Quote:

The contestant should always switch doors. Without switching doors she has a 1/3 chance of winning. Switching doors gives her a 2/3 chance of winning.

Her chance of winning was 1 in three to begin with—and after Monty opens a door, the chance of her winning is still just one in three.

Here's another way of thinking about it. Let's say you pick door #1. You've divided the doors into two sets:

{1}{2,3}

Set {1} has a 1/3 chance of holding the winning door. Set {2,3} has a 2/3 chance of holding the winning door. You now decide to switch from set {1} to set {2,3}. You now have a 2/3 chance of holding the winning door. Monty opens one of the doors in set {2,3}. Set {2,3} still has a 2/3 chance of holding the winning door, but now you know which of {2,3} is not the winning door. You choose the other door in set {2,3} and you have a 2/3 chance of having the winning door.

A computer simulation of the game using a random number generator should convince any skeptics. (See http://www.macchiato.com/humor/monty_hall_skeptics.htm for Java code that executes the Monty Hall problem.)

A proper use of Bayes' Theorem is required to mathematically solve this problem. Bayes' Theorem tells us how to properly update the odds after we are given new information. The result is not always intuitive. (See http://astro.uchicago.edu/rranch/vkashyap/Misc/mh.html)

The a priori probability that the prize is behind door X, P(X) = 1/3

The probability that Monty Hall opens door B if the prize were behind A,
P(Monty opens B|A) = 1/2

The probability that Monty Hall opens door B if the prize were behind B,
P(Monty opens B|B) = 0

The probability that Monty Hall opens door B if the prize were behind C,
P(Monty opens B|C) = 1

The probability that Monty Hall opens door B is then
p(Monty opens B) = p(A)*p(M.o. B|A) + p(B)*p(M.o. B|B) + p(C)*p(M.o. B|C)
= 1/6 + 0 + 1/3 = 1/2

Then, by Bayes' Theorem,

P(A|Monty opens B) = p(A)*p(Monty opens B|A)/p(Monty opens B)
= (1/6)/(1/2)
= 1/3
and
P(C|Monty opens B) = p(C)*p(Monty opens B|C)/p(Monty opens B)
= (1/3)/(1/2)
= 2/3
In other words, the probability that the prize is behind door C is higher when Monty opens door B, and you SHOULD switch!