-=Dark_Angel=-, P.I.12 Jul 2004

You should have just used the pair (D, H), and avoided all the unnecessary remarks about which doors are empty, making the proofe easier to follow. You also barely explained the use of Bayes' theorem. I have made a more readable version for my own benefit.

An element of the probability space is a pair (D, H) where D and H are elements of {A, B, C}. D tells you which door has the prize. H tells you which door the host opens. For example, the pair (C, B) describes the situation in which door C has the prize and the host opened door B.

You initially choose door C. It clearly has a 1 in 3 chance of containing the prize. Now the host may open either A or B with equal likelihood, making P(C, A) = P(C, B) = 1/3 * 1/2 = 1/6.

Choosing C rules out (A, C) and (B, C). Since before any doors are opened all doors as just as likely to contain the prize, P(C, A) + P(C, B) = P(A, B) = P(B, A) = 1/3 = 2/6.

The probability space is therefore:

State --------- Probability
(A, B) -------- (2/6)
(B, A) -------- (2/6)
(C, A) -------- (1/6)
(C, B) -------- (1/6)

The host then opens door B. (This rules out (C, A) and (B, A)). The probability now that the prize is behind C can be found using Bayes' Theorem:

P(prize is behind C | host opens B) =
P(prize is behind C & host opens B) / P(host opens B) =
P((C, B) / (P(C, B) + P(A, B)) =
(1/6) / (1/6 + 2/6) =
1/3

Therefore by sticking you have only a 1/3 chance of winning the prize. Therefore by switching to A you have a 2/3 chance of winning.

This can be further shown by finding the chance that the prize is behind A using Bayes' theorem:

P(prize is behind A | host opens B) =
P(prize is behind A & host opens B) / P(host opens B) =
P((A, B) / (P(C, B) + P(A, B)) =
(2/6) / (1/6 + 2/6) =
2/3