The "Three Doors" mathematical problem

[De_Bunk]

Im sure some of you have come across this before..but i was fascinated by the debate surrounding this.

Somehow, both sides of the discussion seem to make sense....

Heres the question.

You've got three doors..

Behind two of them is a goat, the other a car

You choose one door to open....but you dont open it yet.

Then, someone opens one of the other doors to reveal a goat.

You are left with two closed doors

But before you open your chosen door, you are given the choice to switch to the other unopened door and open that one instead.

Do you stay with your original choice of door or do you switch

Would switching from your original choice increase the odds of you finding the car...

( i know this is probably old maths problem...but it done my head in..)

If you want the link to the simulation...ask..

DB

[pgwenthold]

Why not ask Marilyn?

[De_Bunk]

Pgwenthold...

I half guessed it was old..

But it still done me in...

I didnt believe it until i ran the simulator..


:)

Never mind...

DB

[LostAngeles]

Ask Cecil. And then everyone else...

http://www.straightdope.com/classics/a3_189.html

[De_Bunk]

LostAngeles...

See above...

:)

DB

[ceptimus]

Google on 'Monty Hall puzzle'


See you in a few months time. :)

[Paul C. Anagnostopoulos]

Here's the best way I've found to think about it:

You select your door, but don't open it.

Monty offers you both the other doors in exchange for your choice. You'd take that deal in a heartbeat, right?

Well, that's what he's done by showing you which of the other two doors wouldn't be worth anything in that deal.

~~ Paul

[Humphreys]

The answer is really simple.

The straight dope guy is wrong. It doesn't matter if Monty knows which door contains the prize or not.

Think of it like this. Separate the doors into two piles. Put the door you picked in one pile, and the other two doors in a separate pile. Now, Monty cannot pick the door you chose, so he is left with the two doors in the other pile to choose from.

Now at this stage, we will both agree that it is more likely the correct door is not the one you chose, but one of the other two (there are two doors to one, so it's more likely).

So, when Monty removes one of the doors in the other pile, the one left is more likely to be the one than the one you chose, so you should always switch.

The way this puzzle tricks your brain is the fact people think Monty is picking any random door from the three available - he isn't (he can't pick the door you chose, only one of the other two).

If Monty doesn't know which door contains the prize, it's just as likely he'll pick the correct door to open as it is that he'll pick the wrong door. When he picks the wrong door, you know for a fact the other door is most likely the right one.

In short, definitely switch doors no matter what!

[sorgoth]

...I still don't get it.

[Rolfe]

My brain came out of my ears on this one some years ago. In the process I tried the following "reductio ad absurdum".

Suppose there were 100 doors, 99 goats and one car. You choose, but don't open the door. Monty selects a door, and it's a goat. Do you want to switch? But it goes on. Monty keeps selecting doors, and they're always goats. In the end the only doors still shut are the one you chose at random at the beginning and the single other one Monty has not yet opened.

Will you switch now?

I'm not sure if this illuminates much, but please feel free to discuss.

Rolfe.

[Batman Jr.]

Another really easy way to think about it:

There are two cases you must consider: 1) You initially selected the right door. 2) You initially selected the wrong door. If you first selected the right door, then in switching doors you would always lose, right? If you first selected the wrong door, then in switching doors you would always win. The probability of first selecting the wrong door is 2/3. This means that switching gives you a 2/3 chance of getting the right door!

Let's construct a scenario to test this out and say the prize is behind door #2.

1) What happens if you pick door #1? Monty will open up door #3. If you switch, you win.

2) What happens if you pick door #2? Monty will either open up door #1 or #2. Switching here will make you lose.

3) What happens if you pick door #3? Monty will open up door #1. If you switch, you will again win.

There existed two cases where you won and only one where you lost!

[De_Bunk]

Heres the link to one of the better simulators...

You can let it run, at hi-speed, a defined number of times to see the odds...

http://www.grand-illusions.com/simulator/montysim.htm

DB

[Rolfe]

Yes, that was my other "reductio ad absurdum".

There are three possibilities.

1. You selected the car, unknowingly. 1 out of 3 probability. Monty perforce must select a goat. And you will lose by switching.

2. You selected a goat, unknowingly. 2 out of 3 probability. So there is now a 50/50 chance Monty will reveal the car. Isn't there?

2 (a). Monty reveals a goat. You win by switching.
2 (b). Monty reveals the car. The game is up.

In this scenario, it's an even break whether you switch or not, because you only have two of the three possibilities still running, and one of these is a win and the other a lose.

But is 2 (b) actually a realistic option? Does it ever happen?

That's the "reductio ad absurdum". You run the trial ten times, and observe whether Monty ever does reveal the car. If he does, then you gain nothing by switching. But if he never reveals the car, then you know that he is deliberately avoiding it. In this scenario, you will win more often than you lose if you always switch.

So, you always switch. If he's not deliberately avoiding the car, then you gain nothing, but you lose nothing either. If he is, then you definitely gain. So you never lose by switching.

I think. Like I said, my brain came out of my ears some time ago.

Rolfe.

[sorgoth]

Originally posted by Batman Jr.


There are two cases you must consider: 1) You initially selected the right door. 2) You initially selected the wrong door. If you first selected the right door, then in switching doors you would always lose, right? If you first selected the wrong door, then in switching doors you would always win. The probability of first selecting the wrong door is 2/3. This means that switching gives you a 2/3 chance of getting the right door!


I GET IT! Thank you!

So since the chances of getting the wrong door are 2/3, so if you DID, and he gets rid of one for sure, and then you switch, you have a 2/3 chance of getting it right! Haha!

[BPSCG]

Originally posted by sorgoth
So since the chances of getting the wrong door are 2/3, so if you DID, and he gets rid of one for sure, and then you switch, you have a 2/3 chance of getting it right! Haha! Exactly. Your chances of having picked the car door were 1/3. The chances that the car was behind ONE of the other doors was 2/3. Monty's opening one of the other doors doesn't change those odds. All it tells you is that the remaining door is the one that has the 2/3 chance of being the winner.

[BillHoyt]

Originally posted by Humphreys
The answer is really simple.

The straight dope guy is wrong. It doesn't matter if Monty knows which door contains the prize or not...
The way this puzzle tricks your brain is the fact people think Monty is picking any random door from the three available - he isn't (he can't pick the door you chose, only one of the other two).

If Monty doesn't know which door contains the prize, it's just as likely he'll pick the correct door to open as it is that he'll pick the wrong door. When he picks the wrong door, you know for a fact the other door is most likely the right one.

In short, definitely switch doors no matter what!
Right conclusion, but dead wrong analysis, Humphreys.

First off is the evidence from the game show, which I have to guess you've never seen. Monty always revealed a goat door. Why? Because there is no drama in doing otherwise. There's nothing left to choose. The contestant knows she's lost. By revealing a goat door, the drama is increased, and the contestant still has a chance and a seemingly difficult choice to make.

Now for the analysis:
(Assume your explanation is right...)
o There is only one correct door.
o The contestant chooses one to start the game. 1/3 probability she is right.
o The chance one of the other doors is correct is 2/3.
o Monty chooses one of these at random.
o The chance he chose the prize is 1/3, but the game is now over, isn't it?
o The chance he didn't choose the prize is also 1/3, but, given that, the chance you should switch to the other door is 1/2.
o No advantage in switching!

(Assume now that Monty knows...)
o There is only one correct door.
o The contestant chooses one to start the game. 1/3 probability she is right.
o The chance one of the other doors is correct is 2/3.
o Monty selects (with knowledge) a goat.
o That selection, being done with certainty, is not part of the probability picture. Effectively, the 2/.3 probability has now jumped to the door the contestant did not originally choose.
o Switch, d***it, switch!

[Walter Wayne]

The assumption is that the probability of him revealing a false door is independant of which door you chose. The way the puzzle is phrased, we don't know whether he does this everytime or if he intentionally tries to throw people off.

IIRC correctly, on the game show Monty always revealed one. So in that particular case, always switch choices.

Walt

[wipeout]

I'm sure someone will eventually mix the Monty Hall problem with something in quantum mechanics to create the ultimate puzzle for screwing with people's heads. :D

[Batman Jr.]

I'll explain the alternate case where Monty Hall doesn't know which door has the car:

In order for switching to work, you will first have to pick a wrong door and Monty will have to choose the other wrong door. There are six cases:

1) Right and Wrong#1

2) Right and Wrong#2

3) Wrong#1 and Right

4) Wrong#1 and Wrong#2

5) Wrong#2 and Wrong#1

6) Wrong#2 and Right

Notice that we now count right-wrong#1 and right-wrong#2 as two cases. This is because Monty can now pick the right door when we first pick the wrong door. In the other scenario, since Monty knew where the car was, the probability of him picking the other wrong door after you having picked the first wrong door was always 1. We had to count right-wrong#1 and right-wrong#2 as one case before because the probability of each happening is (1/3)(1/2). The probability of the other two cases was (1/3)(1)—Monty had a 100% probability of picking the wrong door, remember?—and since each case must have an equal chance of happening (right-wrong#1+right-wrong#2=(1/3)(1/2)+(1/3)(1/2)=(1/3)1) and both right-wrong#1 and right-wrong#2 have the same outcome, we grouped them into one case. But here, all of the cases become (1/3)(1/2) as Monty has only a 50% of getting the other wrong in the case of first picking one of the wrongs.

However, we know Monty at this point in the game revealed the goat! So, we have to eliminate cases 3 and 6 from above. We are left with 1, 2, 4, and 5. Switching works when both Monty and you selected wrong doors. Which instances match this criterion? Why, 4 and 5. That means the probability of a switch working is an unbiased 1/2!

[garys_2k]

Originally posted by Batman Jr.
Another really easy way to think about it:

There are two cases you must consider: 1) You initially selected the right door. 2) You initially selected the wrong door. If you first selected the right door, then in switching doors you would always lose, right? If you first selected the wrong door, then in switching doors you would always win. The probability of first selecting the wrong door is 2/3. This means that switching gives you a 2/3 chance of getting the right door!

Let's construct a scenario to test this out and say the prize is behind door #2.

1) What happens if you pick door #1? Monty will open up door #3. If you switch, you win.

2) What happens if you pick door #2? Monty will either open up door #1 or #2. Switching here will make you lose.

3) What happens if you pick door #3? Monty will open up door #1. If you switch, you will again win.

There existed two cases where you won and only one where you lost!
This explanation did it for me, thank you. I lost hair over this problem a few years ago and never quite "got it." Now I do.

[Batman Jr.]

I'm glad I can help! ;)

[T'ai Chi]

I've tried to make this clear as possible. Let me know if it works. :)

Assume, without loss of generality, that the prize is behind door 1.

An outcome from the game can be kept tract of by the 4-tuple (a,b,c,d). The letter a is the door you initially choose, b the door Monty opens, c the door you stay with, and d will be the outcome, that is, WIN or LOSE.

NOT SWITCHING DOORS

Our sample space, that is, the set of all possible outcomes, is:

S = {(1,2,1,WIN), (1,3,1,WIN), (2,3,2,LOSE), (3,2,3,LOSE)}

Therefore, P(LOSE) = P(2,3,2,LOSE) + P(3,2,3,LOSE).
(where P(grapefruit) means 'the probability of grapefruit')

The probability associated with initially choosing door 2 is the same as the probability of initially choosing door 3, and that is 1/3. However, when you do this, you lose, and you can lose in two distinct ways, so P(LOSE) = 2/3.

So P(WIN) = 1 - P(LOSE) = 1 - 2/3 = 1/3.

SWITCHING DOORS

S = {(2,3,1,WIN), (3,2,1,WIN), (1,2,3,LOSE), (1,3,2,LOSE)}

Because we only win when we switch to door 1, it means we only win if we initially choose door 2 or door 3. When we initially choose a door, we have a 1/3 probability of doing such. Therefore, our probability of winning is in part based on the probability of choosing door 2 or door 3.

So P(WIN) = P(2,3,1,WIN) + P(3,2,1,WIN).

P(WIN) = 2/3.

[Humphreys]

Originally posted by BillHoyt
Right conclusion, but dead wrong analysis, Humphreys.

I don't think so.

Originally posted by BillHoyt
First off is the evidence from the game show, which I have to guess you've never seen.

I'll admit I've never seen the show, but I've seen the puzzle described lots of times.

Originally posted by BillHoyt
Monty always revealed a goat door. Why? Because there is no drama in doing otherwise. There's nothing left to choose. The contestant knows she's lost. By revealing a goat door, the drama is increased, and the contestant still has a chance and a seemingly difficult choice to make.

Yes. But it wouldn't make any difference once this stage was reached whether he originally knew or not.

Originally posted by BillHoyt
Now for the analysis:
(Assume your explanation is right...)
o There is only one correct door.
o The contestant chooses one to start the game. 1/3 probability she is right.
o The chance one of the other doors is correct is 2/3.
o Monty chooses one of these at random.
o The chance he chose the prize is 1/3, but the game is now over, isn't it?
o The chance he didn't choose the prize is also 1/3, but, given that, the chance you should switch to the other door is 1/2.
o No advantage in switching!

Monty's choice is still restricted to the doors you didn't choose. So, how can it make any difference whether he knows what door contains the prize and chooses it deliberately, or he doesn't know what door contains the prize but picks it by chance anyway!

There is no difference. Remember, he doesn't have a free choice of doors - he can't pick the same door you picked.

When we are given the option to swich, we are already past the stage where Monty chooses a door. We presume he must have picked a goat door, otherwise the game would be over already.
That's why it makes no difference at this stage whether Monty has knowledge or not.

How can it make a difference whether he picks a goat door deliberately, or by chance? The result is the same. Still switch!

Originally posted by BillHoyt
(Assume now that Monty knows...)
o There is only one correct door.
o The contestant chooses one to start the game. 1/3 probability she is right.
o The chance one of the other doors is correct is 2/3.
o Monty selects (with knowledge) a goat.
o That selection, being done with certainty, is not part of the probability picture. Effectively, the 2/.3 probability has now jumped to the door the contestant did not originally choose.
o Switch, d***it, switch!

But what if Monty selected a goat without knowledge? It makes no difference once this stage is reached! (if it is indeed reached).

But I could be wrong about all this (although this has never happened before)...

[Humphreys]

Actually, I'm starting to realize why I may be wrong about this now. I can't quite understand yet why it would make a difference whether Monty picked a goat door deliberately, or by chance.

But I can see that if we had 100 raffle tickets, and you picked one, then we removed 98 of the others by chance, and they were all losers - it would do us no good switching.

So you are right Bill, I think, but can you explain better why?

I'll think about it more, anyway.

[Humphreys]

Ha, don't worry, I understand it now. I was wrong - it does matter whether Monty has knowledge or not. The whole puzzle only works if we make that assumption.

Thanks for the correction, Bill.

[metacristi]

De Bunk

Im sure some of you have come across this before..but i was fascinated by the debate surrounding this.

Somehow, both sides of the discussion seem to make sense....

Heres the question.

You've got three doors..

Behind two of them is a goat, the other a car

You choose one door to open....but you dont open it yet.

Then, someone opens one of the other doors to reveal a goat.

You are left with two closed doors

But before you open your chosen door, you are given the choice to switch to the other unopened door and open that one instead.

Do you stay with your original choice of door or do you switch

Would switching from your original choice increase the odds of you finding the car...

( i know this is probably old maths problem...but it done my head in..)

If you want the link to the simulation...ask..

DB



My solution to the Monty Hall problem:

Let

A,B,C =doors
'X'=1=the car is behind door 'X'
'X'=0=there is nothing [or the goat] behind door 'X' ; where 'X'=A,B or C


There are possible 4 distinct events after choosing our door,let it be A,(where the underlined doors are those shown later as having the goat [or nothing in some variants of the problem] behind):

event1: A=1 AND B=0 AND C=0 ;P1=1/6

event2.: A=1 AND B=0 AND C=0 ;P2=1/6

event3: A=0 AND B=1 AND C=0 ;P3=1/3

event4: A=0 AND B=0 AND C=1 ;P4=1/3

We have:

P[event1]=P1=(1/3)*(1/2)*1=1/6

P[event2]=P2=(1/3)*(1/2)*1=1/6

P[event3]=P3=(1/3)*1*1=1/3

P[event4]=P4=(1/3)*1*1=1/3


Therefore the probabilities to have chosen a door having or not the car behind are respectively:

P[the car behind]=P1+P2=2*(1/6)=1/3

P[nothing behind]=P3+P4=2*(1/3)=2/3

From here it's clear why switching increase the chances to find something behind the door from 1/3 to 2/3.

[Paul C. Anagnostopoulos]

Man, all this math! By opening one of the other doors and showing you a goat, Monty has effectively given you both other doors. Easy pie.

~~ Paul

[CurtC]

BillHoyt, you're right that it comes down to whether Monty Hall always shows you a goat-door. But I watched the show lots of times, and the fact is, he didn't. Sometimes, he would just reveal the door that the contestant had picked, right or wrong. Offering the switch was *not* something he did every time.

So now, you have to figure in his motivation in giving you this choice. Is it done to keep you from winning? In that case, never switch. Is it done to make the show more exciting? Then it probably is 50-50. Is it done randomly? Switch. Is it done to give away more merchandise? Switch.

If anyone denies that the motivation of the host is what makes the difference, consider this scenario. You're walking down a city street and come upon a street hustler offering you a game of three-card monte. You put your money down, and the hustler moves the cards around thoroughly, at a speed you can't keep up with. You pick your card. The hustler, instead of revealing the one you picked, turns over a non-winning card, and offers you the choice to switch your guess to the remaining one. Should you switch? If you do, you're a fool.

[aerosolben]

I'm glad to see no one here is insisting on the p = 1/2 solution. Everytime this problem comes up, there are always people who adamantly insist the probability is 1/2. Trying to explain the idea is infuriating, as the straight dope link demonstrates.

I dislike the two child problem even more, because it depends so much on phrasing: "the first child is a girl" gives 1/2 chance of a male sibling, while "at least one child is a girl" gives 2/3 chance of a male sibling.

[Doc Dish]

Originally posted by De_Bunk Heres the link to one of the better simulators...

You can let it run, at hi-speed, a defined number of times to see the odds...

http://www.grand-illusions.com/simulator/montysim.htm

DB

I just let it run 20000 times! The results are (drumroll please):

Keep choice: 10000 times Wins: 3319 cars (33%) Losses: 6681 goats (67%)
Change choice: 10000 times Wins: 6714 cars (67%) Losses: 3286 goats (33%)

[athon]

What if I prefer the goat?

Athon

[Doc Dish]

Originally posted by athon: What if I prefer the goat?

Athon

You should seek professional assistance ;)

[T'ai Chi]

I personally had a hard time thinking about it until I actually wrote out the sample space.

[Number Six]

Originally posted by Walter Wayne
The assumption is that the probability of him revealing a false door is independant of which door you chose. The way the puzzle is phrased, we don't know whether he does this everytime or if he intentionally tries to throw people off.

IIRC correctly, on the game show Monty always revealed one. So in that particular case, always switch choices.

Walt

That is it. It is best to switch _only if_ Monty's choice to reveal a door to you is independent of whether or not you picked the right door. If you know beforehand that Monty is going to ask you if you want to switch, then you switch. But if you don't know that he's going to ask and then he asks, you can't say whether it's best to switch or not because you don't have enough information to come to a conclusion.

[Badger]

Originally posted by athon
What if I prefer the goat?

Athon

Then it's a win-win situation. Imagine how many goats you could get for a car!

[athon]

Originally posted by Badger


Then it's a win-win situation. Imagine how many goats you could get for a car!

Why would I want goats for my car? That's not where I want to keep them.

;)

Athon

[Underemployed]

I am going to apply for government funding to let me sit and watch all the recordings of this monty hall show to provide a rigourous analysis of real-world outcomes. Anyone want to co-author who has a big-screen TV?

[Rolfe]

Originally posted by Number Six
But if you don't know that he's going to ask and then he asks, you can't say whether it's best to switch or not because you don't have enough information to come to a conclusion. If you don't have that information, then overall it's still best to switch. If he isn't deliberately avoiding revealing the car, then you still sit on the 50/50 odds, so it's not going to make your chances any worse.

Given that there is some chance that he is deliberately avoiding the car, and in that case your odds are 2/3 in favour of switching, then always switch. You won't decrease your chances, and depending on the game he's playing (which you don't know for sure), you may increase your chances.

Rolfe.

[Matabiri]

Originally posted by Rolfe
My brain came out of my ears on this one some years ago. In the process I tried the following "reductio ad absurdum".

I came up with another reductio ad absurdum a while ago which I think works better:

Imagine you're playing 100 simultaneous games. You pick 100 doors. If you opened those doors now, you'd win, on average, 33 cars, right.

The host opens 100 doors which you didn't pick, revealing 100 goats. How would the act of opening those doors raise the number of cars you win from 33 to 50? It doesn't.

[Tanja]

Originally posted by Matabiri


I came up with another reductio ad absurdum a while ago which I think works better:

Imagine you're playing 100 simultaneous games. You pick 100 doors. If you opened those doors now, you'd win, on average, 33 cars, right.

The host opens 100 doors which you didn't pick, revealing 100 goats. How would the act of opening those doors raise the number of cars you win from 33 to 50? It doesn't.

If the host reveals 100 doors with goats, then by changing your choice, instead of getting the 33 cars you chose right the first time, you get the 66 (67) cars you DID NOT choose right the first time.

Edited to add: I am not sure what you actually meant by your example, that the odds go up or that they don't?

[Rolfe]

My "hundred doors" example was to show that it did matter whether or not Monty was deliberately avoiding the car.

If you chose at random, you'd have a 1 in 100 chance of having got the car. Assuming you didn't, and Monty was opening doors at random, the overwhelming probability (98 out of 100 I think) is that he would reveal the car at some point during all the subsequent door-opening.

If the car is still hidden when only two doors are still shut, then what do you think? Either you hit lucky with your original guess (1 in 100), or Monty by chance has left the car till the last door (another 1 in 100), or he is deliberately avoiding the car. The last is by far the most probable scenario, and it's intuitively obvious that you'd switch.

The other example I had was also to show if he is avoiding the car. Just run the three-doors trial a number of times. 33% of the time Monty should reveal the car, if he's just opening either of the doors at random. In that case, switching won't help you win more cars, but it won't spoil your record either - it's a 50/50 shot. If Monty never reveals the car, he's deliberately avoiding it, and in that case you benefit from switching by a 1/3 to 2/3 ratio. So the answer is, switch. If he's not avoiding the car, then it makes no difference, if he is, then you improve your odds. You switch to take advantage of the latter case if it applies, as there's no down-side if it doesn't.

In the online version quoted by the earlier poster, it's obviously set up so that Monty never reveals the car.

Rolfe.

[garys_2k]

IIRC there were several variants of the game. Sometimes one of the prizes was "Monty's Cookie Jar," which contained an unknown amount of cash. It could be a goat (very little money), or the grand prize, and he would reveal the fact that that jar was behind one of the doors from time to time. I do remember him once showing that the revealed jar was the grand prize.

Another variant had three prize levels. One would be the goat, the other some fairly nice furniture (mid-level prize), the other the grand prize. He would sometimes reveal the goat or one of the more valuable prizes, but I can't remember if he'd ever reveal the grand prize or just the mid-level one.

[Matabiri]

Originally posted by Tanja


If the host reveals 100 doors with goats, then by changing your choice, instead of getting the 33 cars you chose right the first time, you get the 66 (67) cars you DID NOT choose right the first time.

Edited to add: I am not sure what you actually meant by your example, that the odds go up or that they don't?

That's exactly what I meant. It was supposed to demonstrate that swapping is the right strategy. I.e. that the odds don't change.

[CurtC]

Rolfe wrote:
If you don't have that information, then overall it's still best to switch. If he isn't deliberately avoiding revealing the car, then you still sit on the 50/50 odds, so it's not going to make your chances any worse.

Given that there is some chance that he is deliberately avoiding the car, and in that case your odds are 2/3 in favour of switching, then always switch. You won't decrease your chances, and depending on the game he's playing (which you don't know for sure), you may increase your chances.I'm not sure that I understand your argument, but it's easy to imagine a scenario where switching is the wrong strategy. Let's say that Monty only offers the switch if you guessed the correct door already. In this case, you lose 100% of the time by switching.

[Number Six]

Originally posted by Rolfe
If you don't have that information, then overall it's still best to switch. If he isn't deliberately avoiding revealing the car, then you still sit on the 50/50 odds, so it's not going to make your chances any worse.

Given that there is some chance that he is deliberately avoiding the car, and in that case your odds are 2/3 in favour of switching, then always switch. You won't decrease your chances, and depending on the game he's playing (which you don't know for sure), you may increase your chances.

Rolfe.

If his choice of whether to reveal a goat to you is not independent of your initial choice then his choice of whether to reveal a goat to you depends on whether your initial choice is correct. And if his choice of whether to reveal a goat depends on whether your inital choice is correct, you would have to know _how_ it depends on your initial choice to make an informed decision.

Suppose you don't know he's going to reveal a door and then you pick Door A and then he reveals that behind Door B is a goat. Should you switch? We can't say. Maybe he was only going to reveal a goat if your initial choice was correct, in which case you should not switch. Maybe he was only going to reveal a goat if your inital choice was incorrect, in which case you should switch. But the point is, if you don't know then you can't say which course of action is best.

[Rolfe]

Originally posted by Number Six
Maybe he was only going to reveal a goat if your initial choice was correct, in which case you should not switch.OK, I see, it's another variant once you've recognised that Monty's motives are important.

The way I was originally told it, was that he always opens a door and invites you to switch if you choose. But if you don't know that one for sure then yes, it's moot.

Well, I think we've successfully trashed the theory that it doesn't make any difference whether Monty knows what's behind what door, or has any motivation to his actions! :D

Rolfe.

[Interesting Ian]

Originally posted by CurtC
BillHoyt, you're right that it comes down to whether Monty Hall always shows you a goat-door. But I watched the show lots of times, and the fact is, he didn't. Sometimes, he would just reveal the door that the contestant had picked, right or wrong. Offering the switch was *not* something he did every time.

So now, you have to figure in his motivation in giving you this choice. Is it done to keep you from winning? In that case, never switch. Is it done to make the show more exciting? Then it probably is 50-50. Is it done randomly? Switch. Is it done to give away more merchandise? Switch.

If anyone denies that the motivation of the host is what makes the difference, consider this scenario. You're walking down a city street and come upon a street hustler offering you a game of three-card monte. You put your money down, and the hustler moves the cards around thoroughly, at a speed you can't keep up with. You pick your card. The hustler, instead of revealing the one you picked, turns over a non-winning card, and offers you the choice to switch your guess to the remaining one. Should you switch? If you do, you're a fool.

Ummm . . no.

First of all if he doesn't offer any switch then there is no further discussion.

So let's consider the scenario where he does offer a switch. Either the host knows or doesn't know which door the prize is behind.

Let's consider he doesn't know and consequently reveals the door with the prize behind. Obviously you switch to that door (if you're allowed!).

Let's consider he doesn't know and consequently reveals the door with the goat behind. It is very clear that it doesn't matter whether the contestant switches or not. You have half a chance of sticking or switching to the other unopened door.

Now let's consider the scenario where the host does know. He reveals the door with the prize behind. Obviously you switch to that door (if you're allowed).

Let's consider he does know and consequently reveals the door with the goat behind. Now this is the complex one. If the host always uses his knowledge of what lies behind the doors to choose the door with the goat behind it, then obviously it will be a very wise idea for the contestant to switch, because the probability goes up from 1/3rd to 2/3rds of a chance of winning the prize. But if the host, despite his knowledge of what lies behind the doors, just picks a door arbitrarily (no preference), then again switching will be a half of a chance.


But in whatever scenario, switching does not make you a fool because switching can never make it less likely you would win than if you had stuck with the original door.

In fact we would not know if the hustler knows which door the prize is behind. So all we can say is that by switching the probability of getting the prize will be from 50% - 66.7%. Determining the precise percentage will be a question of psychology.

[Interesting Ian]

Originally posted by Number Six


That is it. It is best to switch _only if_ Monty's choice to reveal a door to you is independent of whether or not you picked the right door. If you know beforehand that Monty is going to ask you if you want to switch, then you switch. But if you don't know that he's going to ask and then he asks, you can't say whether it's best to switch or not because you don't have enough information to come to a conclusion.

No, it's always best to switch because although it might only be 50/50 chance whether you switch or not, the point is you don't know that. The chance of winning the prize is anything from 50% to 67% if you switch. Since you cannot determine the precise percentage, it is wise to switch.

[Rolfe]

Originally posted by Interesting Ian
But in whatever scenario, switching does not make you a fool because switching can never make it less likely you would win than if you had stuck with the original Unless, as Number Six just pointed out, Monty only offers you the opportunity to switch if he knows you've already picked the winning door.

Once you allow Monty's motives to come into consideration, as you must, you do face that one too unless it is specifically excluded from the original scenario.

However, if the switch is always offered irrespective of whether or not the winning door was initially selected, then yes, that statement above is correct.

Rolfe.

[Interesting Ian]

Originally posted by Rolfe
Unless, as Number Six just pointed out, Monty only offers you the opportunity to switch if he knows you've already picked the winning door.

Once you allow Monty's motives to come into consideration, as you must, you do face that one too unless it is specifically excluded from the original scenario.

However, if the switch is always offered irrespective of whether or not the winning door was initially selected, then yes, that statement above is correct.

Rolfe.

Damn damn damn! Just typed out my detailed reply and my computer restarted all by itself just as I finished! :mad: Months since it's done that. Maybe bitTorrent got something to do with it. Can't be bothered to type out all again. Anyway . .


If he offers the switch then this makes it more likely as a general policy that either he always offers the switch or more often offers the switch than otherwise. This then still makes it a wiser choice to switch (notwithstanding that reasoning from a single case is dodgy!).

But you're right. If people are familiar with this problem then they will have a propensity to switch. The host can use this knowledge and only offer the switch when you originally pick the correct door. Damn!

Unless you know that he's going to think like this then deliberately not switch when offered LOL All psychological games :)

[CurtC]

Rolfe wrote:
The way I was originally told it, was that he always opens a door and invites you to switch if you choose.I've been hearing this puzzle for about 15 years, and I don't think I've ever heard it explicitly stated that the host is required to open a door. The puzzle is solvable only if that stipulation is added, and in that case it's the 2/3 answer.

[Rolfe]

Originally posted by CurtC
I've been hearing this puzzle for about 15 years, and I don't think I've ever heard it explicitly stated that the host is required to open a door. The puzzle is solvable only if that stipulation is added, and in that case it's the 2/3 answer. Well, it's been a long time and I no longer have the original format. I do know that it was discussed as if he always offered a switch.

However, it's still, as II said, either 2/3 or 50/50, depending on Monty's intentions. It's the fact that you can't decrease your chance in that scenario that makes it "always switch", so that if he's deliberately avoiding the car you get the improved chance.

Rolfe.

[rastamonte]

Originally posted by Interesting Ian

In fact we would not know if the hustler knows which door the prize is behind. So all we can say is that by switching the probability of getting the prize will be from 50% - 66.7%. Determining the precise percentage will be a question of psychology.

When you say hustler, you must mean the three-card monte hustler. If so, you are wrong. He DOES know which card is the winning card. And he only offers you a switch if you have already chosen the winning card.

[Interesting Ian]

Originally posted by rastamonte


When you say hustler, you must mean the three-card monte hustler. If so, you are wrong. He DOES know which card is the winning card. And he only offers you a switch if you have already chosen the winning card.

I've already admitted my mistake. Read my post above :rolleyes:

[Number Six]

Originally posted by Interesting Ian


No, it's always best to switch because although it might only be 50/50 chance whether you switch or not, the point is you don't know that. The chance of winning the prize is anything from 50% to 67% if you switch. Since you cannot determine the precise percentage, it is wise to switch.

If you don't know the mechanism Monty used to decide to open a door with a goat then you have no more information about whether the door you chose was the correct door than you did when you intially chose it.

If you know beforehand that Monty is going to reveal a door with a goat to you then it's best to switch. Otherwise, you can't know whether it's best to switch unless you can read Monty's mind.

[Number Six]

Originally posted by Interesting Ian

If he offers the switch then this makes it more likely as a general policy that either he always offers the switch or more often offers the switch than otherwise.

If you don't know the mechanism Monty uses to decide whether to reveal a door with a goat to you then you don't know whether his offer to switch makes it more likely that it's a general policy.

[Walter Wayne]

Originally posted by Rolfe
Well, it's been a long time and I no longer have the original format. I do know that it was discussed as if he always offered a switch.

However, it's still, as II said, either 2/3 or 50/50, depending on Monty's intentions. It's the fact that you can't decrease your chance in that scenario that makes it "always switch", so that if he's deliberately avoiding the car you get the improved chance.

Rolfe. Yes you can decrease your chances by switching depending on Monty's knowledge and intention.

Lets suppose Monty intentions are to prevent you from winning the car. The simplest way for him to do this would be to accept your initial guess if you are wrong, and give the option to switch if you guessed right. This would mean that if he reveals a goat and gives you the option to switch, you have picked the correct door, and have a 100% chance to win if you stick with your guess, and 0% if you change.

More generally let us suppose that Monty's decision to reveal a goat is dependant on whether you chose right or not. Let's say:
p_c represents the chance Monty reveals a goat if you choose correctly first time
p_i represents the chance Monty reveals a goat if you choose incorrectly first time

Then the chance that you choose correctly and Monty reveals a goat is:
P(correct & goat) = p_c/3
Then the chance that you choose incorrectly and Monty reveals a goat is:
P(incorrect & goat) = 2p_i/3

Thus the odds of him revealling a goat are
P(goat)=P(correct & goat)+P(incorrect & goat)=p_c/3+2p_i/3

And then the chance of your initial guess being incorrect given that he reveals a goat is
P(incorrect given goat) = P(incorrect & goat)/P(goat)
P(incorrect given goat) = (2p_i/3)/(p_c/3+2p_i/3)
P(incorrect given goat) = 2p_i/(p_c+2p_i)

If the chance of your initial guess being incorrect given that he reveals a goat is greater than 0.5 you should switch.

Walt

[Rolfe]

Originally posted by Walter Wayne
Yes you can decrease your chances by switching depending on Monty's knowledge and intention....Yes, yes, yes, I acknowledged that above. Maybe I worded it badly, but that post was predicated on Monty always opening one of the other two doors, even if you've originally guessed wrong.

Curt C said on that case it was a 2/3 advantage to switch, and I commented that this was only the case if he always deliberately revealed a goat - if he's opening either of the two remaining doors at random and sometimes reveals the car, there's no advantage.

Now, if you were trying to tell me how switching would decrease your chance of winning in a scenario where Monty ALWAYS opens one of the two doors you didn't choose and offers you the chance to switch to the third door, then I think you're going to have to run me through it again....

It's getting late....

Rolfe.

[Walter Wayne]

I see a misinterpreted the term "deliberately avoiding the car". I thought avoiding given in away as opposed to avoiding revealing it.

Damn it, I made a long winded post for nothing. :)

Walt

[plindboe]

Wow, this simply puzzle has surely been explained to death. I vote for Batman Jr. as the tutor offering the best explanations. :D