Monty Hall Problem

[Yaotl]

Good god (or gods), two more pages since I went home? I'm going to make a solemn vow never to discuss the MHP with anyone ever again.

[CurtC]

Paul and Bill, I think we all agree that the probability will be very different depending on whether the host always offers the switch, or offers it only if you picked right initially, etc., in other words, his strategy. And we all agree that if he was constrained to offer the switch without regard to your initial pick, that the answer is 2/3 in favor of switching.

Universal agreement on these, in my experience with this problem, is quite an accomplishment.

The only thing we seem to still disagree on is whether, in the problem statement as commonly stated, it's reasonable to assume that Monty's offer to switch would have been granted independent of your first guess. I have to admit that I can't understand the position that the assumption is warranted. Since you two are the ones making it, could you defend it? I'm curious how the reasoning would go. (I'm explicitly not inviting TeaBag420 to respond, because his m.o. so far has been to spew insults and exasperation with no substance.)

[Dr Adequate]

If Monty Hall had made a practice of offering the switch only when the first guess was right, that would have been a fairly poor strategy. And the answer would be "never switch". And no-one would be in the least interested in the problem.

The Monty Hall Problem is as stated. He always unveils one goat. He can't have a "strategy" as such. It doesn't matter whether this was true of the actual Monty Hall, or whether it's a scurrilous lie. Maths problems can be fictional.

And yes, the answer is that you should switch.

The shortest argument goes like this. When you picked, there was a 1/3 chance you were right first time. After the unveiling, there's still a 1/3 chance you were right first time --- nothing can change that. This means that there's a 2/3 chance that the car's behind the unselected drawn curtain.

[CurtC]

Originally posted by Dr Adequate
The Monty Hall Problem is as stated. He always unveils one goat.Maybe you can point me to a statement of the problem where he always unveils one goat? I don't think I've ever seen one in the wild.

Let's go back to the street hustler version. Let's say it's a given that this is a fair shell game, that the hustler makes his money because he moves the shells so quickly that anyone has only a 1/3 chance of guessing correctly. You put down your money, make your pick, and then the hustler doesn't reveal your cup, but instead turns over an empty cup and offers you the choice to switch. Should you?

You'd be a fool to switch in this case, I hope that's obvious. And this problem is no different in substance from the Monty Hall problem. The only difference is what your assumed motivations for the game host are.

[Number Six]

I wonder what the original wording was in Ask Marilyn. I don't recall it saying that Monty always switches and also she fooled many people educated in the field of probability and if she had made clear that the switch would always be made then I think fewer people would have been fooled.

In general, I often hear the puzzle posed without it made clear that Monty always switches. It's just "You're on Let's Make
A Deal...you pick a door...Monty opens another door and shows you a goat and asks you if you want to change your choice...should you change your choice?" When it's worded that way there is no correct answer unless you make assumptions.

[BillHoyt]

This is just getting monumentally silly, folks. This is a probability problem, pure and simple, despite all the what-ifs, how-abouts, assumptions, lace, garters and speckled dwarves you try to toss in. Here is a clear statement, from the MathWorld site:

Assume that a room is equipped with three doors. Behind two are goats, and behind the third is a shiny new car. You are asked to pick a door, and will win whatever is behind it. Let's say you pick door 1. Before the door is opened, however, someone who knows what's behind the doors (Monty Hall) opens one of the other two doors, revealing a goat, and asks you if you wish to change your selection to the third door (i.e., the door which neither you picked nor he opened). The Monty Hall problem is deciding whether you do.

The correct answer is that you do want to switch. If you do not switch, you have the expected 1/3 chance of winning the car, since no matter whether you initially picked the correct door, Monty will show you a door with a goat.


MathWorld on Monty Hall Problem (http://mathworld.wolfram.com/MontyHallProblem.html)

[Number Six]

I think MathWorld is wrong then. If Monty is standing there and you pick #1 and then he opens another door and reveals a goat and asks you if you want to change your choice, it only makes sense to change if you assume Monty was going to ask you if you wanted to change no matter which door you chose initially.

OTOH, if Monty is standing there and says to you "I know what is behind the three doors and after you make your choice then I'm going to open another door and reveal a goat and ask you if you want to change your choice" then it does make sense for you to switch.

The description on MathWorld sounds like the first case instead of the second.

[BillHoyt]

Originally posted by Number Six
I think MathWorld is wrong then. If Monty is standing there and you pick #1 and then he opens another door and reveals a goat and asks you if you want to change your choice, it only makes sense to change if you assume Monty was going to ask you if you wanted to change no matter which door you chose initially.

OTOH, if Monty is standing there and says to you "I know what is behind the three doors and after you make your choice then I'm going to open another door and reveal a goat and ask you if you want to change your choice" then it does make sense for you to switch.

The description on MathWorld sounds like the first case instead of the second.

"since no matter whether you initially picked the correct door, Monty will show you a door with a goat."

[Yaotl]

Originally posted by Number Six
I think MathWorld is wrong then. If Monty is standing there and you pick #1 and then he opens another door and reveals a goat and asks you if you want to change your choice, it only makes sense to change if you assume Monty was going to ask you if you wanted to change no matter which door you chose initially.

OTOH, if Monty is standing there and says to you "I know what is behind the three doors and after you make your choice then I'm going to open another door and reveal a goat and ask you if you want to change your choice" then it does make sense for you to switch.

The description on MathWorld sounds like the first case instead of the second.

I don't see a practical difference in your two scenarios.

[BillHoyt]

Originally posted by Yaotl
I don't see a practical difference in your two scenarios.

If Monty only offered the switch choice when you had originally picked the correct door...

[Dr Adequate]

Originally posted by CurtC
Maybe you can point me to a statement of the problem where he always unveils one goat? I don't think I've ever seen one in the wild.
That just is what the "Monty Hall Problem" is. No account of it says "and then the host chooses whether of not to open a door without a goat behind it", they all say along the lines of:
Finalists in a tv game show are invited up onto the stage, where there are three closed doors. The host explains that behind one of the doors is the star prize - a car. Behind each of the other two doors is just a goat. Obviously the contestant wants to win the car, but does not know which door conceals the car.

The host invites the contestant to choose one of the three doors. Let us suppose that our contestant chooses door number 3. Now, the host does not initially open the door chosen by the contestant. Instead he opens one of the other doors - let us say it is door number 1. The door that the host opens will always reveal a goat. Remember the host knows what is behind every door!

The contestant is now asked if they want to stick with their original choice, or if they want to change their mind...
If it's optional, then we have a completely different problem.
Let's go back to the street hustler version. Let's say it's a given that this is a fair shell game, that the hustler makes his money because he moves the shells so quickly that anyone has only a 1/3 chance of guessing correctly. You put down your money, make your pick, and then the hustler doesn't reveal your cup, but instead turns over an empty cup and offers you the choice to switch. Should you?

You'd be a fool to switch in this case, I hope that's obvious. And this problem is no different in substance from the Monty Hall problem. The only difference is what your assumed motivations for the game host are.
You should switch if he does that every time, street hustler or Monty Hall. If you watched, and he only did it sometimes, then you'd get suspicious.

[Number Six]

Originally posted by BillHoyt
"since no matter whether you initially picked the correct door, Monty will show you a door with a goat."

If we knew that beforehand then yes, it is correct to switch. But note that MathWorld only told us that Monty would definitely show us a goat in their explanation of the correct solution. That is, they told us that _after_ they told you that it was best to switch.

In order to determine if it is best to switch we need to know at the time we're making the decision whether Monty was going to reveal a goat no matter what door we chose. If we don't know Monty's motiviations at the time we're making our decision then we can't tell what is best _unless_ we make assumptions about Monty's motivations.

[gnome]

Originally posted by Number Six
If we knew that beforehand then yes, it is correct to switch. But note that MathWorld only told us that Monty would definitely show us a goat in their explanation of the correct solution. That is, they told us that _after_ they told you that it was best to switch.

In order to determine if it is best to switch we need to know at the time we're making the decision whether Monty was going to reveal a goat no matter what door we chose. If we don't know Monty's motiviations at the time we're making our decision then we can't tell what is best _unless_ we make assumptions about Monty's motivations.

I have always favored the version where he always reveals a goat because stated that way, it is a simple puzzle with a non-intuitive answer. It illustrates something curious about probability. I don't think it's a stretch to suppose that the author of the puzzle intended it this way.

[BillHoyt]

Originally posted by Number Six
If we knew that beforehand then yes, it is correct to switch. But note that MathWorld only told us that Monty would definitely show us a goat in their explanation of the correct solution. That is, they told us that _after_ they told you that it was best to switch.

In order to determine if it is best to switch we need to know at the time we're making the decision whether Monty was going to reveal a goat no matter what door we chose. If we don't know Monty's motiviations at the time we're making our decision then we can't tell what is best _unless_ we make assumptions about Monty's motivations.

:rolleyes: Do this experiment. Read your monitor. Now move in closer and read it again. Now closer. Closer still. Notice how there is a point, past which, closer reading is counterproductive? You're beyond that point.

The problem statement doesn't say "maybe" Monty reveals the goat. It says he reveals the goat. Period.

[garys_2k]

I had NO idea this thread would do this! Amazing, it's really amazing.

[CurtC]

Originally posted by Dr Adequate
Now, the host does not initially open the door chosen by the contestant. Instead he opens one of the other doors - let us say it is door number 1. The door that the host opens will always reveal a goat. Remember the host knows what is behind every door!
If it's optional, then we have a completely different problem.But in the example you yourself use here, it is given that the host always reveals a goat. In that case, the correct answer is "switch." I have no quarrel with that one.

I just don't think it's reasonable to assume, the way the problem is usually stated (see the OP of this thread), that Monty is constrained to offer you the switch always. Maybe that's 'cause I used to watch the show as a kid, and I know how freewheeling and impromptu it was. But all you know from most problem statements is that you've been offered a choice this time, from which it's quite a leap of assumption-making to arrive at the conclusion that the game is always played this way.

Cecil Adams put it well in his column on the subject (http://www.straightdope.com/classics/a3_189.html):Your analysis of the game show question is correct, Bobo, only if we make several assumptions: (1) Monty Hall knows which door conceals the prize; (2) he only opens doors that do NOT conceal the prize; and (3) he always opens a door. Assumptions #1 and #2 are reasonable. #3 is not.

...

Cecil is happy to say he has heard from the originator of the Monty Hall question, Steve Selvin, a UCal-Berkeley prof (cf American Statistician, February 1975). Cecil is happy because he can now track Steve down and have him assassinated, as he richly deserves for all the grief he has caused. Hey, just kidding, doc. But next time you have a brainstorm, do us a favor and keep it to yourself.

[Number Six]

If the Monty's decision to reveal a goat to you is unrelated to whether you guessed correctly then you should switch.

That means that if Monty decided to reveal a goat to you before you announced your decision then you should switch.

Or if Monty decided to reveal a goat to you after you announced your decision but the basis on which he made his decision to reveal the goat was independent of whether your choice was right, then you should switch.

But if Monty's decision to reveal a goat to you is related to whether your choice was correct then you have to know _how_ his decision and your choice were related in order to decide if you should switch.

[rppa]

Originally posted by BillHoyt
The problem statement doesn't say "maybe" Monty reveals the goat. It says he reveals the goat. Period.

You're not the first person to say this gives you useful information about how Monty acts in general. But nobody so far has said WHY this is useful.

Please show me your analysis by which this is adequate information to assess the optimum strategy.

[TeaBag420]

Originally posted by Number Six
I wonder what the original wording was in Ask Marilyn.

As I've mentioned several times, I have posted that text, with attribution to Whitaker/Whittaker/whatever on this thread.

I leave it as an exercise to you to find it. It's here somewhere.

On the other hand, if only there were a network of computers where one could search for such information.

[CurtC]

As noted in Marilyn is tricked by a game show host (http://www.wiskit.com/marilyn/gameshow.html), the question as posed to her did NOT say that the rules to the game are that Monty must always show a goat and switch. According to Wikipedia, the question was:Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

By the way, by the time Marilyn published her column with this puzzle, I was already familiar with it - it was a weekly puzzler on Car Talk before that. However, at that time I wasn't yet hip to the subtlety of Monty's motivations.

[Yaotl]

Originally posted by BillHoyt
If Monty only offered the switch choice when you had originally picked the correct door...

I meant that both situations say that it's best to switch.

[Paul C. Anagnostopoulos]

CurtC said:
Maybe you can point me to a statement of the problem where he always unveils one goat? I don't think I've ever seen one in the wild.
Agreed, but since the problem makes no hint at its being conditional, I think it's fair to assume it's not. And, as Dr. A said:
If Monty Hall had made a practice of offering the switch only when the first guess was right, that would have been a fairly poor strategy. And the answer would be "never switch". And no-one would be in the least interested in the problem.
The point is to solve the problem and learn something, right?

garys_2k said:
I had NO idea this thread would do this! Amazing, it's really amazing.
You're too easily amazed. :D If we simply agreed, what in hell would we talk about? We all agree, so we expand the topic to include something we can argue about. We're skeptics, man!

~~ Paul

[Cabbage]

Here's a question for those of you who still claim (incorrectly) that enough information is given in the problem (as stated in the OP or in Marilyn's original column) to conclude that your probability of winning is 2/3 when you switch:

Say you randomly pick a positive integer. What is the probability the number you pick is even?

Now this question seems to be pretty clear in what it's asking. I know that it might not be clear how this relates to the car and goats problem, but if you'll humor me and respond, I promise I'll try to make the connection shortly in the course of the discussion.

[Cabbage]

Oh, and I forgot to mention; please justify your answer.

[RussDill]

Its a lot easier to think about if you think about picking goats instead of cars. Intuitively, you have a 66% chance of picking a goat. Monty has a 100% chance of picking a goat. You also cannot both pick the same door.

What is the probability that you both selected goats? 66%

[epepke]

Originally posted by Cabbage
Here's a question for those of you who still claim (incorrectly) that enough information is given in the problem (as stated in the OP or in Marilyn's original column) to conclude that your probability of winning is 2/3 when you switch.

I admire your tenacity, Cabbage, but seriously, it's a lot easier and more satisfying to win free drinks and meals off of these people. I've found empirically to about a 90% confidence rate that they just don't listen. And so, you can win a free drink or meal nine times out of ten, and who will give you better odds than that?

[Paul C. Anagnostopoulos]

Cabbage said:
Here's a question for those of you who still claim (incorrectly) that enough information is given in the problem (as stated in the OP or in Marilyn's original column) to conclude that your probability of winning is 2/3 when you switch.
Aaaaarrrggh! All right, I give! Uncle! Uncle! No informal consideration of math problems shall be allowed. All problems will be specified to the degree that they are rendered pedantic and uninteresting.

Say you randomly pick a positive integer. What is the probability the number you pick is even?
This is poorly specified. Is "randomly picking" the same as "picking randomly"? Am I picking from among all positive integers or only a finite subset? Does "even" mean divisible by 2?

~~ Paul

[Cabbage]

Originally posted by Paul C. Anagnostopoulos
Aaaaarrrggh! All right, I give! Uncle! Uncle! No informal consideration of math problems shall be allowed. All problems will be specified to the degree that they are rendered pedantic and uninteresting.
But being specific is a fundamental part of mathematics! The specificity doesn't make the problem uninteresting, it makes the problem solvable. Are you claiming that simply adding the phrase "Mony always opens a door and offers a switch" would make an otherwise interesting problem pedantic and uninteresting?

One problem I've always noticed in discussions between people skilled in math and people with little mathematical experience is that people lacking experience are often unable to ask a mathematical question in a meaningful way.

For example, a friend of mine may come up to me and tell me of some astounding coincidence like, "I started talking to this stranger on the bus today, and it turns out he was from the same hometown as my mother and used to date her! What are the chances of that?" In some cases, they may even take such a coincidence as a meaningful providential intervention.

What are the chances of that? Who can say? I don't know. In order to ask a meaningful probability question, a framework must be established.

What span of time are we looking at? I mean, are you interested in the probability of that happening today, or just at some point throughout your entire lifetime? Was the event significant to you because the guy had formerly dated your mother, or would any mutual connection between you and the stranger have been significant? All such details (not limited to the two specifically listed here) must be laid out in order to establish this framework.

On that note, this is exactly the sort of issue I have with the Monty Hall problem in its typical form (i.e., without specifying that Monty always opens a door and gives a choice).

Now honestly, don't get me wrong, I'm not at all bothered by those who choose to interpret it this way Marilyn interpreted it, and claim odds of winning as 2/3 if you switch. What baffles me is why many people don't even realize that they are, in fact, making an assumption in proposing that solution. That assumption, as has been stated many times before, is that Monty always opens a door and offers a switch. This was never stated in the problem. If you want to argue that you are making the assumption in order to make sense of an otherwise unanswerable problem, fine; let's just be clear that you are in fact making that assumption.
This is poorly specified. Is "randomly picking" the same as "picking randomly"? Am I picking from among all positive integers or only a finite subset? Does "even" mean divisible by 2?
I don't see the difference between "randomly picking" and "picking randomly"; as I understand them, they mean the same thing. Could you clarify the distinction as you understand it?

Yes, you are picking from among all (infinitely many) positive integers, and, of course "even" means divisible by 2.

[Cabbage]

To be fair, I must admit I'm kind of feeling like a playground bully right now for bringing up that problem regarding picking a positive integer at random. I feel I must admit that, like my earlier chess example, it's not analogous to Monty Hall in a probabilistic sense, but in the sense that both are vague.

If anyone wants to continue to discuss it, I'll certainly go along, but the details running around in that problem run much deeper than the Monty Hall problem. To be fair, I should at least post a link (http://en.wikipedia.org/wiki/Probability_axioms) to Kolmogorov's probability axioms first, so that everything is out in the open. These are the standard axioms for a probability function P.

[Paul C. Anagnostopoulos]

Cabbage said:
But being specific is a fundamental part of mathematics! The specificity doesn't make the problem uninteresting, it makes the problem solvable. Are you claiming that simply adding the phrase "Mony always opens a door and offers a switch" would make an otherwise interesting problem pedantic and uninteresting?
It's not the problem that's pedantic, it's this discussion. By all means, though, clarify the standard wording of the problem. You're right, it won't hurt. But to claim that the problem as worded was ambiguous and thus could not serve its purpose to surprise people about their intuitions is ... well ... pedantic. In fact, I daresay that such a clarification would do absolutely no good for the very people the problem is meant to educate.

... let's just be clear that you are in fact making that assumption.
Absolutely clear. All I was saying is that seems like precisely the default assumption, given no mention of any conditions whatsoever.

I don't see the difference between "randomly picking" and "picking randomly"; as I understand them, they mean the same thing. Could you clarify the distinction as you understand it?
:D Just yanking your chain. Actually, though, "randomly picking" has the connotation of picking sometimes and not picking other times, whereas "picking randomly" has the connotation you want. But maybe that's just me.

But continue on. I have no idea why the probability of picking an even integer isn't .5.

~~ Paul

[epepke]

Originally posted by Cabbage
But being specific is a fundamental part of mathematics! The specificity doesn't make the problem uninteresting, it makes the problem solvable. Are you claiming that simply adding the phrase "Mony always opens a door and offers a switch" would make an otherwise interesting problem pedantic and uninteresting?

Moreover, and I think more importantly, skeptics, if they're worth their salt, should be able to detect this kind of missing information.

Because the world is full of hucksters, and the world is full of inconclusive evidence, and skeptics by definition should be on guard against this kind of stuff.

[Paul C. Anagnostopoulos]

Had a known that the fate of the rational world depended upon rigorously specifying the Monty Hall problem in Parade magazine, I would have been much less cavalier about it, and certainly would have used more bold words.

You people are over the top. :bowl:

I love you guys.

~~ Paul

[Cabbage]

Originally posted by Paul C. Anagnostopoulos
It's not the problem that's pedantic, it's this discussion. By all means, though, clarify the standard wording of the problem. You're right, it won't hurt. But to claim that the problem as worded was ambiguous and thus could not serve its purpose to surprise people about their intuitions is ... well ... pedantic. In fact, I daresay that such a clarification would do absolutely no good for the very people the problem is meant to educate.


Absolutely clear. All I was saying is that seems like precisely the default assumption, given no mention of any conditions whatsoever.


:D Just yanking your chain. Actually, though, "randomly picking" has the connotation of picking sometimes and not picking other times, whereas "picking randomly" has the connotation you want. But maybe that's just me.

But continue on. I have no idea why the probability of picking an even integer isn't .5.

~~ Paul
If it's absolutely clear that the 2/3 vs. 1/3 solution does involve making an assumption, then I don't really have a problem with your opinion on the problem. For the record, I agree, it looks to me like whoever wrote the problem probably intended it to be interpreted the way many people do--that Monty always opens a door and offers a switch. There is that subtle ambiguity there, however, and I think it's much preferable to add that statement to the problem for clarity. What can it possibly hurt? The problem is still as tricky as ever, and now it is unambiguous.

In general, poorly worded problems frustrate me, and some textbooks have more than their share. Personal anecdote: When I was a freshman in college taking a physics class, I remember struggling with one particular homework problem. The idea of the problem was: A man throws a ball at a certain velocity (under earth's gravity). How far does the ball travel? The answer was in the back of the book, but I couldn't figure out how to get it. It seemed to me that the height of the man would matter--Why wasn't I given that information? After playing around with it for a while, I finally realized--You had to assume that at the instant the ball was thrown, its initial position was ground level. Who the hell throws and releases a ball at ground level? A perfect example of a poorly worded problem.
Anyway, about the positive integers problem. I have no idea what sort of responses I may have gotten, but here's a quick rundown of the point I was planning on making in a Socratic dialogue:

Poster: Wouldn't it be 1/2? Only half of the integers are even, the other half are odd?

Me: What do you mean, both sets are infinite! How is one infinity twice as large as another? In fact, there is a bijection between them: f(n)=2n gives a 1-1 correspondence between the even integers and all of the positive integers. The two sets are the same size, or cardinality. There are sets with larger cardinalities, but these both have the same cardinality--cardinality aleph-naught.

Poster: Well, what I meant was, look at how they're ordered: 1, 2, 3, 4, .... Every other one is even. That's what I meant when I said half of them are even--every other one.

Me: Ah, I see where you're going with this now! However, what makes you think this order you're accustomed to is relevant? That's an assumption! What if some omnipotent deity took all of the positive integers, tossed them into an infinitely large urn, and then selected one at random? Now you have infinitely many integers, and infinitely many of them are even! Furthermore, they're all scrambled up now! What now?

Poster: Well, I would imagine that each integer has an equally likely chance of coming up, but since it's infinitely many out of infinitely many, I'm not sure how to evaluate that.

Me: I only brought up the omnipotent deity as a thought experiment; if we want to follow along those lines, and assume that each integer has an equally likely chance of coming up, that is yet another assumption! Is that what you wish to do?

Poster: Yes.

Me: Fair enough, we'll work under that assumption and see what happens. So every number is as likely as any other number to be selected? OK, let's call that probability p, where p is some number strictly between 0 and 1.

Poster: OK

Me: Now refer back to the Kolmogorov probability axioms, our probability function must be countably additive. So the probability of picking any number (which should obviously be a full probability of 1) will be the infinite sum of all the probabilities of the individual integers. That is, the countable sum from 1 to infinity of the positive constant p. What is that sum?

Poster: It diverges. It's infinite.

Me: Hmmm...That's a problem isn't it? it was supposed to be just 1. It doesn't make any sense to have infinite probability.
Poster: Well maybe the probability of picking a particular integer isn't positive. Maybe each integer has zero probability of being picked!

Me: An interesting idea, but we've got a problem there, too. Like before, what's the sum from 1 to infinity of the constant zero?

Poster: Zero.

Me: Exactly. Again, it needs to be 1 in order for us to have a probability model.

Poster: What now?

Me: Well, we've reached a dead end with this assumption. The probability of picking an individual integer has to be zero or one, and either possibility leads to an impossibility. Our initial assumption that each integer is just as likely as any other is faulty--that situation can't happen....

[end dialogue]

Anyway, I'm only (once again) trying to illustrate that we should always be clear when we make assumptions and what those assumptions are. In some cases, like this one, it's entirely possible to find out our assumptions don't make any sense. (Obviously that wasn't going to be a problem with the Monty Hall problem, but, in general terms, always be aware of your assumptions).

[TeaBag420]

Originally posted by CurtC
Paul and Bill, I think we all agree that the probability will be very different depending on whether the host always offers the switch, or offers it only if you picked right initially, etc., in other words, his strategy. And we all agree that if he was constrained to offer the switch without regard to your initial pick, that the answer is 2/3 in favor of switching.

Universal agreement on these, in my experience with this problem, is quite an accomplishment.

The only thing we seem to still disagree on is whether, in the problem statement as commonly stated, it's reasonable to assume that Monty's offer to switch would have been granted independent of your first guess. I have to admit that I can't understand the position that the assumption is warranted. Since you two are the ones making it, could you defend it? I'm curious how the reasoning would go. (I'm explicitly not inviting TeaBag420 to respond, because his m.o. so far has been to spew insults and exasperation with no substance.)

If I'm "curt" with you, it's because time is a factor here. Bluck me, motherfower.

[Paul C. Anagnostopoulos]

Haven't we assumed the probability of picking a given integer = 0 in various other conversations we've had here? I thought that was because it was infinitessimally small, so we called it zero. But all the probabilities would still add up to 1. No?

So what do we say when we run this experiment a bunch of times and half the integers are even? Sounds almost supernatural now.

In general, poorly worded problems frustrate me, and some textbooks have more than their share.
I'm with you there, man! I find stupid problems in my kids' math books all the time. You put the Monty Hall problem in my kid's math book, I expect it to be unambiguously worded.

I think I may have just pedanticized myself in the foot.

~~ Paul

[TeaBag420]

Originally posted by epepke
Moreover, and I think more importantly, skeptics, if they're worth their salt, should be able to detect this kind of missing information.

Because the world is full of hucksters, and the world is full of inconclusive evidence, and skeptics by definition should be on guard against this kind of stuff.

Yeah, if only someone had pointed out that the shell game was a con.

Was man nicht kennt, darum muss man steigen.

Or something pretty close to that.

[TeaBag420]

Here, YET AGAIN, for my learning disabled friends (you retarts know who you are) is the original MTP:

"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the other doors, opens another door, say No. 3, which has a goat. He then says to you, 'Do you want to pick door No. 2?' Is it to your advantage to take the switch?"

Guess what? It doesn't ask about the probability. MARILYN CHANGED THE PROBLEM (BTW, FU Mr. Reading Comprehension). A lot of you seem to be saying the problem is defective. WHERE HAVE I HEARD THAT BEFORE?

The answer of course is yes, switch.

More later.

Don't forget the unspoken assumption that the third door is NOT numbered "9".

What if you already had a car but you always wanted a goat?

What if pigs had wings?

[Cabbage]

Originally posted by Paul C. Anagnostopoulos
Haven't we assumed the probability of picking a given integer = 0 in various other conversations we've had here? I thought that was because it was infinitessimally small, so we called it zero. But all the probabilities would still add up to 1. No?
I don't remember ever talking about that here (not that it didn't necessarily happen), but it's been brought up a few times on the Straight Dope board that I post regularly on; do you ever post over there?

Anyway, yes, that's a very interesting question. Like I said, it's impossible in standard probability theory, but I believe it is possible in something called nonstandard probability, where instead of assigning an event a standard probability from the real interval [0,1], you assign the event a nonstandard probability, from the hyperreal (or surreal) interval [0,1]. That would include infinitesimals, so you could say each integer has a certain (nonzero) infinitesimal chance of getting picked. That's about all I know of nonstandard probability theory, though.

So what do we say when we run this experiment a bunch of times and half the integers are even? Sounds almost supernatural now.
There are, of course, (nonuniform) probability functions on the positive integers, but any question like, "What's the probability of picking an even integer?" are unanswerable without specifying a particular model.

[garys_2k]

Should have been:

"Does he always open the door? It makes a difference, you know."

"Yes, he always opens the door."

"Ok."

Instead, it was:

"You know, the OP never said if he always opened the door, and you know you can't answer the question unless you know that, it's unanswerable and so many people make the mistake of assuming he always does but maybe he doesn't and if that's the assumption then they're just wrong because it does make a big difference, you know it might be a con game where he only shows you a door if he knows you picked the good prize and if you didn't he might not give you the choice so in that case you really have to know what he does and the OP didn't tell us but he should have because, well, you people just shouldn't make assumptions like you did because that's just sloppy and I can't stand sloppy thinking and I am just so much better than you other dorks and idiots here because I don't miss the implied assumptions and won't let that sort of thing that always drives me crazy go without putting up a good pedantic argument about it."

See the difference?

Now, what are the odds that a thread like this would attract a pedant like that?

[TeaBag420]

Originally posted by garys_2k
Should have been:

"Does he always open the door? It makes a difference, you know."

"Yes, he always opens the door."

"Ok."

Instead, it was:

"You know, the OP never said if he always opened the door, and you know you can't answer the question unless you know that, it's unanswerable and so many people make the mistake of assuming he always does but maybe he doesn't and if that's the assumption then they're just wrong because it does make a big difference, you know it might be a con game where he only shows you a door if he knows you picked the good prize and if you didn't he might not give you the choice so in that case you really have to know what he does and the OP didn't tell us but he should have because, well, you people just shouldn't make assumptions like you did because that's just sloppy and I can't stand sloppy thinking and I am just so much better than you other dorks and idiots here because I don't miss the implied assumptions and won't let that sort of thing that always drives me crazy go without putting up a good pedantic argument about it."

See the difference?

Now, what are the odds that a thread like this would attract a pedant like that?

Which one are you talking about?

[rppa]

Originally posted by TeaBag420
Here, YET AGAIN, for my learning disabled friends (you retarts know who you are) is the original MTP:

"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the other doors, opens another door, say No. 3, which has a goat. He then says to you, 'Do you want to pick door No. 2?' Is it to your advantage to take the switch?"

Guess what? It doesn't ask about the probability.

I'm happy to see a non-probabilistic analysis.

The answer of course is yes, switch.

Gee, I guess I'm a learning disabled retart, but somehow that doesn't look like an analysis.

Could you amplify just a little? Starting by defining "to your advantage"?

[garys_2k]

Originally posted by TeaBag420
Which one are you talking about?
:D

Yeah, there are several. I was thinking of the leafy garden plant, though. the one with "a short stem and a dense globular head of usually green leaves that is used as a vegetable" (or so says Merriam-Webster).

[Art Vandelay]

Originally posted by CurtC

Cecil Adams put it well in his column on the subject (http://www.straightdope.com/classics/a3_189.html):
Your analysis of the game show question is correct, Bobo, only if we make several assumptions: (1) Monty Hall knows which door conceals the prize; (2) he only opens doors that do NOT conceal the prize; and (3) he always opens a door. Assumptions #1 and #2 are reasonable. #3 is not.
Actually, there's a fourth assumption: that he never opens your own door. Imagine this situation: suppose one of the goats starts braying. Monty Hall says "Well, I guess we know that one is not a car." and opens that door, revealing the goat. Should you switch?

Originally posted by epepke
I admire your tenacity, Cabbage, but seriously, it's a lot easier and more satisfying to win free drinks and meals off of these people. I've found empirically to about a 90% confidence rate that they just don't listen. And so, you can win a free drink or meal nine times out of ten, and who will give you better odds than that?
If you're relying on empiricism, don't you have an expectation of 3/10 of a free meal? 90% don't listen, but of those, one third will luck out and win anyway.

[TeaBag420]

Originally posted by epepke
I admire your tenacity, Cabbage, but seriously, it's a lot easier and more satisfying to win free drinks and meals off of these people. I've found empirically to about a 90% confidence rate that they just don't listen. And so, you can win a free drink or meal nine times out of ten, and who will give you better odds than that?

How long does it take you to convinince them that you have a bigger penis, um, I mean, that they have failed to solve the problem. It's been days here and statistically I would bet that someone has a bigger penis than you.

Being right and getting paid are two different things, Swinging Dick. I smell a liar.

[69dodge]

Originally posted by Paul C. Anagnostopoulos
Haven't we assumed the probability of picking a given integer = 0 in various other conversations we've had here? I thought that was because it was infinitessimally small, so we called it zero. But all the probabilities would still add up to 1. No?No, but you can do a similar thing with real numbers.

You can randomly pick a real number between, say, 3 and 8. In this case, the probability of any single number is 0. But just looking at the probability of single numbers misses the main idea here. The probability of picking 6.5 is 0; the probability of picking 12.5 is also 0. So then what does it mean to say that we're picking a number between 3 and 8?? It means we can say things like, the probability is 1/5 that the number is between 6 and 7 but the probability is 0 that it's between 12 and 13. The meat is in the probability of intervals, not in the probability of individual points.So what do we say when we run this experiment a bunch of times and half the integers are even?What experiment? How would you actually go about generating random integers?

You can generate random bits easily enough: just flip a coin. But how do you turn them into a random integer?

You can turn them into a random real number between 0 and 1 by sticking a radix point in front of them. (To get a number between 3 and 8, just multiply by 5 and then add 3.) Of course, you'll have to flip the coin forever if you want to get a specific number; but each successive flip narrows the range down, so that only a finite number of flips are needed to decide whether the number is between 6 and 7, or between 12 and 13. And if you use this method to generate a real number between 3 and 8, the probability is indeed 1/5 that it will be between 6 and 7, and indeed 0 that it will be between 12 and 13.

[Paul C. Anagnostopoulos]

Good point, dodge. We can generate random integers between 1 and n, but then each has a probability of 1/n and half are even. But we cannot generate random integers between 1 and infinity, so it's difficult to run the experiment.

Why can we assume the probability is zero for picking real numbers, but not for integers? The set has to be uncountably infinite?

Cabbage, we assumed a probability of zero during a conversation started in one of Interesting Ian's threads. It's this gargantuan monster, I think:

http://forums.randi.org/showthread.php?s=&threadid=43483&highlight=probability

~~ Paul

[Interesting Ian]

Originally posted by rppa
You will never settle it once and for all.

The answer depends on Monty's rules of behavior.



Indeed. One needs to ask 2 questions.

a) Does Monty know which door the car is behind.

b) If he does, then what's his game? In other words what's he trying to make you do. Also pertinent here would be the question of whether he knows you are familiar with this question, and familiar with all possible responses etc. It's a psychological game.


Want to create even more arguments? Here's another one that has the net bitterly divided:

You meet a woman with her son. She tells you she has two children. What is the probability the other child is a boy? [/B]

Just over half at a guess. I lack the info to give a specific response though.

[rppa]

Originally posted by Paul C. Anagnostopoulos
Good point, dodge. We can generate random integers between 1 and n, but then each has a probability of 1/n and half are even. But we cannot generate random integers between 1 and infinity, so it's difficult to run the experiment.

Why can we assume the probability is zero for picking real numbers, but not for integers? The set has to be uncountably infinite?

Yes, sort of. It's a little more technical than that, I think. The key thing is that the integral of the probability density over the set of possible values has to be equal to 1. That is, the probability that X lies somewhere in the set of possibilities better be 1.

Maybe the easiest way to say it is this: the sum of a countable number of 0's is 0. But the sum of an uncountable number of 0's can be nonzero.

On the other hand, if you wanted a non-uniform distribution over all integers, that would be possible. For instance suppose you wanted
P(X=n) = a/n^2, n=1, 2, ...

Since the sum (1/n^2) from 1 to infinity is some finite value (called zeta(2)), setting a = 1/zeta(2) makes this add up to 1.

[Interesting Ian]

Originally posted by epepke
As I said, I have empirical evidence that this cannot be explained. It just bounces off people's foreheads. That's why I prefer to win drinks off of them.

However, with full expectation that few people will read this, and those who read it will fight it, I feel comfortable in saying the following. Nothing in the original question to vos Savant indicated that Monty is in any way obligated to offer a choice. For all you know, he could only offer you the choice if you had picked the curtain with the car. Which means that a strategy to pick the other curtain would fail 100% of the time.



Yes absolutely.



But people are just so terribly impressed with themselves for understanding what is essentially the kind of math that should reasonably be expected from a 14-year-old that they work terribly hard to avoid the obvious conclusion. Which means they suck at being skeptics. Which is why I don't feel bad for winning free drinks off of them and humiliating them in front of their peers. Because skeptics should think about these things. [/B]

So apparently I don't suck at being a skeptic :D

I know I know . .logical fallacy LOL

[Interesting Ian]

Originally posted by Dr Adequate


The Monty Hall Problem is as stated. He always unveils one goat.

But is this happenstance or because of an obligation? :p

[Interesting Ian]

Originally posted by Dr Adequate

The shortest argument goes like this. When you picked, there was a 1/3 chance you were right first time. After the unveiling, there's still a 1/3 chance you were right first time --- nothing can change that. This means that there's a 2/3 chance that the car's behind the unselected drawn curtain. [/B]

Wrong, further information can alter the probability; indeed this monty problem is highly unusual in that it doesn't. If for instance Monty's choosing of the door was random, but yet still revealed a goat, this would mean your original choice has increased in probability from 1/3rd to 1/2 of being the car.

[Interesting Ian]

Originally posted by Cabbage


Say you randomly pick a positive integer. What is the probability the number you pick is even?

[/B]

Randomly pick a positive number?? How pray would you do that?? How indeed could you do that even with the help of a genuine random process in the world??

[Interesting Ian]

Originally posted by Paul C. Anagnostopoulos
Haven't we assumed the probability of picking a given integer = 0 in various other conversations we've had here? I thought that was because it was infinitessimally small, so we called it zero. But all the probabilities would still add up to 1. No?

~~ Paul

Yeah, that huge argument I had with everyone else apart from about 3 others who agreed with me (dodge69, vorticity, Cecil). Something about me saying a finite sequence of random numbers will eventually come across any string of numbers you care to specify. That it is guaranteed to occur in a finite string.

Almost everyone said I was stupid and that only an infinite string of numbers could guarantee it :) People never did realise that they were wrong and said that I don't understand because I packed in Maths at the age of 16 LOL

Anyway, that involved a lot about 1 over infinity having a probability of 0, but nevertheless not being logically impossible; not to mention the distinction I made between infinite and unlimited :D

[Interesting Ian]

WOW!! That thread's still here! :D

http://forums.randi.org/showthread.php?s=&threadid=36384&highlight=unlimited+infinite

[Vorticity]

Originally posted by Interesting Ian
WOW!! That thread's still here! :D

http://forums.randi.org/showthread.php?s=&threadid=36384&highlight=unlimited+infinite
Leaping, creeping Jeebus, let's not get into THAT again...

[rppa]

Originally posted by Interesting Ian
Yeah, that huge argument I had with everyone else apart from about 3 others who agreed with me (dodge69, vorticity, Cecil). Something about me saying a finite sequence of random numbers will eventually come across any string of numbers you care to specify. That it is guaranteed to occur in a finite string.

If the finite string has length n, you can in principle calculate the probability that a given string of length m never occurs. That probability is nonzero.

I don't want to open up what was obviously a huge can of worms, but look: Pick any length-n string (n 0's for instance) that doesn't contain your length-m string. The probability of that particular length-n string occurring is 1/10^n. That's small, but it isn't zero. And there are a great many strings that don't contain your target string, each of them having a probability of 1/10^n of occurring.

Anyway, that involved a lot about 1 over infinity having a probability of 0, but nevertheless not being logically impossible; not to mention the distinction I made between infinite and unlimited :D

I think I see where you're going. This gets into the kind of arguments I have with people who can't understand how you can say that there are infinitely many integers, but all of them are finite. I think you're coming down on the side of the angels with that one. There is indeed a mathematical distinction between infinite (the cardinality of the set of integers for instance) and finite but unbounded (the magnitude of any particular integer).

So if what you're saying is "if you wait long enough, the probability is 1 that the length-m will occur in finite time", then you are correct. That's different from picking one particular length-n string.

I'm going to guess that you argued that "probability = 1" means "certain". That's another eternal argument in the math forums. It doesn't. I'll bet the term "almost certain" came up in that discussion.

Sorry if I've given flashbacks to the other readers.

[Interesting Ian]

Originally posted by rppa
If the finite string has length n, you can in principle calculate the probability that a given string of length m never occurs. That probability is nonzero.

I don't want to open up what was obviously a huge can of worms, but look: Pick any length-n string (n 0's for instance) that doesn't contain your length-m string. The probability of that particular length-n string occurring is 1/10^n. That's small, but it isn't zero. And there are a great many strings that don't contain your target string, each of them having a probability of 1/10^n of occurring.



I think I see where you're going. This gets into the kind of arguments I have with people who can't understand how you can say that there are infinitely many integers, but all of them are finite. I think you're coming down on the side of the angels with that one. There is indeed a mathematical distinction between infinite (the cardinality of the set of integers for instance) and finite but unbounded (the magnitude of any particular integer).

So if what you're saying is "if you wait long enough, the probability is 1 that the length-m will occur in finite time", then you are correct. That's different from picking one particular length-n string.

I'm going to guess that you argued that "probability = 1" means "certain". That's another eternal argument in the math forums. It doesn't. I'll bet the term "almost certain" came up in that discussion.

Sorry if I've given flashbacks to the other readers.

I was never talking about an upfront specified finite number. So maybe we agree there.

Yes I think probability = 1 is "certainty". In order for it not to be certain there has to be a possibility of it not being certain. Logical possibility is not relevant. A logical possibility can not occur ie it cannot be a possible future event. It can only have occurred in retrospect.

[Yahweh]

Originally posted by hgc
I know most of you are familiar with this one already, but we must settle it once and for all...

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?
I wrote a program to test this problem, and keeping your original door limits your odds to about 1/3 and switching bumps you up to 1/2.

Edit to add: Scratch that. I reread the problem again, and I saw I made a mistake. I assumed that one of the doors which can be eliminated was the door which was chosen. I've edited my code below (edits appear in green), and I'm getting the correct answer of 2/3.

Private Sub CommandButton1_Click()
Dim SwitchYes As Long
Dim SwitchNo As Long
Dim StayYes As Long
Dim StayNo As Long

Dim Doors(3) As Boolean
Dim Choice(3) As Boolean

Dim MyDoors As Integer
Dim MyDoors2 As Integer
Dim MyChoice As Integer
Dim MyChoice2 As Integer

Dim SwitchTrials As Long
Dim StayTrials As Long

Dim I As Long
Dim J As Long

Randomize

For J = 0 To 100000
For I = 0 To 3
Doors(I) = False
Choice(I) = False
Next I

MyDoors = Int(Rnd * 3) + 1 'This is the grandprize door
MyChoice = Int(Rnd * 3) + 1 'This is the original choice

Doors(MyDoors) = True
Choice(MyChoice) = True

If Int(Rnd * 2) + 1 = 1 Then '<=== Changing some of these values can affect
'switch 'whether you want the program to switch everytime,
For I = 0 To 3 'stay everytime, or decide at random whether to
Choice(I) = False 'switch door or stay.
Next I

Do
'This eliminates one of the goat doors. Correction: This eliminates a goat door of the two remaining unchosen doors.
MyDoors2 = Int(Rnd * 3) + 1
If MyDoors2 <> MyDoors And MyDoors2 <> MyChoice Then
Exit Do
End If
Loop

Do
'This chooses one of the remaining doors
MyChoice2 = Int(Rnd * 3) + 1
If MyChoice2 <> MyChoice And MyChoice2 <> MyDoors2 Then
Choice(MyChoice2) = True
Exit Do
End If
Loop

If Choice(MyDoors) = True Then
SwitchYes = SwitchYes + 1
Else
SwitchNo = SwitchNo + 1
End If

SwitchTrials = SwitchTrials + 1
Else
'stay
If Choice(MyDoors) = True Then
StayYes = StayYes + 1
Else
StayNo = StayNo + 1
End If

StayTrials = StayTrials + 1
End If
Next J

MsgBox "Switch Trials (" & SwitchTrials & "):" & vbNewLine & _
"Switch Doors And Won Car: " & SwitchYes & " / " & Left(SwitchYes / (SwitchTrials + 1), 6) & vbNewLine & _
"Switch Doors And Won Goat: " & SwitchNo & " / " & Left(SwitchNo / (SwitchTrials + 1), 6) & vbNewLine & _
vbNewLine & _
"Stay Trials (" & StayTrials & "):" & vbNewLine & _
"Stayed And Won Car: " & StayYes & " / " & Left(StayYes / (StayTrials + 1), 6) & vbNewLine & _
"Stayed And Won Goat: " & StayNo & " / " & Left(StayNo / (StayTrials + 1), 6)
End Sub


Here are the graphical results:

[epepke]

Originally posted by Interesting Ian
So apparently I don't suck at being a skeptic :D

No sarcasm; you're doing a pretty good job here.

[epepke]

Originally posted by TeaBag420
How long does it take you to convinince them that you have a bigger penis, um, I mean, that they have failed to solve the problem.

About 20 minutes.

It's been days here and statistically I would bet that someone has a bigger penis than you.

Maybe. I'm a good sport, and I'll pay up if I lose. This, however, hasn't happened yet.

Being right and getting paid are two different things, Swinging Dick. I smell a liar.

You should get out more often, then. And it's Bwana Dik, not Swinging Dick.

[epepke]

Originally posted by Paul C. Anagnostopoulos
Had a known that the fate of the rational world depended upon rigorously specifying the Monty Hall problem in Parade magazine, I would have been much less cavalier about it, and certainly would have used more bold words.

You people are over the top. :bowl:

I love you guys.

~~ Paul

Paul, I realize this is just an attempt at a cheap put-down on your part, but I think this is fair dinkum considering how much press this conundrum got in the Skeptical Inquirer, pushed as it was by that guy (I'm terrible at names) small brunette, physician in the Tampa Bay area, had two white Trans-Ams in the 1990s, Gary Posner maybe? Anyway, whether the name is right or not, he had a column in SI and talked about it, and since then, getting the 2/3 answer has become a shibboleth of having a "Skeptic" tattoo on your forehead, which is mostly if not completely unrelated to being skeptical.

I wasn't the one who elevated it to Shibboleth Status, but I do note that when I subscribed to SI, I saw only one letter to the editor about it that displayed real skeptical thinking.

[Cabbage]

Originally posted by garys_2k
:D

Yeah, there are several. I was thinking of the leafy garden plant, though. the one with "a short stem and a dense globular head of usually green leaves that is used as a vegetable" (or so says Merriam-Webster).
(Looks around...:confused: ) I don't see anyone around here fitting that description. My leaves are usually red.






;)

Seriously, though, it looks like I've managed to partially hijack this thread from Monty Hall into a more general probability discussion; that wasn't my intention, but maybe that's a good thing...

(reads some of the recent posts)

...or maybe not.

Anyway, I think I've said all I can possibly say about Monty; anymore and I would simply be repeating myself (though in reality I probably reached that stage three pages ago). If you still don't agree with me, I don't guess I'm gonna change that.

Regarding picking an integer at random vs. a real number at random, yes, of course there's no mechanism by which this experiment can be carried out; it's a thought experiment, but certainly a worthwhile one, with all sorts of applications, I'm sure.

To have a uniform probability distribution on a set, the set does need to be either finite or uncountable; a countable set is where such a notion breaks down.

In the uncountable case, there may still be some "limitations". For example, take a uniform probability distribution on the real interval [0,1], Lebesgue measure, say.

So you can say things like (as has been mentioned before), the probability of your number being less than 1/3 is 1/3, or the probability of it being between 3/7 and 5/7 is 2/7, or that it has a probability of 0 of being in the Cantor set.

The probability of a particular number being picked is zero. Probability zero does not mean "impossible", which is demonstrated by the simple observation that while every particular number has zero probability of being picked, some number has to be picked.

Similarly, probability one is not synonymous with "certain".

Anyway, the "limitation" I was referring to is that there are some subsets of the real interval [0,1] for which it doesn't make sense to ask what the probability of our number being in that set is. Such sets are not in the domain of our probability function. I believe the axiom of choice is required to guarantee the existence of such sets.

[T'ai Chi]

Nice simulation Yahweh.

I'll repeat my analysis (http://www.martini.nu/justin/lmad.htm) which agrees.

[Art Vandelay]

Originally posted by Interesting Ian
Wrong, further information can alter the probability; indeed this monty problem is highly unusual in that it doesn't. If for instance Monty's choosing of the door was random, but yet still revealed a goat, this would mean your original choice has increased in probability from 1/3rd to 1/2 of being the car. And if he reveals that the door that you originally chose has a car behind it, you'd be a fool to switch.

[Paul C. Anagnostopoulos]

epepke said:
Paul, I realize this is just an attempt at a cheap put-down on your part,
Indeed, I was just yanking various chains. Also, it was an opportunity to use the bowling smilie, for reasons that now escape me.

but I think this is fair dinkum considering how much press this conundrum got in the Skeptical Inquirer, pushed as it was by that guy (I'm terrible at names) small brunette, physician in the Tampa Bay area, had two white Trans-Ams in the 1990s, Gary Posner maybe?
You can remember that he had two white Trans Ams, but not his name? You sound like me! [I wonder if the plural should be Transes Am?]

Anyway, whether the name is right or not, he had a column in SI and talked about it, and since then, getting the 2/3 answer has become a shibboleth of having a "Skeptic" tattoo on your forehead, which is mostly if not completely unrelated to being skeptical.
Ah, now this is interesting. I had not picked up on this becoming some sort of skeptical right of passage. Certainly, in that context, blithely assuming the conditions of the problem without explicit clarification would be a mistake.

In the interest of being a "true skeptic" and not a "skeptical buffoon," :D I will retract my statement that all this concern over the precise wording of the Monty Hall problem is pedantic. All future statements of the problem should be worded precisely. However, I stick with my claim that that will not help most people see the light, because it is not any possible ambiguity that is the barrier to understanding.

~~ Paul

[gnome]

Originally posted by Interesting Ian
Yeah, that huge argument I had with everyone else apart from about 3 others who agreed with me (dodge69, vorticity, Cecil). Something about me saying a finite sequence of random numbers will eventually come across any string of numbers you care to specify. That it is guaranteed to occur in a finite string.

Almost everyone said I was stupid and that only an infinite string of numbers could guarantee it :) People never did realise that they were wrong and said that I don't understand because I packed in Maths at the age of 16 LOL

Anyway, that involved a lot about 1 over infinity having a probability of 0, but nevertheless not being logically impossible; not to mention the distinction I made between infinite and unlimited :D

Hey, except for me that said it was not guaranteed to appear even IN an infinite string. But I'm not getting into it again, for my comments read the thread.

[gnome]

Originally posted by Paul C. Anagnostopoulos
Good point, dodge. We can generate random integers between 1 and n, but then each has a probability of 1/n and half are even. But we cannot generate random integers between 1 and infinity, so it's difficult to run the experiment.

Actually, it's easy to pick a random integer between 1 and infinity.

Pick a random decimal number n between 0 and 1 (most random number generators do this) and your random integer is 1/n, rounded off.

[Robin]

Originally posted by TeaBag420
So you offer two possibilities, both of which say it's a math problem, but previously you said it's not a math problem. Care to jack, um, I mean explain?

Perhaps you would care to "firetrucking" read what I said.

There are two possibilities both of which are math problems. But if you don't know which it is, then it is not a math problem.

It is a case, as I have said numerous times, of clarifying the question.

In fact 99% of the debate in this forum is generated by not clarifying the question.

(edited for spelling)

[TeaBag420]

Originally posted by rppa
I'm happy to see a non-probabilistic analysis.

The answer of course is yes, switch.

Gee, I guess I'm a learning disabled retart, but somehow that doesn't look like an analysis.

Could you amplify just a little? Starting by defining "to your advantage"?
If you want a car, you maximize your chances of getting a car. If you prefer a goat (not that there's anything wrong with that) think about how many goats you could fu...uh, get in exchange for a car. Either way, you win. That's "to your advantage" Johnny Shortbus.

[rppa]

Originally posted by TeaBag420
Originally posted by rppa
I'm happy to see a non-probabilistic analysis.

The answer of course is yes, switch.

Gee, I guess I'm a learning disabled retart, but somehow that doesn't look like an analysis.

Could you amplify just a little? Starting by defining "to your advantage"?

If you want a car, you maximize your chances of getting a car. If you prefer a goat (not that there's anything wrong with that)

No, no, that's fine. OK, so "to my advantage" means "maximizing the probability that I will get a car."

Now that's a well-defined mathematical problem. It's simple to solve if we have a model for what the probability of getting a car is under various scenarios.

But you claim that the answer to this maximization problem does not require that knowledge. All it requires is the knowledge that in the game you just played, where you don't know what's behind your door, Monty opened a door containing a goat.

So please enlighten me. Without making any more assumptions about Monty's behavior, tell me what the probability of getting a car is if I switch, the probability if I don't switch, and why switching is better. Because what you've written, basically "obviously the answer is to switch" still doesn't look like an analysis to me.

[rppa]

Originally posted by gnome
Actually, it's easy to pick a random integer between 1 and infinity.

Pick a random decimal number n between 0 and 1 (most random number generators do this) and your random integer is 1/n, rounded off.

Yes, this is valid. Just not uniformly distributed. It's easy to come up with probablity distributions over all the integers between 1 and infinity, and procedures for using them to pick a random integer. It's just not possible to define the uniform distribution on [1,infinity].

In your distribution, there is a 33% chance that you will come up with a number between 1/1.5 and 1, which means that there is a 33% chance that the number you pick will be equal to 1.

[Robin]

Epepke, why are you appointing yourself the arbitrer of true skeptical thinking when you got it wrong in the first place yourself?

You said "Actually, the ultimate result is that it's whatever Monty feels like doing at the time." which is of course wrong.

It may or may not depend on Monty's feelings, this is not clear in the question.

I fail to see why those who make one assumption are 'tards and those who make another assumption are true skeptics.

[Robin]

Yahweh revealed his code to solve the problem, let me reveal the code which I think defines the problem:


<PRE>
BOOLEAN monty_hall_problem(PARAM host_strategy, PARAM contestant_strategy,
BOOLEAN host_must_offer_choice, BOOLEAN host_might_reveal_car)
{
Integer winningdoor=nextRandomInteger(3);
Integer firstguess=nextRandomInteger(3);
Integer revealed_prize;
Integer finalguess=firstguess;
if (host_must_offer_choice OR
host_decides_to_offer_choice(winningdoor,
firstguess,host_strategy))
{
revealed_prize=openDoor(winningdoor,firstguess,
host_might_reveal_car);
if (revealed_prize==winningdoor) return LOSE;
if (contestant_switches(contestant_strategy) )
finalguess=anythingbut(firstguess,revealed_prize);
}
if (finalguess==winningdoor) return WIN;
return LOSE;
}
</pre>


Now if I were to ask what was the probability of this function returning WIN given that the contestant_switches function always returns TRUE then you would not be able to answer. It would depend on the results of the functions that are not given.

The true critical thinker would ask for the missing information to be supplied and not jump at one conclusion or the other.

I am prepared to confess I jumped at one conclusion, are the "it depends on the intentions of the host" crowd prepared to admit they jumped at the other?

(Incidentally, here are the results again of the program when it is run: )<table border=1>
<tr><td align=center colspan=5>Host Never Reveals Prize</td></tr>
<tr><td rowspan=2 valign=center>Contestant Strategy</td><td align=center colspan=4>Host Behaviour</td></tr>
<tr><td>Bad Monty</td><td>Fair Monty</td>
<td>Good Monty</td><td>Rules say offer change</td></tr>
<tr><td>1. Never switch even if offered</td>
<td>0.33</td><td>0.33</td><td>0.33</td><td>0.33</td></tr><tr><td>2. Toss a coin if offered</td>
<td>0.17</td><td>0.42</td><td>0.66</td><td>0.50</td></tr><tr><td>3. Always switch if offered</td>
<td>0.00</td><td>0.50</td><td>1.00</td><td>0.67</td></tr><tr><td align=center colspan=5>Host May Reveal Prize</td></tr>
<tr><td rowspan=2 valign=center>Contestant Strategy</td><td align=center colspan=4>Host Behaviour</td></tr>
<tr><td>Bad Monty</td><td>Fair Monty</td>
<td>Good Monty</td><td>Rules say offer change</td></tr>
<tr><td>1. Never switch even if offered</td>
<td>0.33</td><td>0.33</td><td>0.33</td><td>0.33</td></tr><tr><td>2. Toss a coin if offered</td>
<td>0.17</td><td>0.33</td><td>0.50</td><td>0.33</td></tr><tr><td>3. Always switch if offered</td>
<td>0.00</td><td>0.33</td><td>0.67</td><td>0.33</td></tr></table>

[Paul C. Anagnostopoulos]

gnome said:
Actually, it's easy to pick a random integer between 1 and infinity.

Pick a random decimal number n between 0 and 1 (most random number generators do this) and your random integer is 1/n, rounded off.
If the machine has f fraction bits in its widest floating-point format, the largest number you're going to select with this algorithm is about 2^f.

~~ Paul

[epepke]

Originally posted by Robin
Epepke, why are you appointing yourself the arbitrer of true skeptical thinking when you got it wrong in the first place yourself?

You said "Actually, the ultimate result is that it's whatever Monty feels like doing at the time." which is of course wrong.

It may or may not depend on Monty's feelings, this is not clear in the question.

I fail to see why those who make one assumption are 'tards and those who make another assumption are true skeptics.

I already said that I've found empirically that it's impossible to explain this, and your response is a case in point. You're really grasping at very tiny straws so that you can believe I am wrong. Believe what you like. P.T. Barnum said, "never try to smarten up a chump." He was right.

I don't think of myself as an arbiter of skeptical thinking. I think of myself as a lifeform who occasionally preys on skeptics, taking advantage of their gullibility.

[Robin]

Epepke
P.T. Barnum said, "never try to smarten up a chump." He was right.


Thanks for the advice, I won't try to convince you anymore. I forgot it was "miss the point" week around here.

[Robin]

Ok apart from the self-confessed chumps out there does anybody else have trouble with this concept? One assumption here is no better than the other. Gnome had it right in the first place. One group are saying that the result depends on what the host wants - which assumes that the host does have a choice. Another group says 0.667 which assumes the host did not have a choice.

Of course it is quite interesting to explore the possibilities but critical thinking consists of clarifying the question, not going off and dreaming possible interpretations. As I have said if you can assume nothing then the possibilities of the original question to Marilyn are endless.

[Cabbage]

Robin, I see where you're coming from now. You're breaking up the problem into two cases:

1. Where Monty is forced by the rules to always offer a switch, and

2. Where the rules are open and Monty may or may not offer a switch.

You correctly claim that you're making an assumption if you assume case 1 or case 2.

The point I've been trying to make all along, though, is that from my perspective, I'm not making any assumption. I claim that the problem is vague and unsolvable not because I'm assuming case 2, but because I don't have enough information to distinguish whether we are in case 1 or case 2 in the first place.

That sums up my position.

One final thing. For those that think this position is overly pedantic, I'll remind you that many math "brain teasers" are founded on some detail or quirk in the language. For example, the old problem, "I have two coins in my pocket totalling 55 cents. One of them is not a nickel. What are the two coins?"

[Robin]

You're breaking up the problem into two cases:

1. Where Monty is forced by the rules to always offer a switch, and

2. Where the rules are open and Monty may or may not offer a switch.

You correctly claim that you're making an assumption if you assume case 1 or case 2.
But I further claim that saying "it depends on what the host wants" is assuming case 2 because in case 1 the host has no room for strategy.

However you are right in saying it is "vague and unsolvable".

Maybe it is closer to that other famous old brain teaser:

How do you get a condom on an elephant? (Hint, take the "e" 's out of "elephant" and the "f" out of "weigh").

[Cabbage]

Originally posted by Robin
But I further claim that saying "it depends on what the host wants" is assuming case 2 because in case 1 the host has no room for strategy.

However you are right in saying it is "vague and unsolvable".

Maybe it is closer to that other famous old brain teaser:
I see your point about the claim, "It depends on what the host wants." It would be better to claim, "It depends on the rules of the game, and whether or not Monty is required to offer the switch."

I don't think I've heard that one before; how do you get a condom on an elephant?

[Robin]

Originally posted by Cabbage
I don't think I've heard that one before; how do you get a condom on an elephant?

To be honest I don't know if this one works in the USA or anywhere outside GB and Australia. I will leave it for a while in case anyone else wants to get it.

You've got to use the hint: (Hint, take the "e" 's out of "elephant" and the "f" out of "weigh").

The strange thing is that overly analytical people never get it, they go on for ages going "don't tell me, I'll work it out"

But most people just point out to you the obvious point about the hint, see the trick and start laughing.

[TeaBag420]

Originally posted by Robin
To be honest I don't know if this one works in the USA or anywhere outside GB and Australia. I will leave it for a while in case anyone else wants to get it.

You've got to use the hint: (Hint, take the "e" 's out of "elephant" and the "f" out of "weigh").

The strange thing is that overly analytical people never get it, they go on for ages going "don't tell me, I'll work it out"

But most people just point out to you the obvious point about the hint, see the trick and start laughing.

First, you get an Australian homo puffy boy.....the sort who is disposed to manipulate large mammals to a state of turgidity.

Homo says what?

[gnome]

How about this take on the general problem?

You have two possible situations... that Monty always offers a switch (in which case you improve your chances by switching)... or that Monty sometimes will not. (In which case your chances remain the same if you switch).

Without even knowing the likeihood of either possibility...

By switching, you are doing no worse (if Monty sometimes does not offer a switch), and possibly doing better (if Monty always offers a switch). So it's still better to switch.

:D

[epepke]

Originally posted by gnome
How about this take on the general problem?

You have two possible situations... that Monty always offers a switch (in which case you improve your chances by switching)... or that Monty sometimes will not. (In which case your chances remain the same if you switch).

Maybe Monty only gives you the choice if your original pick was the curtain with the car. You don't know that he doesn't, and it's unspecified in the problem. Where does that 50% go then?

Yes, yes. I know. I've determined empirically that there is a high likelihood that this concept is impossible to convey. And I'm missing the point, or I'm over the top, or I'm a goofball, or I'm wrong for suggesting it. Have a nice day, chumps.

[Number Six]

The likelihood that Monty will offer you the switch isn't relevant. What is relevant is whether his offer is independent of whether your initial choice was correct.

If his offer is independent of whether your initial choice was correct then you should switch.

If his offer is not independent of whether your initial choice was correct then you can't determine whether you should switch unless you know _in what way_ his offer depends on your initial choice.

Maybe he wants to make you lose, in which case he only offers the switch if your initial choice is correct. Maybe he wants you to win, in which case he only offers the switch if your initial choice is incorrect. Maybe he has decided beforehand to offer you the switch with probably X if your initial choice is correct and with probability Y if your choice is incorrect. (Incidentally, the first two sentences of this paragraph are special cases of that). If you don't know what that probability is then you can't say whether it's best to switch.

[PogoPedant]

Stick this in you Java Virtual Machine and run it!

In the app below we have two actors: Andy and Bob. Andy never switches and Bob always switches. Because of this, I've cheated a bit: Monty offers Bob the switch, but does not offer the same to Andy. Allthough Andy and Bob are playing with the same doors, they are ignorant of the other player's actions. Both players can win the car; it will count as a win for both.

If you're clueless as to what to do with the code below, I guess I could post a runnable, but I couldn't be bothered making one right now...

Note that the app will take a wee while to run, and you'll only see the currently running game.



import java.util.*;
import javax.swing.*;
import java.awt.*;

/**
* @author Pogo
*
*/
public class MainMonty extends JFrame implements Runnable {

private int nsw;//never switches: Andy

private int asw;//always switch: Bob

private Random rand;

private static final int MAX = 1000000;

private int getMonty(int car, int bob) {
if (car != 0 && bob != 0)
return 0;
if (car != 1 && bob != 1)
return 1;
if (car != 2 && bob != 2)
return 2;
return 42;
}

private int bobSwitch(int monty, int bob) {
if (monty == 0 && bob == 1)
return 2;
if (monty == 0 && bob == 2)
return 1;
if (monty == 1 && bob == 0)
return 2;
if (monty == 1 && bob == 2)
return 0;
if (monty == 2 && bob == 0)
return 1;
if (monty == 2 && bob == 1)
return 0;

return 42;
}

private String g(boolean g) {
if (!g)
return "Goat";
else
return "Car";
}

public void run() {
nsw = 0;
asw = 0;
rand = new Random();

setSize(300, 200);
JTextArea text = new JTextArea();
JScrollPane scroll = new JScrollPane(text);
scroll.setPreferredSize(new Dimension(1000, 800));
getContentPane().add(scroll);
setDefaultCloseOperation(EXIT_ON_CLOSE);
show();

for (int i = 0; i < MAX; i++) {
text.setText("");
boolean[] doors = { false, false, false };
int c = rand.nextInt(3);
doors[c] = true;
text.append("Game " + i + " on!\nDoors:\n" + g(doors[0]) + ", "
+ g(doors[1]) + ", " + g(doors[2]) + ".\n");

int andy = rand.nextInt(3);
int bob = rand.nextInt(3);
int monty = getMonty(c, bob);

text.append("Andy picks door " + (andy + 1)
+ " and Bob picks door " + (bob + 1) + "\n");
text.append("Monty opens door " + (monty + 1) + " for Bob\n");
bob = bobSwitch(monty, bob);
text.append("Bob switches to " + bob + "\n");

if (doors[andy]) {
text.append("Andy wins!\n");
nsw++;
}
if (doors[bob]) {
text.append("Bob wins!\n");
asw++;
}
if (i % 1000 == 0) {
System.gc();
setTitle(MAX - i + " games left");
}
}

setTitle("finished!");
text.append("andy won " + nsw + " out of " + MAX + " games.\n");
text.append("bob won " + asw + " out of " + MAX + " games.\n");

}

public static void main(String[] args) {
System.out.println("Playing the Monty Hall!");
new MainMonty().run();
}
}

[PogoPedant]

Aaaaa!! Just noticed that code has been offered before... Ignore my ugly Java hack.. :(

[Art Vandelay]

Originally posted by Robin
But I further claim that saying "it depends on what the host wants" is assuming case 2 because in case 1 the host has no room for strategy.
No, because which case holds depends on what the hosts wants. Are you saying that the idea that the host has free will is an "assumption"?

[epepke]

Originally posted by Number Six
[B]The likelihood that Monty will offer you the switch isn't relevant. What is relevant is whether his offer is independent of whether your initial choice was correct.

That is incisive. The probabilistic argument assumes independent variables.

I tried to explain this the third of four times I tried to explain the Monty Hall problem, and it didn't work then. It's nice to see someone grok it.

[richardm]

Originally posted by Iconoclast
Oh God, it's happening again.

Nine pages and counting :D

[hgc]

Originally posted by TeaBag420
First, you get an Australian homo puffy boy.....the sort who is disposed to manipulate large mammals to a state of turgidity.

Homo says what? The origin of Teabag's name (http://www.urbandictionary.com/define.php?term=teabagging&f=1)

[rppa]

Originally posted by Robin
I am prepared to confess I jumped at one conclusion, are the "it depends on the intentions of the host" crowd prepared to admit they jumped at the other?

Since you posted a table proving that it depends on the intentions of the host (giving probability as a function of intentions of the host), I'm confused as to what you could mean by this statement.

What conclusion is being jumped to by saying "the probabilities depend on the intentions of the host"?

[garys_2k]

Originally posted by hgc
The origin of Teabag's name (http://www.urbandictionary.com/define.php?term=teabagging&f=1)
Youch! That would force me to get a new ID if it applied.

[CurtC]

And I suppose the "420" part refers to this (http://www.phish.net/faq/420.html).

[rppa]

The recent flap about the "Read Books, Get Brain" ad campaign in New York really made me feel old and out of touch.

http://www.reuters.com/newsArticle.jhtml?type=oddlyEnoughNews&storyID=6731922

[RussDill]

Originally posted by Paul C. Anagnostopoulos
Good point, dodge. We can generate random integers between 1 and n, but then each has a probability of 1/n and half are even. But we cannot generate random integers between 1 and infinity, so it's difficult to run the experiment.

Why can we assume the probability is zero for picking real numbers, but not for integers? The set has to be uncountably infinite?

Cabbage, we assumed a probability of zero during a conversation started in one of Interesting Ian's threads. It's this gargantuan monster, I think:

http://forums.randi.org/showthread.php?s=&threadid=43483&highlight=probability

~~ Paul

Random integer = (coin flip 1) + (coin flip 2) * 2^1 + ...(coint flip n) * 2^(n - 1)


Only the first coin flip has a bearing on whether or not the integer is even or odd, but you can do you infinite coin flips to come up with your number.

[69dodge]

Originally posted by RussDill
Random integer = (coin flip 1) + (coin flip 2) * 2^1 + ...(coint flip n) * 2^(n - 1)

Only the first coin flip has a bearing on whether or not the integer is even or odd, but you can do you infinite coin flips to come up with your number.All integers are finite. But this process will, with probability 1, yield an infinitely large "integer". So it doesn't quite do what you want it to.

[RussDill]

Originally posted by 69dodge
All integers are finite. But this process will, with probability 1, yield an infinitely large "integer". So it doesn't quite do what you want it to.

Its no accident that with probability 1 it picks an infinitely large integer. This is true anytime you randomly pick an integer.

[Robin]

Originally posted by Art Vandelay
No, because which case holds depends on what the hosts wants. Are you saying that the idea that the host has free will is an "assumption"?

No I was saying that the idea that the rules allow the host room for strategy is an assumption. We must all assume that the host will follow the rules. If the second choice is part of the game then what the host wants is irrelevant.

It all depends on whether you read the question as a statement of how the game is played or as just one scenario.

[Robin]

Originally posted by rppa
Since you posted a table proving that it depends on the intentions of the host (giving probability as a function of intentions of the host), I'm confused as to what you could mean by this statement.

What conclusion is being jumped to by saying "the probabilities depend on the intentions of the host"?

The conclusion that the host did not have to offer a choice.

Again I would have to ask people to think of the difference between rules and strategy. The rules delineate what game is being played, strategy delineates how it is played.

Strategy can be altered during the game, rules cannot.

[69dodge]

Originally posted by RussDill
Its no accident that with probability 1 it picks an infinitely large integer. This is true anytime you randomly pick an integer.There's no such thing as an infinitely large integer; that's why I put it in quotes. All integers are finite.

[Art Vandelay]

Originally posted by Robin
No I was saying that the idea that the rules allow the host room for strategy is an assumption. We must all assume that the host will follow the rules. If the second choice is part of the game then what the host wants is irrelevant.
But the rules depend on what the host wants.

It all depends on whether you read the question as a statement of how the game is played or as just one scenario.
Exactly.

Originally posted by RussDill
you can do you infinite coin flips to come up with your number.
No, you can't.

PaulAll future statements of the problem should be worded precisely. However, I stick with my claim that that will not help most people see the light, because it is not any possible ambiguity that is the barrier to understanding.
I think that there are wordings of the problem that make it even more difficult to not get the right answer. For instance, suppose it were worded as follows:

"You are shown three doors. Behind one of them is a car, behind the others, goats. You ask the host 'Do you know where the car is?' The host says that he does not. You then ask 'Could you look behind the second and third doors, and then make a statement of the form "If the car is behind one of these doors, it is behind door..." '. The host agrees, and a few seconds later announces 'If the car is behind one of these two doors, it is behind door number two'. Assuming that the host is honest, which door should you pick?"

[Robin]

Art Vandelay
But the rules depend on what the host wants


Now there is a big assumption!

[Robin]

Originally posted by Number Six
The likelihood that Monty will offer you the switch isn't relevant. What is relevant is whether his offer is independent of whether your initial choice was correct.

If his offer is independent of whether your initial choice was correct then you should switch.

If his offer is not independent of whether your initial choice was correct then you can't determine whether you should switch unless you know _in what way_ his offer depends on your initial choice.

Maybe he wants to make you lose, in which case he only offers the switch if your initial choice is correct. Maybe he wants you to win, in which case he only offers the switch if your initial choice is incorrect. Maybe he has decided beforehand to offer you the switch with probably X if your initial choice is correct and with probability Y if your choice is incorrect. (Incidentally, the first two sentences of this paragraph are special cases of that). If you don't know what that probability is then you can't say whether it's best to switch.

But all of this assumes first that Monty is not obligated to make the offer under the rules of the game.

I think that we all understand the conditions that would hold if the host is free to offer or not offer the switch. I have already put a table of a number of the conditions that might hold.

But what some people fail to undertand is that all of that is irrelevant if the host is obligated to offer the switch. So this matter must be settled first before any host intentions become relevant.
(Edited to remove silly rude comment)

[Art Vandelay]

Originally posted by Robin
Now there is a big assumption! The idea that the host is not effectively a robot is a big assumption?

[Robin]

Originally posted by Art Vandelay
The idea that the host is not effectively a robot is a big assumption?

Well first off the idea that the host makes the rules is a big assumption.

I think we established earlier that the host cannot cheat. In as much as something is part of the rules then the host is effectively a robot.

If you are saying that the host has the free will to ignore the rules of the game or to arbitrarily vary them then you are adding yet another layer of ambiguity.

[hgc]

Originally posted by Art Vandelay
The idea that the host is not effectively a robot is a big assumption? I thought the idea that Monty is a robot (single scenario defines the rules) was a reasonable assumption. But too many people were able to manufacture scenarios of host behavior. I am willing to reword the puzzle to remove ambiguity about Monty's intentions (which don't exist), but I still think that people who didn't get it as worded need a good slice of Occam.

[Robin]

Originally posted by hgc
I thought the idea that Monty is a robot (single scenario defines the rules) was a reasonable assumption. But too many people were able to manufacture scenarios of host behavior. I am willing to reword the puzzle to remove ambiguity about Monty's intentions (which don't exist), but I still think that people who didn't get it as worded need a good slice of Occam.

This is how I understood your original question and you did provide a clarification so I think that inasmuch as we are debating the question put by you, rather than some other version, then the answer is clear.

[Robin]

Incidentally for those who are interested in my schoolyard puzzle earlier:

"There is no 'f' in weigh"

And there is no *****' way we will all agree on this one.:)

[69dodge]

Originally posted by Art Vandelay
I think that there are wordings of the problem that make it even more difficult to not get the right answer.More difficult, yes. But still not impossible. Watch me. :D

(Actually, I had already planned to make the point I'll make, even before I read your post. But then when I read it, it seemed like the perfect hook.)For instance, suppose it were worded as follows:

"You are shown three doors. Behind one of them is a car, behind the others, goats. You ask the host 'Do you know where the car is?' The host says that he does not. You then ask 'Could you look behind the second and third doors, and then make a statement of the form "If the car is behind one of these doors, it is behind door..." '. The host agrees, and a few seconds later announces 'If the car is behind one of these two doors, it is behind door number two'. Assuming that the host is honest, which door should you pick?"Door 3 definitely has a goat, so it's right out. I'd pick door 2, because it certainly is no worse than door 1 and it might be better.

However, we still do not have enough information to assign a definite probability of 2/3 to door 2. The probability that it hides the car might be anywhere from 1/2 to 1. Here's why. If the host sees the car, he has to tell you where it is. But if he sees two goats, he has a choice about what to say. Perhaps he decided, before looking, that if he sees two goats he'll say, "door 2." Then the probability is 1/2. Or, perhaps he decided that if he sees two goats, he'll say, "door 3." Then the probability is 1. If he chooses one or the other of the two statements at random when he has a choice, the probability that door 2 hides the car depends on the probability that he'll choose one statement over the other. Only if he chooses uniformly between them (i.e., the two are equally likely) will the probability be 2/3 that door 2 hides the car.

[Robin]

We should briefly address the "I use this to win drinks off people so it must be true" test of an argument.

There are no details about how this happens but presumably someone in a bar agrees to gamble for a drink without first asking the rules of the game.

This appears not so much a lack of skepticism as a lack of sobriety, a condition not unknown in bars.

In fact it is not difficult - "I bought the last round, don't you remember?" also works, as does exclaiming "Hey that's mine", at which someone replies "What's yours?" and everybody says "A schooner of 'New' thanks" (or whatever their favourite tipple is).

You probably know the general rule about gambling but maybe someone can tell me who said it:

"A man will approach you with a pack of cards and bet you $20 that the Jack of Hearts will jump out and squirt whipped cream in your ear. Don't take the bet, or you will end up $20 poorer with an earful of whipped cream".

(edit)Hey I broke the 10 - do I get a cookie or something?(/edit)

[Art Vandelay]

Originally posted by Robin
Well first off the idea that the host makes the rules is a big assumption.
Well, I said the rules "depend" on the host, which just means that he has some say. And how is that a big assumption? Is the idea that Letterman has some control over the content of his monologue a big assumption? Heck, there's nothing in the problem that actually says there are rules. As far as the problem says, there are no rules other than the whim of the host.

hgcI thought the idea that Monty is a robot (single scenario defines the rules) was a reasonable assumption.I really don't see how that's a reasonable assumption. If it is, then two description of the same exact scenario would result in completely different answers.

1. "Three doors, yadda yadda. The host opens one of the doors you didn't pick and reveals a goat. Should you switch?"

2. "Yadda, yadda. The host opens door number two, and reveals a goat. Should you switch?"

3. "The host opens one the doors that you didn't pick and shows you what's behind it. How should you react?"

These three descriptions are of the exact same events. A person actually experiencing them would not be able to tell which was the "right" description; only a person who isn't actually faced with the choice, and is merely responding to a hypothetical, would see a difference. What sense does it make for a person actually on the show to not know what to do, but someone reading a second-hand account to know?

Here's another puzzle:
"There are one hundred doors, 60 cars and 40 goats. You pick door number twelve. The host opens all the doors except number twelve and one other one, revealing 39 goats. The host then allows you another choice, this time between the two remaining doors. Does it matter which one you pick?

I don't think this is just a nitpick. A lot of people make decisions based on what they see as being the situation, rather than thinking about whether their viewpoint is skewed.

but I still think that people who didn't get it as worded need a good slice of Occam.
Didn't Occam counsel against making assumptions, no matter how "obvious" they are?

69dodgeHowever, we still do not have enough information to assign a definite probability of 2/3 to door 2. The probability that it hides the car might be anywhere from 1/2 to 1. Here's why. If the host sees the car, he has to tell you where it is. But if he sees two goats, he has a choice about what to say. Perhaps he decided, before looking, that if he sees two goats he'll say, "door 2." Then the probability is 1/2.Interesting point. But I don't see how this is any more valid than in the original case. Is that what you meant by "I had already planned to make the point I'll make"?

I'll have think about this a bit more. On one hand, no flaw in your reasoning jumps out at me. On the other hand, "Ask this question of the host, then switch" is clearly a better strategy than simply picking and sticking.

[Robin]

Art Vandelay
Heck, there's nothing in the problem that actually says there are rules. As far as the problem says, there are no rules other than the whim of the host.


Well you said it not me. If you think that it is a reasonable interpretation of the problem that the contestant is on a game show with no rules whatsoever except the whim of the host.

I am sure that any activity without rules qualifies as a game but as of now I officially lose interest in this silly debate.

[CurtC]

Originally posted by Robin
Well you said it not me. If you think that it is a reasonable interpretation of the problem that the contestant is on a game show with no rules whatsoever except the whim of the host.Did you ever see the show Let's Make a Deal? That's exactly what it was. The whim of the host, a very colorful character, Monty Hall.

[hgc]

Originally posted by Robin
...

(edit)Hey I broke the 10 - do I get a cookie or something?(/edit) Behind one of these 3 doors is a cookie...

[hgc]

Originally posted by Art Vandelay
...

hgc

I really don't see how that's a reasonable assumption. If it is, then two description of the same exact scenario would result in completely different answers.

1. "Three doors, yadda yadda. The host opens one of the doors you didn't pick and reveals a goat. Should you switch?"

2. "Yadda, yadda. The host opens door number two, and reveals a goat. Should you switch?"

3. "The host opens one the doors that you didn't pick and shows you what's behind it. How should you react?"

These three descriptions are of the exact same events. A person actually experiencing them would not be able to tell which was the "right" description; only a person who isn't actually faced with the choice, and is merely responding to a hypothetical, would see a difference. What sense does it make for a person actually on the show to not know what to do, but someone reading a second-hand account to know?Gadzooks! I just fell out of my chair!

You give 3 descriptions of scenarios that could conceivably the be same events, but only 1 of them comports the description I gave in the initial puzzle. So what's the point of saying that the other 2 could be the same event? I didn't say the host opens door #2. I did say that the host opens one of the doors you didn't pick. The difference is everything. I left unspecified those details which could change case to case, so that my description would fit with all the possible paths the game might follow, such as what door you picked first and what door Monty reveals. I specified the details that are consistent for every possible path -- Monty reveals a goat. I tried to just precise enough so that a puzzler would infer from the specifics and lack thereof what the rules of the game are -- without being too pedantic in trying to cut off every avenue of expansion beyond the point of solving a probability puzzle.

Here's another puzzle:
"There are one hundred doors, 60 cars and 40 goats. You pick door number twelve. The host opens all the doors except number twelve and one other one, revealing 39 goats. The host then allows you another choice, this time between the two remaining doors. Does it matter which one you pick?That's too hard. What's the answer?
...
Didn't Occam counsel against making assumptions, no matter how "obvious" they are?Not exactly. Occams suggests not to manufacture any more entities than are necessary to hypothesize and explanation. Your manufactured scenarios, 2 and 3, are not required.

[Art Vandelay]

Originally posted by hgc
You give 3 descriptions of scenarios that could conceivably the be same events, but only 1 of them comports [with] the description I gave in the initial puzzle.
What do you mean? All of them comport with your description.

I didn't say the host opens door #2.
You didn't say he doesn't.

That's too hard. What's the answer?
According to your logic, it doesn't make a difference.

Not exactly. Occams suggests not to manufacture any more entities than are necessary to hypothesize and explanation. Your manufactured scenarios, 2 and 3, are not required. But their negations are not required either. You are hypothesizing that Monty makes a habit out of doing this stuff. I say that it is possible to exclaim the events without reference to such an assumption, so Occam favors my point of view.

[hgc]

With another Monty Hall thread already going strong, I first thought it best to leave this one safely buried. But then the masochist in me couldn't let this go unanswered.Originally posted by Art Vandelay
What do you mean? All of them comport with your description.No, no, no, no, no. Convinced? OK, I'll go into details:

Your #2 - "Yadda, yadda. The host opens door number two, and reveals a goat. Should you switch?" I don't say that the host opens Door 2. I say the host opens one of the other doors. The difference is everything. I was not specific about the door because that is a variable from my scenario to other possible scenarios. I didn't ask "should you switch?" I ask what your chances are of getting the car if you should switch. Just so there's no distraction about whether the car is what you want, or the goat that lays the golden egg or whatever.

Your #3 - "The host opens one the doors that you didn't pick and shows you what's behind it. How should you react?" I don't leave out that he reveals the goat; I specify that he reveals the goat. I don't ask generally how you should react. I ask what your chances are of getting the car if you switch your choice.

You didn't say he doesn't [open door #2].I explained that already. I didn't say it because it's not part of the problem. The problem doesn't depend on which door number is which. It works the same way for doors 1, 2, 3 or doors a, b, c or doors green, blue, yellow; in any order in any scenario.
According to your logic, it doesn't make a difference.Sorry, but I didn't attempt to solve your puzzle. I hope it wasn't crucial to making your point.
But their negations are not required either. You are hypothesizing that Monty makes a habit out of doing this stuff. I say that it is possible to exclaim the events without reference to such an assumption, so Occam favors my point of view. No, only required by pedants. All others needn't grow a garden of weedy suppositions and red herrings.

[gnome]

Originally posted by CurtC
Did you ever see the show Let's Make a Deal? That's exactly what it was. The whim of the host, a very colorful character, Monty Hall.

The question is nearly pointless as a probability puzzle if Monty can do whatever he wants. That makes it entirely reliant on his intentions and unsolvable.

So is it unreasonable to assume that in repetitions, the scenario will be the same? If you make any other assumption than that, there's no satisfactory answer at all.

[Mason]

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?

Perhaps I'm mistaken, but the problem is pretty obviously a question of probability, not a question of psychology. It is a word problem to help illustrate a very basic level of probability, not an exercise in psychology.

Again, it is a math problem in the math forum to show basic math functions.

It is not a philosophy problem in the philosophy forum to show elements of psychology.

For example, "If a train leaves from New York at 12:00 traveling at 100 miles per hour..." Is a common word problem that we all see in early math text books. The problem is designed to illustrate certain aspects of math. We can sit around all day and ask "What if the conductor only says he's going 100 miles per hour, but in fact speeds up once he gets out of the city limits? The problem can't be solved because we don't know what the conductor will do once we lose sight of him!"

Hogwash.

It's a simple problem, anyone who is making more of it is just being obtuse for the sole purpose of being obtuse.

[BillHoyt]

I think hgc nailed it with "grow a garden of weedy suppositions and red herrings."

[billydkid]

It took me a little while to grasp this - what is sometimes referred to as The Monty Hall Problem. It has been around for a long time in various forms. It is one thing to know it, but it is something else to actually grasp it intuitively. The most obvious illustration is using an example with a million doors. You are essentially choosing between one door and all of the rest. When looked at that way it is very obvious. Still, virtually nobody I have described the problem to gets it. They can't let go of the 50/50 idea because there are two doors at the end. I have never been great at brain teasers, but it bothers me that it took some effort to wrap my mind around this. I'm more of a spacial kind of guy.

[pgwenthold]

It took me a little while to grasp this - what is sometimes referred to as The Monty Hall Problem. It has been around for a long time in various forms. It is one thing to know it, but it is something else to actually grasp it intuitively. The most obvious illustration is using an example with a million doors. You are essentially choosing between one door and all of the rest. When looked at that way it is very obvious. Still, virtually nobody I have described the problem to gets it. They can't let go of the 50/50 idea because there are two doors at the end. I have never been great at brain teasers, but it bothers me that it took some effort to wrap my mind around this. I'm more of a spacial kind of guy.

There are some who can't grasp the math, but the most vocal (and legitimate) objections come from those who claim there are certain assumptions that are made in solving the problem that are not justified, such that there is no correct answer.

If the problem is set up properly, generally to reflect the spirit in which it is intended, then sure, your answer works fine. However, the devil is in the details, and they technically matter.

[Mark6]

It took me a little while to grasp this - what is sometimes referred to as The Monty Hall Problem. It has been around for a long time in various forms. It is one thing to know it, but it is something else to actually grasp it intuitively. The most obvious illustration is using an example with a million doors. You are essentially choosing between one door and all of the rest. When looked at that way it is very obvious. Still, virtually nobody I have described the problem to gets it. They can't let go of the 50/50 idea because there are two doors at the end. I have never been great at brain teasers, but it bothers me that it took some effort to wrap my mind around this. I'm more of a spacial kind of guy.

I can think of three reasons people have such hard time with Monty Haul problem.

1. Conditional probability is something most people never learn. If you understand conditional probability, it is much easier.

2. A lot of people, including mathematically educated ones, tend to think of "probability of an event" as some objective quantity of said event, whereas in reality it is just a measure of one's ignorance. A tossed coin has "traditionally" 50% of landing on either side. But if you knew precisely the coin's velocity, rotation, aerodynamic properties, air density, and air currents, you would be able to tell with "heads" or "tails" 100% certainty. If you had only some of that information, you could say "80% chance of heads" (for example). As your knowledge of a situation changes, "probability of event" changes. And in Monty Haul you receive additional information when the host opens a door and shows you the goat.

3. The situation is incredibly contrived. Where else, outside a game show, you have to make a decision based on a partial information, from a knowledgeable source, who is deliberately withholding part of information? It is just not a common occurrence.

[IMST]

Here's another puzzle:
"There are one hundred doors, 60 cars and 40 goats. You pick door number twelve. The host opens all the doors except number twelve and one other one, revealing 39 goats. The host then allows you another choice, this time between the two remaining doors. Does it matter which one you pick?


With this puzzle you've been shown 59 cars and 39 goats, right?

If I'm reading that right, you had a 60% chance of guessing right to begin with, so you retain that percentage by staying. Switching would give you a 40% chance of getting the car since all probabilities have been collapsed to those two doors.

Of course, this assumes the host deliberately chose to leave one car and one goat uncovered.

[Skeptic Guy]

The Marquis de Carabas answer: "With that many goats, everyone is a winner."

[gnome]

:eek: Not this thread again! NO!!!! WHY, GOD, NO!!!!

http://www.applefritter.com/images/zombie_1-9316_640x480.jpg

We must encase ourselves in an underground bunker and defend it with our best weapons!

http://www.skepticalcommunity.com/phpbb2/uploads/monkeyshow.jpg

I'M RUNNING THIS MONKEY SHOW NOW, FRANKENSTEIN!!

[OnlyTellsTruths]

I'm pretty sure I arrived at a decent explanation of both sets of odds in this related topic:

http://forums.randi.org/showthread.php?t=108001

http://forums.randi.org/showthread.php?postid=3507440#post3507440

[lionking]

I didn't get it until I read the three card analogy in the wikipedia entry on the problem. I can see why so many say 50/50.

[billydkid]

:eek: Not this thread again! NO!!!! WHY, GOD, NO!!!!

http://www.applefritter.com/images/zombie_1-9316_640x480.jpg

We must encase ourselves in an underground bunker and defend it with our best weapons!

http://www.skepticalcommunity.com/phpbb2/uploads/monkeyshow.jpg

I'M RUNNING THIS MONKEY SHOW NOW, FRANKENSTEIN!!
Come on, it's not the end of the world. You know, I could have started a NEW thread. Count your blessings. On the other hand, SOMETHING compelled you to open it.

[Gravy]

With this puzzle you've been shown 59 cars and 39 goats, right?

If I'm reading that right, you had a 60% chance of guessing right to begin with, so you retain that percentage by staying. Switching would give you a 40% chance of getting the car since all probabilities have been collapsed to those two doors.

Of course, this assumes the host deliberately chose to leave one car and one goat uncovered.What do you think is the probability of getting a prompt response from the person who made that post four years ago? :)

[IMST]

What do you think is the probability of getting a prompt response from the person who made that post four years ago? :)

Wow.

Saw an active thread, never occurred to me to see if the last page contained posts from '04.

Always amazing to see these threads brought back from the dead like that.

ETA: Art Vandelay's last post: 08/07. I'ma change doors please Monty.

[lionking]

I just tried this problem on my family, and they all gave the intuitive 50/50 answer. So I used the three cards, and most still didn't - except my wife who would acknowledge that she is the least (formally) educated of the lot. She saw it immediately.

[Rolfe]

Once upon a time (perhaps in the recesses of this thread), I thought I had my brain around this.

Now I may be remembering wrong, but I thought the conclusion I came to was, switch, whatever, given the problem as set out.

A. If Monty knows where the car is and is deliberately avoiding it, then you improve your chances.

B. If Monty is as ignorant as you are, then the chances stay at 50:50.

If you have no idea which scenario Monty is running it makes sense to switch. Because B is the worst case scenario (unless the game is truly bent), and in that one, while you don't gain by switching, you don't lose either.

By sticking, you are rejecting the possibility of improving your chances if A is in fact the scenario, and getting no possible benefit out of it.

By switching, you allow yourself to take advantage of the possibility of scenario A being the situation, without exposing yourself to a decrease in your chances.

Thus, if the rules are not specified, and A is within the bounds of possibility, then by definition switching is the sensible thing to do.

Now I'm waiting for everyone to point out the three-year-old post where someone proved me wrong about this.

Rolfe.

PS. Billydkid, I approached it thinking about 100 doors and one car. A million does rather seem like overkill.

[TheBoyPaj]

You're right. You should always switch.

[GodMark2]

Once upon a time (perhaps in the recesses of this thread), I thought I had my brain around this.

Now I may be remembering wrong, but I thought the conclusion I came to was, switch, whatever, given the problem as set out.

A. If Monty knows where the car is and is deliberately avoiding it, then you improve your chances.

B. If Monty is as ignorant as you are, then the chances stay at 50:50.

If you have no idea which scenario Monty is running it makes sense to switch. Because B is the worst case scenario (unless the game is truly bent), and in that one, while you don't gain by switching, you don't lose either.

By sticking, you are rejecting the possibility of improving your chances if A is in fact the scenario, and getting no possible benefit out of it.

By switching, you allow yourself to take advantage of the possibility of scenario A being the situation, without exposing yourself to a decrease in your chances.

Thus, if the rules are not specified, and A is within the bounds of possibility, then by definition switching is the sensible thing to do.

Now I'm waiting for everyone to point out the three-year-old post where someone proved me wrong about this.

Rolfe.

PS. Billydkid, I approached it thinking about 100 doors and one car. A million does rather seem like overkill.

C: Monty may be evil, and thus only open a second door if he already knows you picked the car first, but not offering you the choice when you pick a goat.

Include that possibility in the mix, and switching (when offered) is not necessarily the best choice. With only a single instance from which to determine Monty's motive and the rules of the game, you don't have enough information.

[GreyICE]

Once upon a time (perhaps in the recesses of this thread), I thought I had my brain around this.

Now I may be remembering wrong, but I thought the conclusion I came to was, switch, whatever, given the problem as set out.

A. If Monty knows where the car is and is deliberately avoiding it, then you improve your chances.

B. If Monty is as ignorant as you are, then the chances stay at 50:50.

If you have no idea which scenario Monty is running it makes sense to switch. Because B is the worst case scenario (unless the game is truly bent), and in that one, while you don't gain by switching, you don't lose either.

By sticking, you are rejecting the possibility of improving your chances if A is in fact the scenario, and getting no possible benefit out of it.

By switching, you allow yourself to take advantage of the possibility of scenario A being the situation, without exposing yourself to a decrease in your chances.

Thus, if the rules are not specified, and A is within the bounds of possibility, then by definition switching is the sensible thing to do.

Now I'm waiting for everyone to point out the three-year-old post where someone proved me wrong about this.

Rolfe.

PS. Billydkid, I approached it thinking about 100 doors and one car. A million does rather seem like overkill. You ignore the fact that A and B are the same scenario. EXACTLY the same. If he opens it at random and happens to hit the goat, its exactly the same if he knew he hit the goat.

If he opens it at random and hits the car, you ought to know your chances of getting the car. DUH

[Rolfe]

C: Monty may be evil, and thus only open a second door if he already knows you picked the car first, but not offering you the choice when you pick a goat.

Include that possibility in the mix, and switching (when offered) is not necessarily the best choice. With only a single instance from which to determine Monty's motive and the rules of the game, you don't have enough information.


Yes, I remember that possibility being discussed earlier. I think I was including that in the category of "bent" games, though I accept that's debatable.

Rolfe.

[Rolfe]

You ignore the fact that A and B are the same scenario. EXACTLY the same. If he opens it at random and happens to hit the goat, its exactly the same if he knew he hit the goat.

If he opens it at random and hits the car, you ought to know your chances of getting the car. DUH


I'm going to bed now. Somebody else can have a pop at this one.

Rolfe.

Hint: Think about the hundred doors and one car, and whether or not you'd switch if the choice were still on offer when you were down to only two doors left shut - your door and one other.

[GreyICE]

I'm going to bed now. Somebody else can have a pop at this one.

Rolfe.

Hint: Think about the hundred doors and one car, and whether or not you'd switch if the choice were still on offer when you were down to only two doors left shut - your door and one other.
Rolfe, you're not getting it. Obviously you switch. But your scenario B isn't a scenario. It doesn't matter whether Monty opened the door at random or with foreknowledge, as long as he hit a goat.

He reveals the same information either way. It's your lack of understanding that makes you think that they're different scenarios.

The only way it doesn't influence the outcome is if the event is 50:50 going in, and you lose if it hits the car.

The interesting aspect of the Monty Haul problem is that you can use the outcome of one completely random event to influence the outcome of another event.

It's the same thing with the two kids problem - the knowledge of one random event gives you better odds on guessing the outcome of the other random event, despite the fact the two are not linked.

[ravdin]

I probably don't have anything new to add to this discussion. This came up with a group of my coworkers several years ago. We argued about it for a long time and I came up with the best arguments I could- (e.g. 100 doors, walking them through NOT switching several times to show that it really does only work for them 1 out of 3 times).

My friend then came up with the best argument I've heard yet, which was this: if you still don't believe that you should switch, meet us in the alley after work for a special game. Bring a roll of 20 dollar bills with you.

By the way, if you're not convinced by the 100 door argument, I'll play that game with you any time you like. I'll even give you 50 to 1 odds in your favor!

[gnome]

The ambiguity is in the problem statement, that doesn't indicate whether monty will show you a goat after your choice every time. If he will, then switching is a great idea. If he might not, then it isn't.

If you want the problem to have it's classic answer, you must phrase it that he would always show you a goat.

[CurtC]

You ignore the fact that A and B are the same scenario. EXACTLY the same. If he opens it at random and happens to hit the goat, its exactly the same if he knew he hit the goat.

No, if he is opening a door randomly, then your chances aren't improved by switching (it's a 50/50 choice at that point). The "2/3 by switching" answer works only if the host knows where the goat is and purposely avoids it.

[jsfisher]

No, if he is opening a door randomly, then your chances aren't improved by switching (it's a 50/50 choice at that point). The "2/3 by switching" answer works only if the host knows where the goat is and purposely avoids it.


Exactly right.

1/3 of the time, Monty choosing at random will reveal the car. Game over.
1/3 of the time, Monty will reveal a goat and your door has a goat.
1/3 of the time, Monty will reveal a goat and your door has a car.

50/50 with random Monty switching will help.

[GreyICE]

No, if he is opening a door randomly, then your chances aren't improved by switching (it's a 50/50 choice at that point). The "2/3 by switching" answer works only if the host knows where the goat is and purposely avoids it.

No it's not. As soon as he opens a door with a goat behind it, you gain certain information about your choice. If he opens a door with a car behind it, you also gain information about your choice.

While the randomness does not increase your chances of winning at the OUTSET of the game, in the middle of the game, if a door with a goat is opened, it behooves you to switch.

The interesting part of the Monty Haul problem isn't that it increases your chances if he opens a door with a goat - it's that you can obtain information about unknown members of a random system given information about other members of the system even though every bit of information was obtained randomly. Meaning despite the fact that Monty opens the door randomly, once he's opened the door, the information you received from the door opening randomly informs you about other parts of the random system. This means that you can obtain information even if there's no Monty - i.e. no one knows how the system works.

This becomes more interesting with the baby problem (where the outcome is less deterministic and seemingly more random). There you can also see how the game works with no Monty - by gaining the information that one child is a son, you learn statistical chances about the other child's gender despite the fact that no one knows how gender is determined.

[Herzblut]

No, if he is opening a door randomly, then your chances aren't improved by switching (it's a 50/50 choice at that point). The "2/3 by switching" answer works only if the host knows where the goat is and purposely avoids it.
This is impossible. Synaptic structures in his brain have absolutely nothing to do with your winning chances in this phase of game. I mean, by what kind of magic shall there be any kind of relationship?

The winning chance of your initial choice is still 1/3 and the probability that the Ferrari is behind one of the two other doors is still 2/3. That's what counts.

[GreyICE]

This is impossible. Synaptic structures in his brain have absolutely nothing to do with your winning chances in this phase of game. I mean, by what kind of magic shall there be any kind of relationship?

The winning chance of your initial choice is still 1/3 and the probability that the Ferrari is behind one of the two other doors is still 2/3. That's what counts. Finally!

Thanks.

This has some amazingly counter intuitive implications.

[Herzblut]

This has some amazingly counter intuitive implications.
What you mean? The daughter/son questions?

[GreyICE]

What you mean? The daughter/son questions?

Yeah, not the monty haul one (since now I'm confused if I'm remembering correctly, I just charted it out, and I remember that the odds increase might be from the behavioral change, since it's refusing to show up again. Also, I hate that problem. Did I mention that I hate that problem?).

Edit: No, done charting. It's conclusively 50/50 if the door is opened at random. Even if you learn the outcome. It's only if the host is benevolent. Goddamn it Monty Haul. Even if you don't care if the host opened the door at random, it turns out the universe does.

[Vorpal]

The following are probabilities of winning the game.
(1) Monty knows where the car is
Reveals random goat|Keep*: 1/2|Switch: 1/2
Goat, not the player's|Keep: 1/3|Switch: 2/3
(2) Monty doesn't know where the car is, and Monty revealing the car counts as an automatic win:
Reveals random door|Keep: 5/9|Switch: 2/3
Not the player's|Keep: 2/3|Switch: 2/3
(3) Monty doesn't know, and Monty revealing the car counts as an automatic loss:
Reveals random door|Keep: 2/9|Switch: 1/3
Not the player's|Keep: 1/3|Switch: 1/3

P.S. The disagreement with CurtC and jsfisher is apparently because they're talking about different conditions than those who object to them. If Monty (or the player, for fairness) picks a door to be revealed at random that is not the player's first guess, then they are correct in that keeping or switching doesn't matter. However, it is interesting that in all cases, switching does no worse than keeping.

Edit: Clarification: (*) this 1/2 probability switches if Monty's pick was the player's pick, otherwise keeps. If keep regardless of outcome, it is 1/3, although this is a very silly thing to do.

[lionking]

Has anyone read the OP and discussions following it? Monty opens the door and a goat is behind it. Switching results in you having two cards in the game rather than one. You have a 2/3 chance of winning.

[Herzblut]

Has anyone read the OP and discussions following it? Monty opens the door and a goat is behind it. Switching results in you having two cards in the game rather than one. You have a 2/3 chance of winning.
Right. Per definition the opened door reveals a goat, as a matter of fact not condition. How this fact has been established, randomly or by choice doesn't matter obviously. It's given fact.

In case the opened door might reveal the car as well, that's a totally different story.

[AntiTelharsic]

Switching will make you lose only if you had originally picked the door with the car. The probability that you originally picked the door with the car is 1/3. Therefore switching will only make you lose 1/3 of the time. Therefore switching makes you win 2/3 of the time. This is like spending 11 pages arguing about whether or not 1 + 2 = 3.

[lionking]

Switching will make you lose only if you had originally picked the door with the car. The probability that you originally picked the door with the car is 1/3. Therefore switching will only make you lose 1/3 of the time. Therefore switching makes you win 2/3 of the time. This is like spending 11 pages arguing about whether or not 1 + 2 = 3.
Very well put and welcome to the forum.

[Herzblut]

Switching will make you lose only if you had originally picked the door with the prize. The probability that you originally picked the door with the prize is 1/3. Therefore switching will only make you lose 1/3 of the time. Therefore switching makes you win 2/3 of the time. Does this really need 11 pages?
I understand an ambiguity here. After the moderator has opened the door, where are we?

What is the probability P our choice wins the price

-1- and the moderator has revealed a goat
-2- under the condition the moderator has not revealed the car (but the goat) although he might have, because he has chosen the door randomly?

The second option makes the result of the moderator's choice conditional, and hence leads to a conditional probability of P=1/2, which is not my understanding of the problem.

[AntiTelharsic]

Yeah, the problem as usually stated has Monty revealing a goat always; much simpler than fussing around with all the possibilities otherwise.

[Worm]

I finally got this when I saw the illustration on Wikipedia combining the doors

http://en.wikipedia.org/wiki/Monty_hall_problem#Combining_doors

This would probably be even more graphic in the '100 doors'

[Michael C]

Has anyone read the OP and discussions following it? Monty opens the door and a goat is behind it. Switching results in you having two cards in the game rather than one. You have a 2/3 chance of winning.

Right. Per definition the opened door reveals a goat, as a matter of fact not condition. How this fact has been established, randomly or by choice doesn't matter obviously. It's given fact.

In case the opened door might reveal the car as well, that's a totally different story.

No, it does make a difference if Monty knows where the car is when he chooses the door.

- If Monty knows where the car is, and always chooses to show the goat, the chance that the car is behind the contestant's chosen door stays at 1/3 and the chance that the car is behind the other remaining door is 2/3.

- If Monty chooses a door at random, the game will stop if the door reveals a car. If the door reveals a goat, the situation for the two remaining doors becomes symmetrical: there is no reason to suppose that the car has more chances of being behind one or other of the two doors.

[Herzblut]

- If Monty chooses a door at random, the game will stop if the door reveals a car.
The door doesn't reveal a car per definition.

Edit: 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.

[Rolfe]

C: Monty may be evil, and thus only open a second door if he already knows you picked the car first, but not offering you the choice when you pick a goat.

Include that possibility in the mix, and switching (when offered) is not necessarily the best choice. With only a single instance from which to determine Monty's motive and the rules of the game, you don't have enough information.


This is the only possible scenario in which switching would be a bad idea. Monty will only pull the stunt of opening another door to reveal a goat if he knows you have already picked the car. This is what I meant when I said, unless the game is rigged. Though whether you'd call that rigging, I'm not sure.

The rest is as Anti-Telharsic says.

I'm still trying to figure out what GreyICE means, and I think I understand. Would anyone else care to comment on his point that even if Monty is opening either of the other two doors at random, once you know that the door he did actually open has a goat, then the odds still favour switching?

Rolfe.

[Michael C]

I'm still trying to figure out what GreyICE means, and I think I understand. Would anyone else care to comment on his point that even if Monty is opening either of the other two doors at random, once you know that the door he did actually open has a goat, then the odds still favour switching?

If we play many times the version where Monty doesn't know where the car is and opens a door at random, it works like this:

In 1/3 of the games, the car is behind the contestant's chosen door. In 1/3 of the games, the car is behind the door that Monty chooses. In 1/3 of the games, the car is behind the door that neither of them chose. The important thing here is that when the car is revealed behind the door that Monty chooses, the game doesn't go on. We only get to the situation where the contestant has a choice between the two doors in 2/3 of the games.

This is the essential difference between this version of the game and the original one where Monty knows where the car is and chooses to reveal a goat: in the original case Monty can ensure that every game comes to the point where the contestant has a choice between the two doors. This means that if we play 300 games, all 300 of them will get to the point where the contestant has a choice between two doors. Since the contestant has chosen the door with the car in approximately 100 of the games, in the remaining 200 games the car will be behind the other closed door. Obviously he should swap doors.

In the version where Monty chooses at random, 1/3 of the games stop without the contestant getting the chance to choose between the two remaining doors. If we play the game 300 times, only 200 (approximately!) will get to the point where the contestant has the choice between the two doors. In 100 of these games, the contestant has already chosen the door with the car; in 100 of them the car is behind the other door. The contestant's chances of winning are 1/2: he can swap doors but it won't change his chances.

[Herzblut]

In the version where Monty chooses at random, 1/3 of the games stop without the contestant getting the chance to choose between the two remaining doors. If we play the game 300 times, only 200 (approximately!) will get to the point where the contestant has the choice between the two doors. In 100 of these games, the contestant has already chosen the door with the car; in 100 of them the car is behind the other door. The contestant's chances of winning are 1/2: he can swap doors but it won't change his chances.
That's wrong. You play the game only 200 times, 100 experiments are invalid and have to be disregarded because they don't correspond to the specified setup.

[Rolfe]

If we play many times the version where Monty doesn't know where the car is and opens a door at random, it works like this:

[....snip explanation of the two scenarios....]


Yes, yes, yes, I understand all that.

GreyICE appeared to me to be saying that in the version where Monty opens one of the other two doors at random, on the occasions when he reveals a goat rather than the car, there is still some advantage to changing.

Here is what I said.

Now I may be remembering wrong, but I thought the conclusion I came to was, switch, whatever, given the problem as set out.

A. If Monty knows where the car is and is deliberately avoiding it, then you improve your chances.

B. If Monty is as ignorant as you are, then the chances stay at 50:50.

If you have no idea which scenario Monty is running it makes sense to switch. Because B is the worst case scenario (unless the game is truly bent), and in that one, while you don't gain by switching, you don't lose either.

By sticking, you are rejecting the possibility of improving your chances if A is in fact the scenario, and getting no possible benefit out of it.

By switching, you allow yourself to take advantage of the possibility of scenario A being the situation, without exposing yourself to a decrease in your chances.

Thus, if the rules are not specified, and A is within the bounds of possibility, then by definition switching is the sensible thing to do.


I have already acknowledged GodMark2's additional point regarding the "bent game". However, here is what GreyICE said.

You ignore the fact that A and B are the same scenario. EXACTLY the same. If he opens it at random and happens to hit the goat, its exactly the same if he knew he hit the goat.

If he opens it at random and hits the car, you ought to know your chances of getting the car. DUH


And again....

Rolfe, you're not getting it. Obviously you switch. But your scenario B isn't a scenario. It doesn't matter whether Monty opened the door at random or with foreknowledge, as long as he hit a goat.

He reveals the same information either way. It's your lack of understanding that makes you think that they're different scenarios.

The only way it doesn't influence the outcome is if the event is 50:50 going in, and you lose if it hits the car.


I'm trying to understand if he has a point here. I'm seeing "DUH" and "you're not getting it", and "your lack of understanding", so I'm trying real hard to figure out if he might possibly be right.

I mean, arrogant posturing doesn't necessarily mean a poster is wrong, no?

Rolfe.

[billydkid]

I'm going to stretch this out even just a little further just to irritate Gnome. It's not about assumptions or anything like that. It doesn't matter what Monty knows or doesn't know. The description is this - there are 3 closed doors - 1 with a prize and 2 with crap. You pick one of the doors. Monty (and it doesn't even have to Monty) opens one of the other doors to reveal crap. And the question is are you better off switching to the remaining unchosen door or keeping your original door. You are better off switching. People bring in all sorts of irrelevancies which do nothing but confuse the issue. At first blush it seems counterintuitive, but it is not.

[Rolfe]

OK, you're agreeing with GreyICE. Would you care to elaborate?

Rolfe.

[Michael C]

That's wrong. You play the game only 200 times, 100 experiments are invalid and have to be disregarded because they don't correspond to the specified setup.

Yes, you can consider it like that if you wish. It doesn't change the result: of the 200 times you actually play the game, in 100 of them the car is behind the door you chose and in 100 of them the car is behind the other closed door. The probability of winning is 1/2.

[Michael C]

I'm going to stretch this out even just a little further just to irritate Gnome. It's not about assumptions or anything like that. It doesn't matter what Monty knows or doesn't know. The description is this - there are 3 closed doors - 1 with a prize and 2 with crap. You pick one of the doors. Monty (and it doesn't even have to Monty) opens one of the other doors to reveal crap. And the question is are you better off switching to the remaining unchosen door or keeping your original door. You are better off switching. People bring in all sorts of irrelevancies which do nothing but confuse the issue. At first blush it seems counterintuitive, but it is not.

No. Consider:

There are three doors, A, B and C. One door has a prize, the other two have crap.

The probability of the prize being behind door A is 1/3. So is the probability of it being behind door B or door C.

At random, you choose door A. You open it and find crap. What is the probability of the prize being behind door B now? What is the probability of the prize being behind door C?

[CurtC]

I'm going to stretch this out even just a little further just to irritate Gnome. It's not about assumptions or anything like that. It doesn't matter what Monty knows or doesn't know. The description is this - there are 3 closed doors - 1 with a prize and 2 with crap. You pick one of the doors. Monty (and it doesn't even have to Monty) opens one of the other doors to reveal crap. And the question is are you better off switching to the remaining unchosen door or keeping your original door. You are better off switching. People bring in all sorts of irrelevancies which do nothing but confuse the issue. At first blush it seems counterintuitive, but it is not.

This is exactly wrong. There's all kinds of misinformation swirling around here, so
Everyone please pause and read the following.

Sometimes the problem is stated such that Monty always shows you a goat before giving you the choice to switch. If it's stated that way, there is no argument that you'll win 2/3 of the time by switching.

However, usually when I see it stated, it's not explicit in constraining Monty to always offer the switch, all you know is that in this case, your only data point, he did offer the switch.

IN THIS CASE THERE IS NOT ENOUGH INFORMATION TO ARRIVE AT AN ANSWER WITHOUT KNOWING MONTY'S REASONS FOR OFFERING THE SWITCH.

Let's take three assumptions that we can make:

A: He knows where the prize is, will always reveal a goat door, and offer you the choice to switch. We've already determined that you'll win 2/3 of the time by switching.

B. Monty doesn't know where the prize is, he just happened to pick one door which was a goat, and now he's giving you the choice. In this case, if you simulate it yourself with pen and paper, you'll quickly see that your chances are not improved by switching: it's a 50/50 proposition.

C. Monty knows where the prize is, but wants you to lose the game, therefore he offers the choice to switch only for contestants who picked right the first time. In this case, obviously, you need to stick with your original guess (you'll lose 100% of the time by switching).

I hope it's clear that the host's motivation DOES matter, and that simply finding yourself in the situation where he has offered a switch is not enough information without knowing more about how Monty plays the game.

[boooeee]

OK, you're agreeing with GreyICE. Would you care to elaborate?

Rolfe.


Actually, I think GreyICE recanted in post 399.

[sol invictus]

Let's take three assumptions that we can make:

A: He knows where the prize is, will always reveal a goat door, and offer you the choice to switch. We've already determined that you'll win 2/3 of the time by switching.

B. Monty doesn't know where the prize is, he just happened to pick one door which was a goat, and now he's giving you the choice. In this case, if you simulate it yourself with pen and paper, you'll quickly see that your chances are not improved by switching: it's a 50/50 proposition.

C. Monty knows where the prize is, but wants you to lose the game, therefore he offers the choice to switch only for contestants who picked right the first time. In this case, obviously, you need to stick with your original guess (you'll lose 100% of the time by switching).

I hope it's clear that the host's motivation DOES matter, and that simply finding yourself in the situation where he has offered a switch is not enough information without knowing more about how Monty plays the game.

Agreed on all counts.

I might add one:

D) Monty doesn't know where the car is, but will always open one of the two doors you didn't pick and will always offer you the chance to switch. In that case you want to switch to the door he opens if it's the car (obviously!), and it doesn't matter whether you stay or switch (to the door he didn't open) if he opens a door to a goat.

Of the 2/3 times he opens a door to a goat, you win half no matter whether you stay or switch (that's CurtC's option B). Of the 1/3 times he opens the door to the car, you win all. So you win 2/3rds of the time in total.

[Herzblut]

Yes, you can consider it like that if you wish. It doesn't change the result: of the 200 times you actually play the game, in 100 of them the car is behind the door you chose and in 100 of them the car is behind the other closed door. The probability of winning is 1/2.
Why? Each time you play, you pick the right door with 1/3 probability. Hence, if you play 200 times you'll probably hit the car around 66 times.

[GreyICE]

Actually, I think GreyICE recanted in post 399.

I recanted because I was explicitly mathematically wrong, and can prove it.

It very much matters WHY Monty opened the door.

[Skeptical Greg]

Agreed on all counts.

I might add one:

D) Monty doesn't know where the car is.....

You can't add that one, and have the game satisfy the original puzzle ..

The puzzle calls for the car not to be revealed when Monty opens the door ...

[CurtC]

Why? Each time you play, you pick the right door with 1/3 probability. Hence, if you play 200 times you'll probably hit the car around 66 times.

We're talking about the case where Monty doesn't know where the prize is, but always opens a door you didn't pick, and in this situation, you just know that he revealed a goat.

If you say you're playing 200 times, then that means that you started choosing 300 times, and in 100 of them, Monty revealed the prize by his random guess. In the remaining 200 trials, it's in your door 100 of them and in the other door 100 of them: therefore it's a 50/50 choice.

Remember, the problem statement says that you're already in the situation where Monty has revealed a goat, therefore those 100 where he revealed the prize don't count for this analysis.

[Rolfe]

I recanted because I was explicitly mathematically wrong, and can prove it.

It very much matters WHY Monty opened the door.


OK. I tried quite hard to see why you said DUH, told me abruptly that it was me who wasn't getting it, and announced that I lacked understanding. I thought, silly me, that you wouldn't have been quite so obnoxious unless you were sure you were right.

So, you agree that I was right all along? Thanks for mentioning it.

I repeat, unless Monty is rigging the game by only offering the switch if he knows the correct door has already been picked, then one should always switch. In one scenario, the switch will not confer any benefit, but neither will it reduce the chance of winning. In the other scenario, the switch will increase the chance of winning.

Thus, if we can exclude the rigging scenario, switching makes sense whatever, because the worst it will do is make no difference, and it might improve your chances.

Rolfe.

[sol invictus]

You can't add that one, and have the game satisfy the original puzzle ..

The puzzle calls for the car not to be revealed when Monty opens the door ...

One of the main problems here is that people don't carefully state the problem, and then go on to disagree because they have different versions in mind. For example, I think most people think the answer is 50/50 because they have something like my version in mind, plus the additional information that the door which was opened had a goat behind it (so CurtC's item B).

[Skeptical Greg]

The problem was stated in the OP ..

If you want to propose a different problem, then you should do so ..

[pgwenthold]

Agreed on all counts.

I might add one:

D) Monty doesn't know where the car is, but will always open one of the two doors you didn't pick and will always offer you the chance to switch. In that case you want to switch to the door he opens if it's the car (obviously!), and it doesn't matter whether you stay or switch (to the door he didn't open) if he opens a door to a goat.


Do you really think they are going to have a game where you can switch to the door Monty opened if he reveals a car?

[GreyICE]

OK. I tried quite hard to see why you said DUH, told me abruptly that it was me who wasn't getting it, and announced that I lacked understanding. I thought, silly me, that you wouldn't have been quite so obnoxious unless you were sure you were right.

So, you agree that I was right all along? Thanks for mentioning it.

I repeat, unless Monty is rigging the game by only offering the switch if he knows the correct door has already been picked, then one should always switch. In one scenario, the switch will not confer any benefit, but neither will it reduce the chance of winning. In the other scenario, the switch will increase the chance of winning.

Thus, if we can exclude the rigging scenario, switching makes sense whatever, because the worst it will do is make no difference, and it might improve your chances.

Rolfe.
Yes, I'm sorry. Have I eaten enough crow for you, oh impenetrable text wall?

I got it conflated with the baby problem, where in fact it doesn't matter, as long as you know whether the person has a son or not. That's a variant on the coin-box problem, rather than on the forced-choice problem like Monty Haul.

[billydkid]

This is exactly wrong. There's all kinds of misinformation swirling around here, so
Everyone please pause and read the following.

Sometimes the problem is stated such that Monty always shows you a goat before giving you the choice to switch. If it's stated that way, there is no argument that you'll win 2/3 of the time by switching.

However, usually when I see it stated, it's not explicit in constraining Monty to always offer the switch, all you know is that in this case, your only data point, he did offer the switch.

IN THIS CASE THERE IS NOT ENOUGH INFORMATION TO ARRIVE AT AN ANSWER WITHOUT KNOWING MONTY'S REASONS FOR OFFERING THE SWITCH.

Let's take three assumptions that we can make:

A: He knows where the prize is, will always reveal a goat door, and offer you the choice to switch. We've already determined that you'll win 2/3 of the time by switching.

B. Monty doesn't know where the prize is, he just happened to pick one door which was a goat, and now he's giving you the choice. In this case, if you simulate it yourself with pen and paper, you'll quickly see that your chances are not improved by switching: it's a 50/50 proposition.

C. Monty knows where the prize is, but wants you to lose the game, therefore he offers the choice to switch only for contestants who picked right the first time. In this case, obviously, you need to stick with your original guess (you'll lose 100% of the time by switching).

I hope it's clear that the host's motivation DOES matter, and that simply finding yourself in the situation where he has offered a switch is not enough information without knowing more about how Monty plays the game.
No, you are wrong. You don't even have to have Monty in the equation. All you need to know is that there are 3 doors, one of which has a prize. You pick one. Your odds of it being behind that door are 1 in 3. Assume none of the other doors are revealed and you are given the chance to pick both the other doors instead (which you are essentially doing even if an empty door is revealed on that side). Your odds are 2 in 3. Or simpy draw an imaginary line between your door and the other two door - the odds are better if you pick the side on which there are two doors. This become more than obvious if you have a million doors. The one you pick on one side and the 999,999 on the other side. The prize is almost certainly on the side with the 999,999 doors. Revealing 999,998 doors on the other side that are empty virtually guarantees that the prize is behind the remaining door. The odds of it be being the door you picked remain 1 in a million. Monty is completely irrelevant to the puzzle. All we know is that a second door is opened revealing no prize and that is all we need to know.

I don't know, maybe I am wrong, but I completely fail to see what the host's intention have to do with the outcome - he has to open a second door and it is a given that the second door will not contain the prize. I fail to see how what is in the host's mind has any bearing on the statistics.

[pgwenthold]

I don't know, maybe I am wrong, but I completely fail to see what the host's intention have to do with the outcome - he has to open a second door


Unfortunately, a lot of times this stipulation is not made. You are only told that 1) he does open a door, and 2) there is a goat behind it. You are not told that he always does this when this game is played, nor that when he does it, that he knows that he will reveal a goat.


and it is a given that the second door will not contain the prize. I fail to see how what is in the host's mind has any bearing on the statistics.

As noted, you are making an assertion that is not always part of the equation: that he always does this and that the door WILL NOT contain the prize. If that is part of the problem, fine. But it is not always the case. What is often (usually?) given is that a) he HAS opened the door (nothing about it being always) and that b) it DID contain a goat (not that it always contains a goat).

If he doesn't always open a door, then you do not know that he only opened it because you guessed right. Therefore, you need to know his motivation.

Second, whether he knows which is the right door or not also matters. If he does not, and is only guessing, then there is a 1/3 chance that he could reveal one with the car. The problem with that is that, as pointed out above, he has just ruined your chance of winning because it means that you did not select the car initially. This is a case where you would have been wise to switch, but because he revealed the car accidentally, you aren't able to make that switch. Thus, it hurts your chances of winning if you switch.

You can convince yourself of this by using cards (ok, I did it with excel, but it becomes pretty obvious pretty quick - if the second door is selected randomly, then the odds of winning in those events when the car is not revealed is 50/50)

So you need to know 2 things: 1) that he is not acting maliciously, and 2) that he knows what he is doing and only reveals a losing door. If you can establish those two things, than you can assuredly say the odds are 2/3.

ETA: Billy, below you will see the problem as presented in the first post of this thread
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?

You can see that the assumptions you have made (he always does this, and he always reveals a goat) are not explicitly noted. This is a common occurence.

[GreyICE]

No, you are wrong. You don't even have to have Monty in the equation. All you need to know is that there are 3 doors, one of which has a prize. You pick one. Your odds of it being behind that door are 1 in 3. Assume none of the other doors are revealed and you are given the chance to pick both the other doors instead (which you are essentially doing even if an empty door is revealed on that side). Your odds are 2 in 3. Or simpy draw an imaginary line between your door and the other two door - the odds are better if you pick the side on which there are two doors. This become more than obvious if you have a million doors. The one you pick on one side and the 999,999 on the other side. The prize is almost certainly on the side with the 999,999 doors. Revealing 999,998 doors on the other side that are empty virtually guarantees that the prize is behind the remaining door. The odds of it be being the door you picked remain 1 in a million. Monty is completely irrelevant to the puzzle. All we know is that a second door is opened revealing no prize and that is all we need to know.

I don't know, maybe I am wrong, but I completely fail to see what the host's intention have to do with the outcome - he has to open a second door and it is a given that the second door will not contain the prize. I fail to see how what is in the host's mind has any bearing on the statistics.

The fallacy of this is revealed if you imagine the following scenario.

You have three doors. You choose one. You have a 1/3 chance of choosing the right door.

You flip a coin, and draw a dot based on a random door, based on that coinflip. Your claim is that switching to the door without the dot improves your chances of winning to 2/3rds.

This is obviously fallacious, because we've gained no new information.

Yet it's exactly the scenario you're proposing that switching is always helpful, except that we don't know if we have a car behind that door or not.

Imagine your million doors. You eliminate 999,998 of them through a process of utterly random chance.

Your chance of having picked the right door in the first place is now 1/2.

What you missed is that by RANDOMLY eliminating 999,998 doors, you're already sitting in an immensely unlikely scenario, of which there are two possibilities remaining - you picked the door with the car, or the door with the car is the other door.

It's when Monty is forced to eliminate those 999,998 doors without revealing the car that the chances become 999,999 in a million that it's behind the other door.

As I said, the oddity only occurs when Monty is forced to change the random car choice into a goat choice, which gives you additional information. A piece of paper, a pen, and 10 minutes will satisfy you on this (it did me).

[CurtC]

The problem was stated in the OP ..

If you want to propose a different problem, then you should do so ..

Here is the problem statement from the OP:

"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?"

Note that this does not say that Monty knows where the car is, only that in this particular case, he revealed a goat. Maybe he doesn't know and sometimes reveals the car which you didn't choose, game over. You just got lucky and he revealed a goat so you get another chance. That is completely compatible with the problem statement in the OP.

[Rolfe]

Yes, I'm sorry. Have I eaten enough crow for you, oh impenetrable text wall?

I got it conflated with the baby problem, where in fact it doesn't matter, as long as you know whether the person has a son or not. That's a variant on the coin-box problem, rather than on the forced-choice problem like Monty Haul.


OK, OK, don't mention it! I just thought since you were so sure, I ought to try to figure out whether you might be right.

Maybe this isn't the place. Tell me about the baby problem (or the coin-box problem). How does it differ from the present example? (Isn't it "Monty Hall"? - there was no name attached to it at all when I first heard of it though, in the mid-1990s.)

Rolfe.

[sol invictus]

I don't know, maybe I am wrong, but I completely fail to see what the host's intention have to do with the outcome - he has to open a second door and it is a given that the second door will not contain the prize. I fail to see how what is in the host's mind has any bearing on the statistics.

Just try it! It will take you less time than it took to write your last post and you'll quickly see why you're wrong. Take CurtC's case A to begin with since that is the problem as it's usually, if not always, intended to be. You can do it with three cards, or three pieces of paper, or whatever you like.

If you want an explanation there are probably hundreds in this thread. I've never understood why this is confusing, so I'm probably not the one to try to give yet another.

[Skeptical Greg]

Not sure what your point is..

I replied to:
One of the main problems here is that people don't carefully state the problem, ...

My reply was that the OP was plainly ( carefully ) stated ..

[CurtC]

My reply was that the OP was plainly ( carefully ) stated ..

If the OP was carefully stated, then the correct answer is "I'm sorry, there is not enough information provided."

If he wanted the "2/3 by switching" answer, he should have more carefully constrained Monty to always revealing a goat.

[Skeptical Greg]

Just try it! It will take you less time than it took to write your last post and you'll quickly see why you're wrong. Take CurtC's case A to begin with since that is the problem as it's usually, if not always, intended to be. You can do it with three cards, or three pieces of paper, or whatever you like.

If you want an explanation there are probably hundreds in this thread. I've never understood why this is confusing, so I'm probably not the one to try to give yet another.

CurtC's case A is not the problem presented in the OP ..

BillyKid's solution is valid for the OP .. Whether the problem is usually presented some other way, or regardless of what assumptions someone chooses to make.

With three closed doors, the chances of picking the car are 1/3 .. ( out of a sufficiently number of tries, the car will be picked 33% of the time )

Once a door is opened, without the car behind it, the chances go to 2/3 that it is behind the remaining door . ( out of a sufficiently number of tries, the car will be picked 66% of the time )

It is very easy to run a computer simulation to show this ..

Or you can do it by blindly pulling marbles out of boxes .. You just have to do it long enough to see the result converge toward 33 or 66% ..

The computer makes it a lot easier ..

[GreyICE]

OK, OK, don't mention it! I just thought since you were so sure, I ought to try to figure out whether you might be right.

Maybe this isn't the place. Tell be about the baby problem (or the coi-box problem). How does it differ from the present example? (Isn't it "Monty Hall"? - there was no name attache to it at all when I first heard of it though, in the mid-1990s.)

Rolfe.

The coin box is weird, conceptually.

You have three boxes, one with two gold coins, one with two silver coins, one with a gold coin and a silver coin.

If you take a random coin out of a random box, and it's gold, what chance is there that the other coin in the box is silver?

Works similar to Monty Hall (yes, I can't spell) only with randomness. Conceptually similar, yet the con man in this case doesn't have to influence the outcome of the event to change the odds of the color of the other coin in the box.

[Herzblut]

Here is the problem statement from the OP:

"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?"

Note that this does not say that Monty knows where the car is, only that in this particular case, he revealed a goat. Maybe he doesn't know and sometimes reveals the car which you didn't choose, game over. You just got lucky and he revealed a goat so you get another chance. That is completely compatible with the problem statement in the OP.
No. The experiment is
-1- you pick a door
-2- Monty opens another door with goat behind
-3- you may switch or not.

Revealing a car in step -2- is not compliant to the rules of the experiment. It's totally irrelevant why Monty opens a goat door, he just does per definition.

If you want to simulate this experiment, say, 200 times by random choice, then you have to re-iterate experiments such that Monty opens a goat door 200 times in a row by pure chance. If he opens a car door, he invalidates the whole simulation and you have to re-start it from the beginning. I recommend to reveal to Monty the location of the car to shorten experimental effort.

[CurtC]

CurtC's case A is not the problem presented in the OP ..No, but it's the problem that the OP intended to present.

BillyKid's solution is valid for the OP .. Whether the problem is usually presented some other way, or regardless of what assumptions someone chooses to make.No, with the problem as it's presented in the OP, all three of my cases (A, B, and C), as well as many others, are reasonable interpretations of the OP, and they all give different probabilities.

It is very easy to run a computer simulation to show this ..I completely agree, although a pencil and paper simulation is equally satisfactory and quicker.

[Skeptical Greg]

If the OP was carefully stated, then the correct answer is "I'm sorry, there is not enough information provided."
Then you are free not to play..


If he wanted the "2/3 by switching" answer, he should have more carefully constrained Monty to always revealing a goat.

The answer will not be 2/3 if a car is revealed , because the game will be over ..
Switching is no longer an option..

But if you continue to play long enough, and discard the null iterations.. ( car revealed ) then the success when switching will approach 66% ...

....And prove, --- that when given the opportunity to switch, switching gives you a 66% chance of winning a car ..

[Rolfe]

Uh, I know I'm repeating myself, but the reason I say switch is that we don't know Monty's motivation.

If we consider the possibility that he only intends to allow the player to switch when he knows the player has already picked the car, this should display quite clearly that his motivation has every bearing on the outcome.

It is up to us to decide whether we think this is what is going on. However, I would call that behaviour cheating, on the host's part. So, I would propose that this possibility may be excluded from the scenarios to be considered, as the very fact of posing the question tends to imply that the game will be fair.

Monty opening either of the two unchosen doors at random, would be considered a "fair game". If that's what he's doing, then switching confers no advantage. You have a 1/3rd chance of being right, so you'd be wrong to switch; a 1/3rd chance of being wrong, so you'd be right to switch; and a 1/3rd chance that Monty would reveal the car when he opened his random door. You know that the third possibility hasn't happened, but the two remaining ones are still of equal probability.

Monty deliberately choosing not to reveal the car (if you've chosen wrongly to begin with), would also be regarded as a "fair game". In this case there is a 1/3rd chance you've chosen correctly and a 2/3rds chance it is behind one of the other doors. If the second of these possibilities is the case, opening the one door of the two remaining to show where the car isn't, transfers the 2/3rds probability on to the remaining door.

Both these scenarios are fair play. It may be more probable that Monty is playing scenario 1, in which case part of the fun will be opening an unchosen door to reveal the car - oh sorry sir, better luck next time, end of game. But even if he is doing that, switching will not make your chance of winning poorer.

It is possible he may be playing scenario 2, in which case switching just doubled your chance of winning.

So switch anyway. What have you got to lose?

Rolfe.

[sol invictus]

CurtC's case A is not the problem presented in the OP ..

It is consistent with it, and it is also the standard Monty Hall problem you can find anywhere on the internet or in print.

BillyKid's solution is valid for the OP .. Whether the problem is usually presented some other way, or regardless of what assumptions someone chooses to make.

It is consistent with the OP, yes.

With three closed doors, the chances of picking the car are 1/3 .. ( out of a sufficiently number of tries, the car will be picked 33% of the time )

Once a door is opened, without the car behind it, the chances go to 2/3 that it is behind the remaining door . ( out of a sufficiently number of tries, the car will be picked 66% of the time )

It is very easy to run a computer simulation to show this ..

Or you can do it by blindly pulling marbles out of boxes .. You just have to do it long enough to see the result converge toward 33 or 66% ..

The computer makes it a lot easier ..

There is no reason to run a simulation - that result is totally obvious once the problem is stated clearly. There are only three cases - why in the world would you need a computer?

[GreyICE]

Then you are free not to play..



The answer will not be 2/3 if a car is revealed , because the game will be over ..
Switching is no longer an option..

But if you continue to play long enough, and discard the null iterations.. ( car revealed ) then the success when switching will approach 66% ...

....And prove, --- that when given the opportunity to switch, switching gives you a 66% chance of winning a car ..

Nope. Lets say you discard the iterations where he reveals a car. Your chances are 50/50, assuming a coin-flipping Monty.

You choose Door 1 (door designations are arbitrary)

Car is behind door 1 (33%)
Door 2 Opened - switching loses
Door 3 Opened - switching loses

Car is behind door 2 (33%)
Door 2 Opened - Start Game over
Door 3 Opened - switching wins

Car is behind door 3 (33%)
Door 2 Opened - start game over
Door 3 Opened - switching wins

The 2/3rds only shows up if we morph the 'start game over' chances into 'switching wins' chances by forcing Monty to open the goat door, at which point it collapses to:

Car is behind door 1 (33%)
Door 2 Opened - switching loses
Door 3 Opened - switching loses

Car is behind door 2 (33%)
Door 3 Opened - switching wins

Car is behind door 3 (33%)
Door 3 Opened - switching wins

That's when the 2/3rds shows up.

[CurtC]

No. The experiment is
-1- you pick a door
-2- Monty opens another door with goat behind
-3- you may switch or not.

Revealing a car in step -2- is not compliant to the rules of the experiment. It's totally irrelevant why Monty opens a goat door, he just does per definition.
What you're missing here is that the OP is just a description of what has happened this one time. The host not knowing where the prize is, and just randomly having picked a goat this time, is completely consistent with the problem in the OP. And if that's the case, the answer is that it's a 50/50 proposition.

[Rolfe]

The coin box is weird, conceptually.

You have three boxes, one with two gold coins, one with two silver coins, one with a gold coin and a silver coin.

If you take a random coin out of a random box, and it's gold, what chance is there that the other coin in the box is silver?

Works similar to Monty Hall (yes, I can't spell) only with randomness. Conceptually similar, yet the con man in this case doesn't have to influence the outcome of the event to change the odds of the color of the other coin in the box.


Curse you, you quoted my post before I fixed the typos!

That's an interesting one I haven't come across before. I may not sleep tonight.

It took me quite some time to get my head round the two-goats-and-a-car thing, as I said, back in the 1990s when it was apparently discussed in a newspaper. I was told the problem by someone who had read it in the newspaper, and simply informed that some people thought the motivation of the host was important, but my informant couldn't see how. I didn't have internet access at the time.

At first I couldn't see why opening a door could possibly change the 1/3rd chance. I swore blind that there was no advantage to switching. However, I then imagined the 100-doors scenario, and realised in fact that there must be an advantage. Assuming the host is deliberately avoiding the car at step 2, the odds for both doors are now combined onto the one remaining door.

So then I declared that of course one should switch! You'll double your chances!

Then my informant said, but you don't know whether or not the host has deliberately revealed a goat or not. And how come his motivation can influence the odds?

I got all confused, but realised that it does, again by reference to the 100-doors version. And come to think of it, there are other games I've heard of where a steady profit can be made from a situation where the other player thinks an outcome is 50/50, but the person controlling the choices can skew the odds in his favour by always making the same decision. You don't win every time, but over repeated games, you always come out on top.

Then I thought, the problem has no solution, if we don't know the host's motivation! No advantage to switching if he's not avoiding the car, advantage if he is!

Then, almost finally, I realised that of course that in itself is the answer. Since in neither scenario are your chances reduced by switching, and there is a greater-than-zero chance that he is playing the deliberate avoidance of the car version, then that in itself tips the odds in favour of switching in your favour.

You can't decrease your chances and you may increase them, so go for it!

That took me about a week to work out, almost entirely on my own.

Then the possibility that Monty is cheating was raised in this thread....

There's always a wrinkle you haven't considered.

Hence my final formulation of the answer, if you can be assured that Monty is not cheating, then switch.

So who knows how long it will take me to figure out the coin boxes.

Rolfe.

[jsfisher]

No. The experiment is
-1- you pick a door
-2- Monty opens another door with goat behind
-3- you may switch or not.

Revealing a car in step -2- is not compliant to the rules of the experiment. It's totally irrelevant why Monty opens a goat door, he just does per definition.

No, it is very relevant.

The experiment is not quite as you stated. It is not a forward-looking thing where we will do this, then Monty Hall will do something, then we will have a choice. Instead, we are looking at what did happen. The difference may be subtle, but it is very important.

Assume for a moment Monty is evil incarnate. His strategy is to only show us a goat if we have picked the door with the car. He will not show us what is behind any door otherwise, nor would he give you an option to switch.

So, we are not at the end, and we have picked a door, Monty has shown us a goat from behind another door, and Monty gives us an option to switch. Do you really want to switch your initial choice?

[GreyICE]

Lol, Rolfe, thanks for making me feel better.

The coin-box puzzle is interesting, the answer is 2/3rds (like switching in Monte Hall) despite the fact that at face value it looks like 1/2

(this is because despite your 50/50 chance of having the 2 gold coin/1 silver 1 gold box, you only have a 50/50 chance of grabbing a gold coin from the 1 silver 1 gold box).

It's easier because the boxes don't need motivation.

The final stage to that trick is set it up, and offer to bet $10 that the coin will be the same color - which is why it's the start of a con game too.

[Skeptical Greg]

..............

The 2/3rds only shows up if we morph the 'start game over' chances into 'switching wins' chances by forcing Monty to open the goat door, at which point it collapses to:
.................

That's when the 2/3rds shows up.

Well really, that's my point ..
Discarding the " car revealed " iterations, is statistically equivalent to " goat is always revealed "... And becomes the implied rule that CurtC is pointing out .. ( fair or not )


CurtC is objecting to the " one round " nature, of the problem as presented ..


The player increasing his chances of winning to 66% by switching when limited to two doors, after starting with 3, is a statistical fact at that point in time, for that iteration of the game, whether the game is ever played again or not ... Or whether he actually wins the car .. In fact he has a 33% of not winning it .. ( :D )

The only way to get a " win 50% of the time " ( 1/2 ) outcome, is to start with 2 doors and pick one..

Computer simulations show this to be true, no matter how you twist it ..

[GreyICE]

Well really, that's my point ..
Discarding the " car revealed " iterations, is statistically equivalent to " goat is always revealed "... And becomes the implied rule that CurtC is pointing out .. ( fair or not )
Not true

CurtC is objecting to the " one round " nature, of the problem as presented ..


The player increasing his chances of winning to 66% by switching when limited to two doors, after starting with 3, is a statistical fact at that point in time, for that iteration of the game, whether the game is ever played again or not ... Or whether he actually wins the car .. In fact he has a 33% of not winning it .. ( :D )

The only way to get a " win 50% of the time " ( 1/2 ) outcome, is to start with 2 doors and pick one..

Computer simulations show this to be true, no matter how you twist it ..
Not true.

Car is behind door 1 (33%)
Door 2 Opened - switching loses
Door 3 Opened - switching loses

Car is behind door 2 (33%)
Door 2 Opened - ?
Door 3 Opened - switching wins

Car is behind door 3 (33%)
Door 2 Opened - ?
Door 3 Opened - switching wins

The scenarios are 33% wins, 33% loses, and 33% ?s. In the forced to pick a goat, the ? becomes wins. If revealing the car becomes an autoloss, then the odds are still 1/3rd to win. If you start the game over, it becomes 50/50. If it's random and you reveal a goat, it becomes 50/50.

[Herzblut]

Lol, Rolfe, thanks for making me feel better.

The coin-box puzzle is interesting, the answer is 2/3rds (like switching in Monte Hall) despite the fact that at face value it looks like 1/2

You mean a third for different color, right?


(this is because despite your 50/50 chance of having the 2 gold coin/1 silver 1 gold box, you only have a 50/50 chance of grabbing a gold coin from the 1 silver 1 gold box).

Hmm. I prefer to calculate. So, I don't have to understand it. :D

[Rolfe]

What you're missing here is that the OP is just a description of what has happened this one time. The host not knowing where the prize is, and just randomly having picked a goat this time, is completely consistent with the problem in the OP. And if that's the case, the answer is that it's a 50/50 proposition.


Curt, you're dictating the rules while appearing not to dictate them. Read your own post again and you'll see what I mean. (Hint: check the word "if".)

What happened that one time cannot, on its own, reveal the entire setup.

Supposing I have a cunningly weighted coin. It will come down heads only 1/3rd of the time. I toss it once, and get heads. I ask you what the probability of that is. You say, 50/50. I say no, the coin is weighted, there was only a 1/3rd chance that was going to happen.

You can't prove me wrong by saying, but it came up heads and that is entirely consistent with what would have happened if you'd used a normal coin!

You can only tell the coin is weighted, and how, by multiple tosses. After you've tossed it 100 times, you'll know whether heads is a 50/50 shot, or just 1/3rd.

So it is with the Monty Hall problem. (I'm so glad it has a name! When I first heard it, it was "the host of a game show in an impoverished South American republic". Calling it "the one about the two goats and the car" can get you some funny looks.)

You can only tell whether or not the coin is weighted by tossing it lots of times. You can only tell which way the Monty Hall choice is set by running the scenario multiple times.

If you do this, and every single time Monty reveals a goat at step 2, then indeed, you double your chances if you switch. If, however, he reveals the car at step 2 1/3rd of the time, then switching confers no advantage.

Please tell me you get it!

Rolfe.

[Herzblut]

Not true.

Car is behind door 1 (33%)
Door 2 Opened - switching loses
Door 3 Opened - switching loses

Car is behind door 2 (33%)
Door 2 Opened - ?
Door 3 Opened - switching wins

Car is behind door 3 (33%)
Door 2 Opened - ?
Door 3 Opened - switching wins

The scenarios are 33% wins, 33% loses, and 33% ?s. In the forced to pick a goat, the ? becomes wins. If revealing the car becomes an autoloss, then the odds are still 1/3rd to win. If you start the game over, it becomes 50/50. If it's random and you reveal a goat, it becomes 50/50.
What are you talking about? You cannot introduce abritrary '?' scenarios contradicting the experiment. Why don't you add a surprise case where a chimpanzee is revealed from time to time? Or a naked blonde? This is excluded in the very same way a car is.

This is how it is:

Car is behind door 1 (33%) -> switching loses
Car is behind door 2 (33%) -> switching wins
Car is behind door 3 (33%) -> switching wins

[Rolfe]

The coin box is weird, conceptually.

You have three boxes, one with two gold coins, one with two silver coins, one with a gold coin and a silver coin.

If you take a random coin out of a random box, and it's gold, what chance is there that the other coin in the box is silver?


The coin-box puzzle is interesting, the answer is 2/3rds (like switching in Monte Hall) despite the fact that at face value it looks like 1/2

(this is because despite your 50/50 chance of having the 2 gold coin/1 silver 1 gold box, you only have a 50/50 chance of grabbing a gold coin from the 1 silver 1 gold box).

It's easier because the boxes don't need motivation.

The final stage to that trick is set it up, and offer to bet $10 that the coin will be the same color - which is why it's the start of a con game too.


OK, going to have to think about this one. While making the supper I got as far as realising that intuitively the answer was 50/50. My thinking was that if I've grabbed a gold coin, then I know I have either the gold/gold box, or the gold/silver box. The silver/silver one is excluded. So if I got the gold/gold box, then the other coin is gold, but if I got the gold/silver box, the other coin is silver. 50/50.

Mmmm, light beginning to dawn. It's not that simple.

Yup, sleep no more, GreyICE doth murder sleep.

Thanks a bundle!

Rolfe.

[Rolfe]

OK, I get it.

Forget the silver/silver box, that's just handwaving.

Two boxes, one with two gold coins and one with one gold and one silver. Pick a coin at random. 75% of the time you get a gold one, 25% of the time you get a silver one.

Break it down.

50% of the time you got a gold coin from the gold/gold box.
25% of the time you got the gold coin from the gold/silver box.
25% of the time you got the silver coin from the gold/silver box.

So, 2/3rds of the times you picked a gold coin, it was from the gold/gold box.

QED.

Phew. MUCH MUCH easier than Monty Bloody Hall.

I will have sweet dreams after all.

Rolfe.

[Herzblut]

OK, going to have to think about this one. While making the supper I got as far as realising that intuitively the answer was 50/50. My thinking was that if I've grabbed a gold coin, then I know I have either the gold/gold box, or the gold/silver box. The silver/silver one is excluded. So if I got the gold/gold box, then the other coin is gold, but if I got the gold/silver box, the other coin is silver. 50/50.

Mmmm, light beginning to dawn. It's not that simple.

Yup, sleep no more, GreyICE doth murder sleep.

Thanks a bundle!

Rolfe.
Your grabbed gold coin is one of the three

GG1 or GG2 (from g-g box) or
GS (from g-s box).

Two out of three .. no, that must suffice.

Edit: Just see you got it quickly and obsoleted my post.

[Herzblut]

If you say you're playing 200 times, then that means that you started choosing 300 times, and in 100 of them, Monty revealed the prize by his random guess. In the remaining 200 trials, it's in your door 100 of them and in the other door 100 of them: therefore it's a 50/50 choice.

Boah, this weird discussion made my head swirl. I didn't recognize that this is complete nonsense. It's a gambler's fallacy at its best.

How often you skip invalid trials is completely irrelevant. It doesn't say anything about your winning chances in regular trials, each of which is an independent trial not correlated to any of the others.

You argue like somebody repeatedly tossing a coin, and guessing heads must be more likely in the future, because you tossed a tail 5 times in a row.

[GreyICE]

What are you talking about? You cannot introduce abritrary '?' scenarios contradicting the experiment. Why don't you add a surprise case where a chimpanzee is revealed from time to time? Or a naked blonde? This is excluded in the very same way a car is.

This is how it is:

Car is behind door 1 (33%) -> switching loses
Car is behind door 2 (33%) -> switching wins
Car is behind door 3 (33%) -> switching wins

Only if Monty can't reveal a car is this true.

If Monty CAN reveal the car, but DOESN'T because of random chance, this becomes false.

You're actually engaging in the gamblers fallacy here, despite the fact that you think you're not. You're linking two events (the random initial choice of the door, and the random second choice of a door) that are no longer linked.

[Herzblut]

Only if Monty can't reveal a car is this true.

If Monty reveals anything but a goat (car, gorilla, Elton John, ..) the trial is invalid and discarded. Monty cannot reveal a car in any valid trial per definition.


If Monty CAN reveal the car, but DOESN'T because of random chance, this becomes false.

Discarding invalid trials (car, Madonna, Fidel Castro ..) makes Monty reveal a goat in every valid trial. Each trial is independent from all others, it's a random process of repeated non-correlated trials.

You're actually engaging in the gamblers fallacy here, despite the fact that you think you're not.

How so?

[Rolfe]

You can only tell whether or not a coin is weighted by tossing it lots of times. You can only tell which way the Monty Hall choice is set by running the scenario multiple times.

If you do this, and every single time Monty reveals a goat at step 2, then indeed, you double your chances if you switch. If, however, he reveals the car at step 2 1/3rd of the time, then switching confers no advantage.

Please tell me you get it!


If indeed the problem is set so that Monty always reveals a goat, then he is choosing to avoid the car, and so transferring all the 2/3rds probability that the car is behind one of the doors you didn't pick, on to the other closed door. Just as is illustrated by the 100-doors-and-one-car version. Switch.

If, however, the problem is set so that Monty picks a door at random, then there is no advantage to switching. In that event, 1/3rd of the times the scenario is run, he reveals the car at step 2.

You can't tell which rule Monty is operating on without repeated runs of the problem, just as you can't tell whether and in what direction a coin is weighted unless you toss it multiple times.

Rolfe.

[jsfisher]

If Monty picks a door at random after we pick a door (pretty miuch at random, too), then there are 6 possible aggregate selection possibilities, all with probability 1/6 = 1/3*1/2.


We pick door #1, and Monty picks #2.
We pick door #1, and Monty picks #3.
We pick door #2, and Monty picks #1.
We pick door #2, and Monty picks #3.
We pick door #3, and Monty picks #1.
We pick door #3, and Monty picks #2.


Without loss of generality, assume the car is behind door #1.

So, now, if we find ourselves staring at a goat and Monty asking us if we want to switch, it must be one of the cases 1, 2, 4 or 6. Cases 3 and 5 have been eliminated because in them, the car was revealed.

The conditional probability of cases 1, 2, 4, and 6 given we can see a goat behind Monty's door is 1/4 for each case.

For cases 1 and 2, if we switch we lose; for cases 4 and 6, if we switch we win.

It is, therefore, a 50/50 proposition.

[Rolfe]

If Monty picks a door at random ....


If.

There you go again, setting conditions that are not in the original scenario.

Nobody stipulated that he was picking his door at random.

Rolfe.

[Vorpal]

This is exactly why I explicitly defined the different cases back in post #400 (http://forums.randi.org/showpost.php?p=3905943&postcount=400).
-- If Monty picks a random goat, which could be behind the player's door, then the probability of winning via switching is indeed 1/2.
-- If Monty picks a random goat, but never the player's door (as per the OP), then this is the standard interpretation of switching giving 2/3 probability. And yes, this is equivalent to the scenario of any goat not behind player's first guess being revealed, regardless of Monty's knowledge (as Monty's knowledge is used only to guarantee that this condition obtains in the first place).

[jsfisher]

If.

There you go again, setting conditions that are not in the original scenario.

Nobody stipulated that he was picking his door at random.

Rolfe.


The conditional was to join in an existing sub-discussion.

That aside, the real point is that we do not know Monty's strategy, but Monty's strategy does affect the final probabilities. Without the game Monty is playing, we cannot conclude anything about the advantage of switching doors.

[GreyICE]

If Monty reveals anything but a goat (car, gorilla, Elton John, ..) the trial is invalid and discarded. Monty cannot reveal a car in any valid trial per definition.


Discarding invalid trials (car, Madonna, Fidel Castro ..) makes Monty reveal a goat in every valid trial. Each trial is independent from all others, it's a random process of repeated non-correlated trials.


How so?
The gambler's fallacy is that you know the outcome of one random event based on another random event.

Like the gambler, you are saying that because Monty's coin came up goats, your coin is due to come up cars.

Lets examine the cases again.

Car is behind door number 1 - 33%
You lose.

Car is behind door number 2, Monty picks door number 3 - 16.5% (33% chance on the car, 50% chance Monty picks door number 3)

Car is behind door number 3, Monty picks door number 2 - 16.5% (same odds)

This leaves a 50/50 gain from switching.

The fallacy is as follows. It is true that if you flip two coins, and one comes up heads, it is 66% likely that the other one is tails. HOWEVER, this is not the probability we are in. MONTY'S specific coin came up heads. The fact that we have knowledge of the outcome of Monty's coin flip in no way influences your coin flip.

The probabilities involved in Monty's specific coin in no way influence your coin. You are still 50% likely to have flipped heads, and 50% likely to have flipped tails.


To approach this from yet another direction, imagine that Monty picks first, and picks a goat at random. The chances of you picking the car are now 50/50.

[gnome]

*sigh*

I'm headed for the mall, and I'm barricading both entrances when I get there. Who else wants to survive the zombie thread?

[Mobyseven]

No, you are wrong. You don't even have to have Monty in the equation. All you need to know is that there are 3 doors, one of which has a prize. You pick one. Your odds of it being behind that door are 1 in 3. Assume none of the other doors are revealed and you are given the chance to pick both the other doors instead (which you are essentially doing even if an empty door is revealed on that side). Your odds are 2 in 3. Or simpy draw an imaginary line between your door and the other two door - the odds are better if you pick the side on which there are two doors. This become more than obvious if you have a million doors. The one you pick on one side and the 999,999 on the other side. The prize is almost certainly on the side with the 999,999 doors. Revealing 999,998 doors on the other side that are empty virtually guarantees that the prize is behind the remaining door. The odds of it be being the door you picked remain 1 in a million. Monty is completely irrelevant to the puzzle. All we know is that a second door is opened revealing no prize and that is all we need to know.

I don't know, maybe I am wrong, but I completely fail to see what the host's intention have to do with the outcome - he has to open a second door and it is a given that the second door will not contain the prize. I fail to see how what is in the host's mind has any bearing on the statistics.

I feel like I should step in here and defend billydkid, because he's actually quite right.

There is no need to include Monty in the problem, and indeed sometimes it can help people understand it better if he is not included. People often project human motives onto Monty (hell, this thread is a prime example) when really he is only even in the problem for one reason - to open up a door that will reveal a goat.

There are a few ways to think about this. First, imagine a situation with no Monty. You have three doors, and you pick one. Another door mysteriously opens and reveals a goat. You are now given the option to stay with your current door, or swap to the other unopened one. You will benefit by swapping in this situation, just as in the original.

Another situation is that there are three doors, and you pick one. No other doors are opened and you are then asked if you would like to stay with your current door, or switch to both of the two remaining doors. This situation is for all purposes identical to the problems stated above - in this situation, however, the reason you should swap becomes much more obvious.

Bringing Monty into the mix really only serves to confuse things most times, given that he is really only there to play one specific part. But to explain his part, you have to give him certain pieces of knowledge, and that makes people give him a motive and a personality, which just confuses the matter. The knowledge that 'Monty' has in the Monty Hall problem isn't actually his at all...he is just an agent of the system, and it's easy to just eliminate him entirely from the problem.

[Mobyseven]

*sigh*

I'm headed for the mall, and I'm barricading both entrances when I get there. Who else wants to survive the zombie thread?

Twelve pages of thread, much of which can be summed up as, "But what if the Monty Hall problem isn't the Monty Hall problem?"

[jsfisher]

There is no need to include Monty in the problem, and indeed sometimes it can help people understand it better if he is not included. People often project human motives onto Monty (hell, this thread is a prime example) when really he is only even in the problem for one reason - to open up a door that will reveal a goat.

You have made an assumption, there. You have done the very thing you said we shouldn't; you've given Monty a motive.

As the problem is normally stated (and was so in the opening post), we cannot assume Monty would (a) always have opened a door, (b) there'd be a goat behind it, and (c) we'd have an opportunity to change our door choice.

All we know is that Monty did open a door this particular time, and there was a goat behind the door this particular time, and we now have an opportunity to change our choice.

Monty must have some strategy. Depending on what it is, the probability of a change in our choice being beneficial to us can be 0 or 1 or anything in between.

There are a few ways to think about this. First, imagine a situation with no Monty. You have three doors, and you pick one. Another door mysteriously opens and reveals a goat. You are now given the option to stay with your current door, or swap to the other unopened one. You will benefit by swapping in this situation, just as in the original.

Did the door open by accident? Was a non-goat door guaranteed to have opened? What was the probability the door would have opened exactly as it did...and that's equivalent to asking about Monty's behavior.

Another situation is that there are three doors, and you pick one. No other doors are opened and you are then asked if you would like to stay with your current door, or switch to both of the two remaining doors. This situation is for all purposes identical to the problems stated above - in this situation, however, the reason you should swap becomes much more obvious.

If you guarantee me you will always follow this strategy when given the option to switch choices, well, then, I have a little game I'd like to play with you. :)

You have a built-in assumption that you would have been given the option to switch no matter what. All you really know is that you have been given the option for the particular case at hand.

Bringing Monty into the mix really only serves to confuse things most times, given that he is really only there to play one specific part. But to explain his part, you have to give him certain pieces of knowledge, and that makes people give him a motive and a personality, which just confuses the matter. The knowledge that 'Monty' has in the Monty Hall problem isn't actually his at all...he is just an agent of the system, and it's easy to just eliminate him entirely from the problem.

Again, you have made an assumption about Monty's behavior. You have assumed that Monty will aways open a door with a goat behind it after you choose. There is nothing in original problem statement to justify that assumption.

[Herzblut]

You have made an assumption, there. You have done the very thing you said we shouldn't; you've given Monty a motive.

No. A purpose, a reason, a role in the game. The same role can be played by a chimpanzee or a mechanical unlocking mechanism.

What you say is complete nonsense. You fantasize about an "always" which is pure imagination etc. etc.

May I ask, what is your profession? Is it distantly related to mathematics?

[bobdroege7]

If I regurgitated the thread about whether an airplane on a conveyor belt would take off would anyone join me?

[Mobyseven]

Again, you have made an assumption about Monty's behavior. You have assumed that Monty will aways open a door with a goat behind it after you choose. There is nothing in original problem statement to justify that assumption.

Holy hell. What do you mean that there's nothing in the original problem statement to justify that assumption? That is the original problem!

Look, every time you play the game, Monty will open a door with a goat behind it. Every time. He will never, ever reveal a car. That's the nature of the problem. Deal with it.

[Mobyseven]

You have made an assumption, there. You have done the very thing you said we shouldn't; you've given Monty a motive.

Also, this confuses me. I never gave Monty a motive. I pointed out that there's only one reason he's included in the system at all...but that's not a motive, that's a description of the system.

[sol invictus]

What you say is complete nonsense. You fantasize about an "always" which is pure imagination etc. etc.

What precisely do you find so difficult to understand about this discussion? The problem, as stated in the OP, simply does not give enough information to conclude that switching gives you a 2/3 chance of winning. You need to make some additional assumptions - which were almost certainly intended by the OP, as they are part of the standard Monty Hall problem - to come to that conclusion. That's all that's being pointed out here.

It's really not very difficult.

[lionking]

Holy hell. What do you mean that there's nothing in the original problem statement to justify that assumption? That is the original problem!

Look, every time you play the game, Monty will open a door with a goat behind it. Every time. He will never, ever reveal a car. That's the nature of the problem. Deal with it.
Spot on moby, this is the intent of the problem and what makes it most interesting. Monty is offering a two-for-one deal, and the odds are 2/3, even if it doesn't appear so initially. Too many others are complicating a very simple problem. If anyone wants to pose a problem where Monty sometimes opens a door with a car because he didn't know what was behind the door, let them do so. In the OP, he opens a door with a goat.

[AntiTelharsic]

Why would Monty open the door with the car behind it and give you the choice to switch from your obvious goat to the other obvious goat? It's ridiculous.

"You've lost. Would you like to switch to the other, equivalently losing position or just stay where you are? It doesn't matter in the least, so I'm not sure why I'm giving you this opportunity..."

The Monty Hall Problem has the contestant choosing one of three doors, where one door conceals a win and each of the other two conceals a loss; the host opening another of the doors, to reveal a loss; and the contestant being given the chance to either stay with his original choice or change his choice to the remaining door. The problem is assumed to be identical if repeated, because there is no information given to suggest otherwise -- if I take part in the scenario, and then you do, and then I do again, there is nothing in the statement of the problem to suggest that those scenarios may differ. This situation gives a 2/3 probability of winning if you switch. If you're contriving circumstances where switching doesn't give you a 2/3 chance of winning, then you're not talking about The Monty Hall Problem as it is usually defined.

If someone mentions "The Monty Hall Problem", three doors, two goats, and a car, this is what they're talking about. People who want to discuss other conditions should specify how their conditions differ from the standard; as long as they do that there should be no confusion.

[lionking]

Better said than me Anti-Telharsic

[Ivor the Engineer]

<snip>

Supposing I have a cunningly weighted coin. It will come down heads only 1/3rd of the time. I toss it once, and get heads. I ask you what the probability of that is. You say, 50/50. I say no, the coin is weighted, there was only a 1/3rd chance that was going to happen.

You can't prove me wrong by saying, but it came up heads and that is entirely consistent with what would have happened if you'd used a normal coin!

You can only tell the coin is weighted, and how, by multiple tosses. After you've tossed it 100 times, you'll know whether heads is a 50/50 shot, or just 1/3rd.

<snip>

Or the experiment was one of the 0.044% performed with a fair coin which produce a result as or more extreme.

[pgwenthold]

Spot on moby, this is the intent of the problem and what makes it most interesting.

This may be the intent, but it usually isn't stated as such.

Interestingly, many posters are saying, "This is what the problem says" but we are the ones who go back and quote the OP and ask, "Where does it say that he always opens a door to reveal a goat?"

Given just the problem, as written in the OP, there is not enough information to answer the question. If you claim it is the "intent" of the problem, then you are basing your answer on more than just the problem, as written in the OP.

[pgwenthold]

Why would Monty open the door with the car behind it and give you the choice to switch from your obvious goat to the other obvious goat? It's ridiculous.


Because Monty doesn't know where the car is?

Nothing says that Monty has to give you the option to switch if he opens the door and it is a car. In fact, it works just fine on a game show: "Let's see what's behind door number 3? Oh, that's the car; sorry, you lose!"



"You've lost. Would you like to switch to the other, equivalently losing position or just stay where you are? It doesn't matter in the least, so I'm not sure why I'm giving you this opportunity..."


Where is the requirement that he has to allow you to switch when you play the game?



The Monty Hall Problem has the contestant choosing one of three doors, where one door conceals a win and each of the other two conceals a loss; the host opening another of the doors, to reveal a loss; and the contestant being given the chance to either stay with his original choice or change his choice to the remaining door. The problem is assumed to be identical if repeated, because there is no information given to suggest otherwise

Unfortunately, there is no information to suggest either. That is why we keep saying, "There is not enough information."

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.

[CurtC]

Holy hell. What do you mean that there's nothing in the original problem statement to justify that assumption? That is the original problem!

Look, every time you play the game, Monty will open a door with a goat behind it. Every time. He will never, ever reveal a car. That's the nature of the problem. Deal with it.

You're not seeing the point. Let's take this mathematically equivalent scenario: you come upon a game of three-card monte - there are three cards on the table, one is an ace. He shows you where the ace is, then scrambles the cards too quickly for you to follow, then stops (let's also say that there is no cheating going on here, it's simply that he can scramble the cards faster than you can follow). You pick a card. The street hustler pauses, then says "hey, this other card over here is NOT the ace, now do you want to switch your guess to the remaining card?"

The question - should you switch? Any second grader can see that in this situation, if you switch, you lose. The only reason he offered the switch is because you picked the ace on your first guess.

Now, can you see any difference in this game and the Monty Hall problem? They're completely equivalent, except for assigning different motivations to the hosts.

[CurtC]

There you go again, setting conditions that are not in the original scenario.

Nobody stipulated that he was picking his door at random.
We all realize that. In fact, that's the point of this further discussion, that it was not specified in the OP how the host would behave, therefore there's not enough information. But we're then taking a set of interesting possibilities for his methods and working the numbers for those cases. These are not answers to the OP, but tangential discussions.

[CurtC]

What are you talking about? You cannot introduce abritrary '?' scenarios contradicting the experiment. Why don't you add a surprise case where a chimpanzee is revealed from time to time? Or a naked blonde? This is excluded in the very same way a car is.

Let's imagine that we take 300 people to play this game, each gets to play it only once so he has no knowledge of how Monty operates. Now let's group them into different sets depending on how they choose. Further, let's say that in this game, Monty doesn't know where the car is, but will always reveal one of the other two doors, and if he reveals a goat, he offers the chance to switch.

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.

Set 2 - picks a goat first, then Monty (by chance) reveals a goat and offers the choice of switching. This group has 100 people in it.

Set 3 - picks a goat first, then Monty (by chance) reveals the car. The game is over, the contestant loses. This group has 100 people in it.

Now the question is described as in the OP. This tells you that we're dealing with someone who is in either set 1 or set 2. The people in set 3 don't count, because they never get to this predicament.

Now, out of those 200 people, the 100 in set 1 would lose by switching, the 100 in set 2 would win by switching. This is a 50/50 proposition, AND IT'S VALID ONLY IN THE CASE WHEN THE HOST PICKS RANDOMLY WITHOUT KNOWING WHERE THE CAR IS.

(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.)

[Mobyseven]

This may be the intent, but it usually isn't stated as such.

If it isn't stated as such, then the person isn't stating the Monty Hall problem. They are mistaken.

Interestingly, many posters are saying, "This is what the problem says" but we are the ones who go back and quote the OP and ask, "Where does it say that he always opens a door to reveal a goat?"

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?

Epic reading comprehension fail. There's nothing wrong with how the OP stated the problem, it is, indeed, entirely correct.

[Mobyseven]

You're not seeing the point. Let's take this mathematically equivalent scenario: you come upon a game of three-card monte - there are three cards on the table, one is an ace. He shows you where the ace is, then scrambles the cards too quickly for you to follow, then stops (let's also say that there is no cheating going on here, it's simply that he can scramble the cards faster than you can follow). You pick a card. The street hustler pauses, then says "hey, this other card over here is NOT the ace, now do you want to switch your guess to the remaining card?"

The question - should you switch? Any second grader can see that in this situation, if you switch, you lose. The only reason he offered the switch is because you picked the ace on your first guess.

Now, can you see any difference in this game and the Monty Hall problem? They're completely equivalent, except for assigning different motivations to the hosts.

Holy mother of Vishnu, how hard is this to understand? You're assigning motives to Monty that he just doesn't have. The scenario you just described, where Monty will only offer a switch if you've already chosen the car, is NOT the Monty Hall problem.

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.

If Monty is trying to cheap contestants then it's not the Monty Hall problem. If Monty is opening doors at random then it's not the Monty Hall problem. If Monty is a vampire, using goats to steal the souls of unsuspecting competitors, then it's not the Monty Hall problem.

[ddt]

Holy mother of Vishnu, how hard is this to understand? You're assigning motives to Monty that he just doesn't have. The scenario you just described, where Monty will only offer a switch if you've already chosen the car, is NOT the Monty Hall problem.

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.

If Monty is trying to cheap contestants then it's not the Monty Hall problem. If Monty is opening doors at random then it's not the Monty Hall problem. If Monty is a vampire, using goats to steal the souls of unsuspecting competitors, then it's not the Monty Hall problem.

That entirely depends upon what you call the "Monty Hall problem". Its usual formulation rests indeed on the rule that the host will always open another door which has a goat behind it.

However, in the real "Let's Make a Deal" show, this was not the case. From a NYT article (http://query.nytimes.com/gst/fullpage.html?res=9D0CEFDD1E3FF932A15754C0A9679582 60&sec=&spon=&pagewanted=print) on it:
Was Mr. Hall cheating? Not according to the rules of the show, because he did have the option of not offering the switch, and he usually did not offer it.

And although Mr. Hall might have been violating the spirit of Ms. vos Savant's problem, he was not violating its letter. Dr. Diaconis and Mr. Gardner both noticed the same loophole when they compared Ms. vos Savant's wording of the problem with the versions they had analyzed in their articles.

"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.


It is obvious that this influences the probabilities involved. It is also obvious that the game host not having a deterministic decision procedure, based only on the choice of the first door by the contestant, for offering the choice to switch or not, makes it impossible to calculate probabilities.

It is also obvious that it is impossible for the contestant to make a right decision if he does not know the decision procedure of the game host. A benevolent host may offer to switch only if you first picked a goat, in which case switching is always good; a malevolent host may offer to switch only if you first picked the car, in which case switching is always bad.

An interesting question in this is still: do you acquire enough information of watching, say, one season of "Let's Make a Deal" to deduce the decision procedure of the game host?

At least thanks to those who resurrected the thread that I learnt this tidbit of information about the show that problem was based on. I haven't seen a single instance of the original show. From Dutch game shows with the same set-up I got the impression that the host always indeed opened another door with a goat behind it. After all, the price of the car is a fraction of the cost of the production of the show and you want people in the majority of cases to go home with a price.

[CurtC]

Holy mother of Vishnu, how hard is this to understand? You're assigning motives to Monty that he just doesn't have. The scenario you just described, where Monty will only offer a switch if you've already chosen the car, is NOT the Monty Hall problem.

I agree, it's not part of the Monty Hall problem. However, the OP itself was not the Monty Hall problem, because it did not specify that Monty is constrained to always showing a non-prize door. At best, it was ambiguous - does the present tense wording mean that's what he always does, or a conversational way of stating what he did this one time? It sounds to me like it's describing only what he did this one time.

My point here is that when I've seen people attempt to present the Monty Hall problem, the statement very rarely includes the constraint on Monty, therefore you don't have enough information. The OP here was an example of that. Now we're having fun imagining different scenarios that are consistent with the OP and give different answers - OK?

If you want to talk only about the classic Monty Hall problem, then be sure to clearly specify how Monty must operate.

[Rolfe]

This may be the intent, but it usually isn't stated as such.

Interestingly, many posters are saying, "This is what the problem says" but we are the ones who go back and quote the OP and ask, "Where does it say that he always opens a door to reveal a goat?"

Given just the problem, as written in the OP, there is not enough information to answer the question. If you claim it is the "intent" of the problem, then you are basing your answer on more than just the problem, as written in the OP.


The question of what he "always" does arises because it is necessary to imagine multiple runs of the problem to identify the probabilities. However, in order to set up your multiple runs, you have to know what the ground rules are.

I could (if I was techie enough) set up a computer simulation so that a goat is always revealed at step 2, and as we know, this will most certainly demonstrate that switching doors will double your chance of getting the car.

I could also set it up so that either of the two unchosen doors are opened at random. And as we know, this will most certainly demonstrate that switching doors has no effect on your chances of getting the car.

(I could also set it up so that any of the three doors is opened at random. Again, this would demonstrate that switching doors has no effect.)

These are the only three possible "rules" I can think of that can reasonably be applied to the game. If the game is deliberately rigged so that the choice to switch is only offered when the contestant has already chosen the car, the puzzle is meaningless. And Monty is a vile cheat.

Holy mother of Vishnu, how hard is this to understand? You're assigning motives to Monty that he just doesn't have. The scenario you just described, where Monty will only offer a switch if you've already chosen the car, is NOT the Monty Hall problem.

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 cheat 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.

If Monty is trying to cheat contestants then it's not the Monty Hall problem....


Well, exactly. (I had to go back and edit this, because I don't agree that you can specify that the doors are not being opened at random in the Monty Hall problem - or you can, but if you do, then most of the surreal amusement goes out of it, see my next post below.)

And we can also exclude scenarios that say, well, Monty may choose a different scenario each time. This is because we are actually being asked about a single example of the game, in isolation. The repetitions are only necessary to demonstrate the odds. Thus by definition we have to set up the simulation with the rule in place which governs the example of the game which we are actually discussing.

Presumably the puzzle was invented by someone. It would be nice to be able to find that person and ask what the intent was. However, it's been around so long that the chances of finding the original inventor must be approaching zero. So, we have what we have. Just a bald description of the scenario.

The fact remains that it is clearly possible to set up the simulation in two ways (I'll exclude the third one above, because I don't think it adds anything to the discussion). Both sets of rules produce our proposed scenario - one door chosen, another opened, and that revealing a goat. We can't tell which set of rules is in force from the initial premise. Nevertheless, the set of rules in force has a huge effect on the outcome. One set - switching doubles your chances of winning; the other - no advantage to switching.

I think the reason the puzzle seems surreal is that it's very difficult to imagine the what Monty is thinking can influence the odds. However, it becomes easier if you imagine it as playing a computer simulation - I understand there are such simulations available on the net.

Are you playing a game where the programmer has set it up with "always a goat"? SWITCH!

Are you playing a game where the programmer has set it up with "random unchosen door is opened"? Don't bother.

And this still applies even if you only play once.

Does this make it any clearer? Now, instead of Monty's human and no doubt capricious mind, we simply have two computer games. It's just that you don't know which one you're playing.

Stop. Think about it. Now please explain to me where I'm wrong.

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 swithcing 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.

Rolfe.

[pgwenthold]

Holy mother of Vishnu, how hard is this to understand? You're assigning motives to Monty that he just doesn't have.

1) No one is assigning any motives to Monty but you
2) We are saying that, given the wording of the problem, we _don't know_ his motives, and in order to come up with an answer requires that we make assumptions that are not inherent in the original problem as it is stated here.

The OP of this thread is readily available to you, and has even been quoted twice in the last two days. Therefore, it should not be hard for you to show us how the problem shows that he does not have these motives.


Notice, you are the one claiming he "doesn't have" motives. We are claiming that we don't know that he doesn't have motives.

[Rolfe]

I agree, it's not part of the Monty Hall problem. However, the OP itself was not the Monty Hall problem, because it did not specify that Monty is constrained to always showing a non-prize door. [....] If you want to talk only about the classic Monty Hall problem, then be sure to clearly specify how Monty must operate.


Well, maybe. However, I've been aware of this problem for a long time, and it has never been presented in that way to me. I first became aware of it in the early to mid 1990s, when it was apparently the subject of much correspondence in the Guardian newspaper. While I only had second-hand exposure to that correspondence, it's quite clear from the nature of the arguments that it was similar to the OP. Only one example described, in the present tense. No information given about whether this was or was not part of a longer run of games, or about how any other runs of the game turned out, or about what underlying rules should be assumed. Indeed, the name "Monty Hall" wasn't even mentioned that I know about. It was "a game show host in an impoverished South American republic".

I think the point is that if you specify the ground rules exactly, it takes just a little bit longer than GreyICE's boxes and coins puzzle to figure out, and one you've done that, it's done. Move on. You completely understand the problem, and its solution. Might have taken you a couple of hours, if you're slow.

However, by not specifying the rules, the problem takes off into a completely different sphere. First you think you have the answer one way, then you realise the answer may be different if you assume a different set of rules, then you sit back, boggled, contemplating the possibility that what the host is thinking can affect the odds attached to your choice.

That's why the problem keeps resurfacing, and why this zombie thread is still going even after six years.

Rolfe.

[sphenisc]

Lol, Rolfe, thanks for making me feel better.

The coin-box puzzle is interesting, the answer is 2/3rds (like switching in Monte Hall) despite the fact that at face value it looks like 1/2

(this is because despite your 50/50 chance of having the 2 gold coin/1 silver 1 gold box, you only have a 50/50 chance of grabbing a gold coin from the 1 silver 1 gold box).

It's easier because the boxes don't need motivation.

The final stage to that trick is set it up, and offer to bet $10 that the coin will be the same color - which is why it's the start of a con game too.

If the probability is 2/3s it's going to be a different colour then this seems an excellent way of losing money fast!!!

[CurtC]

The OP of this thread is readily available to you, and has even been quoted twice in the last two days. Therefore, it should not be hard for you to show us how the problem shows that he does not have these motives.

To be completely fair, I didn't notice it before, but the wording of the OP could possibly be interpreted as saying what Monty always has to do. I didn't read it that way, and I think most people wouldn't either, but the use of the present tense could be ambiguous: "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."

Does the present tense there indicate what he always does? At best, it's ambiguous, but certainly it's not explicit in saying what rules Monty follows.

However, by not specifying the rules, the problem takes off into a completely different sphere. First you think you have the answer one way, then you realise the answer may be different if you assume a different set of rules, then you sit back, boggled, contemplating the possibility that what the host is thinking can affect the odds attached to your choice.
Exactly. I have come across this puzzle many many times, and I don't think I've ever seen it worded to constrain Monty's behavior. This lack of constraint adds an interesting twist to the puzzle which seems to irritate some people.

[ddt]

To be completely fair, I didn't notice it before, but the wording of the OP could possibly be interpreted as saying what Monty always has to do. I didn't read it that way, and I think most people wouldn't either, but the use of the present tense could be ambiguous: "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."

Does the present tense there indicate what he always does? At best, it's ambiguous, but certainly it's not explicit in saying what rules Monty follows.
I, for one, interpret the OP as fixing the rules of Monty's behaviour. If Monty had a choice whether he'd open another door or not, and even one with the car behind it, I had expected a modality in the form of an auxiliary verb (can, may) or a subjunctive :). As the OP is stated, in a present tense indicative, it means how he always behaves.

Disclaimer: English is a second language for me.


Exactly. I have come across this puzzle many many times, and I don't think I've ever seen it worded to constrain Monty's behavior. This lack of constraint adds an interesting twist to the puzzle which seems to irritate some people.
Actually, all the times I've heard the puzzle it was always with the automatic behaviour of the host to open a second door with a goat...

[Mobyseven]

Stop. Think about it. Now please explain to me where I'm wrong.

You aren't. We're in agreement.

I agree, it's not part of the Monty Hall problem. However, the OP itself was not the Monty Hall problem, because it did not specify that Monty is constrained to always showing a non-prize door. At best, it was ambiguous - does the present tense wording mean that's what he always does, or a conversational way of stating what he did this one time? It sounds to me like it's describing only what he did this one time.

My point here is that when I've seen people attempt to present the Monty Hall problem, the statement very rarely includes the constraint on Monty, therefore you don't have enough information. The OP here was an example of that. Now we're having fun imagining different scenarios that are consistent with the OP and give different answers - OK?

If you want to talk only about the classic Monty Hall problem, then be sure to clearly specify how Monty must operate.

:hb: :hb: :hb:

1) No one is assigning any motives to Monty but you

Bugger off. Seriously, this is just starting to piss me off. I haven't assigned any motive to Monty. I have shown what his function in the scenario is. I haven't come anywhere remotely near assigning him a motive, unlike others in the thread.

You say I've assigned him a motive, I say put up or shutup. Quote the part where I assigned him a motive, or concede that I bloody well didn't.

2) We are saying that, given the wording of the problem, we _don't know_ his motives, and in order to come up with an answer requires that we make assumptions that are not inherent in the original problem as it is stated here.

Wrong. I quoted the problem as it was presented, and bolded the relevant part. There's no issue with it outside of your inability to accept what is written in plain English.

The OP of this thread is readily available to you, and has even been quoted twice in the last two days. Therefore, it should not be hard for you to show us how the problem shows that he does not have these motives.

I know the OP is readily available. I quoted it, and bolded the relevant part. Moreover, you're talking out of your arse when you ask me to prove that the problem as stated shows that he does not have motives A through Z. I can't prove that, and while you think you're just being oh-so-clever in 'trapping' me, you're missing the whole bloody point: His motives don't matter one bleedin' scrap. His function in the scenario is clearly defined, and regardless of what imaginary motive you want to toss around the outcome of the scenario won't change. If you decide to assign a motive to Monty that changes his function in the scenario, go right ahead - but if you do that, you're no longer discussing the Monty Hall problem.

And now, a short interlude for your amusement...

Signor: This cheese tastes a bit off.
Signora: But what if it was beef?
Signor: ...but it's not beef. It's cheese.
Signora: But what if it was beef?
Signor: Then I guess it would taste...like beef?
Signora: Ah! So you're saying that this cheese tastes like beef!
Signor: *Applies head to wall vigorously*


fin

Notice, you are the one claiming he "doesn't have" motives. We are claiming that we don't know that he doesn't have motives.

I'm the one pointing out that his motives are worth a rat's fart in a typhoon. It's his function that is key. You can go crazy like a freshman literature student imagining motives for the man, but unless you change his function the outcome is the same - and if you change his function then you aren't discussing the Monty Hall problem anymore.

[CurtC]

I'm the one pointing out that his motives are worth a rat's fart in a typhoon. It's his function that is key.

Whatever.

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

[Michael C]

I, for one, interpret the OP as fixing the rules of Monty's behaviour. If Monty had a choice whether he'd open another door or not, and even one with the car behind it, I had expected a modality in the form of an auxiliary verb (can, may) or a subjunctive :). As the OP is stated, in a present tense indicative, it means how he always behaves.

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?

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.

[Michael C]

The coin box is weird, conceptually.

You have three boxes, one with two gold coins, one with two silver coins, one with a gold coin and a silver coin.

If you take a random coin out of a random box, and it's gold, what chance is there that the other coin in the box is silver?.

The coin-box puzzle is interesting, the answer is 2/3rds (like switching in Monte Hall) despite the fact that at face value it looks like 1/2

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

[Michael C]

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).

[jsfisher]

You and Monty are playing a game of "Guess which door has the car." The only problem is, you don't know the rules for the game. The only thing you know are the events described in the opening post. (You picked a door; Monty revealed a goat behind another door; you now have an option of changing your door choice.)

Suppose these are the actual rules:


You get to pick a door.
If the door you pick has a goat behind it, then the game is over with you winning a goat.
Otherwise (i.e. the door you picked has the car), Monty opens one of the remaining doors to reveal a goat.
You get an option to change your door choice.
The game ends with you getting whatever is behind your final choice.


I am not saying these have to be the rules; they are just offered as a for-instance. Are these rules in any way inconsistent with the conditions stated in the opening post?

If they are not consistent, what is the specific inconsistency. If they are consistent, would you switch door choices if given the opportunity?