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The Monty Hall problem

I want to know what are the various ideas which you might have come across to arrive at the solution of the Monty Hall Problem ?

My idea is to get them all under a single page which would significantly help me in my research work.

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marked as duplicate by Henry, Chris Eagle, Henning Makholm, Hans Lundmark, t.b. Sep 16 '12 at 15:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2 Answers 2

Celtschk gave a very good answer. I cannot say it better. To add to this I suggest that you can take a look a any of several introductory books on probability and/or statistics that discuss the Monty Hall problem. I included it in my "Essentials of Biostatistics" book. But there are many others. Also people become convinced by simulating the game on the computer. Ask the subject to pick a door. Then draw a uniform random number selecting door 1 say if 0<=u<=1/3, door 2 if 1/3

When I worked at Nichols Research Corporation in Costa Mesa California I made a bet with a colleague that 2/3rds was the right answer. I was only able to collect on the bet by convincing him that I was right by doing this simulation. It doesn't take long to see that 1/2 is not the right answer.

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To state the obvious one first: Just calculate the probabilities the straightforward way. This requires no advanced thinking and as long as you just slavishly follow the rules, you'll reach the correct conclusion.

If you don't follow that path, the point of all explanations is to counter the wrong intuition that the last step is an independent new choice.

One way is that if you chose from the beginning to keep your first choice, then of course the probability of winning cannot depend on whether the quiz master opens another door first (and we know in advance that behind that door there will be a goat). So the probability of winning when not changing is the probability of choosing right the first time, which is $1/3$. The probability of winning when changing therefore has to be $2/3$, because you know that the prize is behind one of the doors.

Another way is to note that if you have chosen the wrong door from the beginning, the show master actually has no choice of which door to open. Therefore he gives you the additional information: "If you have chosen wrong, that other door I didn't open contains the price." Again, the probability that you've chosen right is $1/3$, and therefore the probability that the other door contains the prize is $2/3$.

Also note that it is important that the quiz master always opening the empty door is part of the game. To see that, assume the other extreme: The stingy show master who opens the prize door whenever possible, in order to avoid the candidate winning the price. In that case, it's obvious to everyone that if he opens a goat door, the best strategy is not to change, because if one hadn't chosen right from the beginning, he certainly would have opened the prize door instead. This demonstrates clearly that the second choice is clearly affected by the strategy of the quiz master, and therefore argues against the notion that you have to have a probability of $1/2$ for each choice afterwards; of course that argument doesn't provide you with the actual probabilities of the original problem, but it at least makes it plausible that just like the quiz master favouring the prize door makes it more profitable not to change, the quiz master favouring goat doors makes it more profitable to change.

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Very nice answer +1 @celtschk. It is all a matter of clearly thinking out what the conditional probability is for your door given that Monty Hall showed you a door that had a clunker and not the prize. – Michael Chernick Sep 16 '12 at 13:25

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