E). Where it's assumed that the inverse conditional probability P(E|A) is 100%. Then

- the standard case is P(E)=100%,
- the random case is P(E)=2/3,
- the bastard case is P(E)=1/3.

[lenny]

However, once you know that didn't happen, the chances of the car being behind the remaining door are still only 1/3rd, same as the chance of it being behind your original door,

once you know it is not behind one of the unchosen doors, you have additional information. the probability of the prize being behind the open door is now known to be zero, so the chance of it being behind the remaining doors can not still be 1/3 each; there are only two doors left!

one can argue that the probability of the prize being behind "the remaining door" and the probability of "your original door" are equal, but in that case it makes no sense to argue:
the chances of the car being behind the remaining door are still only 1/3rd
no?

[boooeee]

My viewpoint is still slightly different and I don't like talking about what "always" happens. Actually, in a game situation nothing else counts but the evidence probability P(E) in this one game. Formally, if

A= car is behind player's door ("Assumption")
E= another door with goat is opened ("Evidence")

Then, in any case, the probability of A, given E simply is

http://www.randi.org/latexrender/latex.php?$$%20P%28A%7CE%29=%20%5Cfrac%7B%5Cfrac%7 B1%7D%7B3%7D%7D%7BP%28E%29%7D%20$$

depending only on P(E). Where it's assumed that the inverse conditional probability P(E|A) is 100%. Then

- the standard case is P(E)=100%,
- the random case is P(E)=2/3,
- the bastard case is P(E)=1/3.


Nicely put.

[Rolfe]

once you know it is not behind one of the unchosen doors, you have additional information. the probability of the prize being behind the open door is now known to be zero, so the chance of it being behind the remaining doors can not still be 1/3 each; there are only two doors left!

one can argue that the probability of the prize being behind "the remaining door" and the probability of "your original door" are equal, but in that case it makes no sense to argue:

no?


Yes, you're right, I mis-spoke.

Originally, 1/3rd chance of picking correctly first time, 1/3rd chance that it was behind the door that Monty opened, and 1/3rd chance that it's behing the third door.

But after Monty has opened that door (at random), 1/3rd chance on the original door and 1/3rd chance on the remaining door now becomes 50/50 on the two unopened doors.

Sorry.

Rolfe.