# The Monty Hall Dilemma

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This was mentioned in another thread. It is one of my favorite puzzles, because even when you know the answer, it is still confusing. You are a contestant on Let's Make A Deal. Your are shown 3 doors -- behind one of them is a brand new car and behind the other 2 is nothing. You are asked to pick one of the doors. Monty opens a door that you didn't pick and shows you that there is nothing there. Obviously, the car is behind your door or the other door. Then he asks if you would like to change your pick to the other door. What should you do? A. Stick with your first pick. B. Switch to the other door. C. It doesn't matter what you do, there is no advantage to switching.

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Are you wanting someone to post the answer, or are you wanting us to think about an interesting probability problem [smile].

(I'd switch, but I'll leave the reasoning for people to think about...)

I think I read somewhere that a potential source of confusion over this problem was that the original Monty Hall problem (the one on the original gameshow) was structured differently. In the gameshow, there were two contestants, who would pick different doors. Monty Hall would then always go up to one of the contestants doors, open it, and show nothing. The remaining contestant would then have the opportunity to stick with their door, or to switch.

In this case, the problem has changed, as there was a 2/3 chance that the car was behind one of the contestants doors; this doesn't change when Monty opens a door (as he always picks a losing door). So in that case it is best to stick with your door (2/3 versus 1/3).

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trapper zoid, are the 2 contestants teamed up or competing against each other? I'll assume they were competing... but teaming up introduces another interesting paradox:

I've always justified the Monty Hall problem as that the only way to lose with the "switch" strategy is if you originally pick the correct choice, the chance of which is 1/3. So 1 - 1/3 = 2/3.

But playing in a co-op team environment introduces another twist: Ostensibly, with a "don't switch" strategy, the only way to lose is if neither originally picks the correct choice yielding a 2/3 win rate. HOWEVER. This also seems to imply that if both teammates opt for a "always switch" strategy, the only way to win is if you both pick the wrong door, which would happen only 1/3 of the time.

So, single player wins 2/3 with switch strategy, but 2-player co-op only wins 2/3 with the stay strategy... Splain that mister probability!

(Actually, I just realized the answer... but it's a good one to think about!)

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I'm assuming they're competing; the final two contestants on the show.

An interesting challenge to think about is the 4 door Monty Hall problem. This is the same as the original Monty Hall problem as defined by John Bolton, but with a twist; now there are 4 doors. After you pick one door, Monty will open one empty door for you. You now have the option to stay or pick one of the other two doors. Monty will then open another empty door, and give you a second choice to stick or switch.

What is the most likely to win strategy, stick/stick, stick/switch, switch/stick or switch/switch?

Then expand that to the N door Monty Hall problem, for a real challenge!

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I haven't seen the show in a few years but IIRC, it was the player who won the best prize that got to keep it. There were 3 doors, one had a nice prize like a car, the other an ok prize like furniture, and the other was a joke prize, like a farm animal.

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Actually, a detailed explanation for the confusion (and general overview) can be found right here.

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Quote:
 Original post by _goatActually, a detailed explanation for the confusion (and general overview) can be found right here.

Very interesting...

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You chances increase by switching because:

- You initially have a 1/3 chance of picking the car, and thus a 2/3 chance of picking nothing.

- The host must reveal what's behind a door which you did not pick. The host must also pick one of the two remaining doors, such that his choice contains nothing.

- Since you have a 2/3 chance of picking nothing, it is more likely that the resulting two doors will be contain nothing and the car, rather than both containing nothing. Thus, since the host must open the door containing nothing, it is more likely the remaining door will contain the car.

EDIT: A more mathematical explanation:

Probabilities:

- Of you initially picking nothing: p(Y|N) ~= .6
- Of you initially picking the car: p(Y|C) ~= .3
- Of the host opening a door which contains nothing: p(H|N) ~= 1
- Of the remaining door containing the car: p(C) = p(H|N)*p(Y|N)~ = .6

Thus, there is a 2/3 chance that after the host opens a door, the remaining door is the car, whereas there is a 1/3 chance that the door you initially chose is the car. Therefore, by switching you increase your odds from 1/3 to 2/3.

[Edited by - nilkn on August 8, 2005 9:41:24 PM]

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Dammit! Nobody likes my dilemma... I still think it is a good one.

So, specifically, refute this:
In single player MH, switching to a different door after the loser is revealed yields a win probability of 2/3 (we all know this)... But imagine if two teammates each pick a different door. If they both have the "switch after non-winning door revealed" strategy, the only way the pair can win is if they both choose the wrong door. The odds of this occurence is 1/3. Why do the odds of winning with "switch" strategy decrease when teammates are added?

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ajas, what are you talking about? If you have three doors and you and your teammate pick two, there is only one door left for the host to open. Now, there is nothing to switch to.

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