The Monty Hall Problem
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 door X, and the host, who knows what's behind the doors, opens another door, say door Y, which has a goat. He says to you, "Do you want to pick door Z?" Is it to your advantage to switch your choice of doors?
(you can easily google for the many pages that mention this problem; once you thought about it see my comments and help a mathematically challenged brotha out)
posted by cho @ 10:36 PM
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3 Comments:
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I drew a chart and worked out this counterintuitive problem. But can someone explain/translate the below mathematical proof for me:
Let the doors be called X, Y and Z.
Let Cx be the event that the car is behind door X and so on.
Let Hx be the event that the host opens door X and so on.
Supposing that you choose door X, the possibility that you win a car if you then switch your choice is given by the following formula:
P(Hz^Cy) + P(Hy^Cz)
= P(Cy).P(Hz|Cy) + P(Cz).P(Hy|Cz)
= (1/3.1) + (1/3.1)
= 2/3
They're just using regular conditional probability rules - if two events A and B are dependent, the chances of A and B are equal to the probability of A times the probability of B given that A has happened. Say the probability that I punch you in the face is .01, and the probability that you kick me in the balls is .05. Now, if you kick me in the balls, the probability that I punch you in the face goes up to .3 (I might strike you in a different location). So, the probability that you kick me in the balls AND I punch you in the face is .05 * .3, or .015. Fortunately for both of us, this probability is low.
In regard to this problem, there are two scenarios in which you win the car, given that you switch away from Door X - one is if the car is behind Door Y AND the host opens Door Z, the other is the car is behind Door Z AND the host opens Door Y. Breaking out those scenarios using the rule above - the probability that the car is behind door Y (or Z) is 1/3. The probability that the host opens door Z (or Y), given the car is behind door Y (or Z) is 1. So, each scenario has a 1/3 * 1 probabilty, and adding those together gives 2/3.
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