[BruceZ]

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If possible, using actual math AND laymen's terms?

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Laymen's terms 1: You pick door A, and Monty must show you one of the other doors that has a goat, so he shows you door C. Now he might have had a choice of door B or door C and chose C randomly, or he might have been forced to show you door C because door B has the prize. In the first case, where he chooses randomly, he would only pick door C 1/2 of the time, but in the second case, where door B has the prize, he will show you C ALL of the time. Therefore, the second case is twice as likely as the first case, so door B is twice as likely to have the prize as your door A. So you should switch your pick to door B to have a 2/3 chance of winning, rather than staying with door A which has a 1/3 chance of winning.

Laymen's terms 2: Imagine we play this game a million times. For convenience, we will always pick door A, so Monty will show us either door B or door C with a goat. Now let's say we decide that we will never switch our initial choice. How often will we win the prize? Obviously 1/3 of the time. Therefore the prize will be behind the other door 2/3 of the time. Not much to that one, eh?


Math: Again assume we chose door A, and Monty shows us that door C has a goat. P(A | B) means the probability of A GIVEN B.

By the definition of conditional probability:

P(B = prize | Monty shows C = goat) =
P(B = prize AND Monty shows C = goat) / P(Monty shows C = goat)

By Bayes' theorem, this is:

P(B = prize | Monty shows C = goat) =
P(B = Prize)*1 / [P(Monty shows C = goat | B = Prize)*P(B = Prize) + P(Monty shows C = goat | A = Prize)*P(A = Prize)]

= (1/3)*1 / [1*(1/3) + 1/2*(1/3)] = 2/3.

So the probability that door B has the prize is 2/3.

Note that the common error people make is solving for P(B = prize | C = goat) rather than P(B = prize | Monty shows C = goat). The first probability is 1/2, and this leads to the erroneous conclusion that it doesn't matter if we switch or not.