## Which Door Hides the Gold?

The short answer is that if you ignore your host's subsequent actions, and stay with Door A, your chances of success are the same as for someone else making the same door choice, but unaware that Door B was later opened. You both win or lose together. Thus your chances of striking it rich are 1/3 if you stick with Door A, and 2/3 if you switch to Door C.
A rigorous proof of these results follows from use of Bayes' Theorem. Defining the following events:

- A, B, C = the events that the gold lies behind Doors A, B, and C respectively
- A
_{s} = the event that the player selects Door A
- B
_{o} = the event that the host opens Door B

We wish to calculate the probability that the gold lies behind Door A, given that the player selected Door A and the host opened Door B. Symbolically, this is expressed as P[A|A_{s}B_{o}]. Using Bayes' Theorem,

P[A|A_{s}B_{o}] = P[AA_{s}B_{o}] / P[A_{s}B_{o}] = N_{1} / N_{2}, where

N1 = P[AA_{s}B_{o}] = P[B_{o}|AA_{s}] P[A_{s}A] = P[B_{o}|AA_{s}] P[A] P[A_{s}]

since A and A_{s} are clearly independent events. Now the probability that Door B will be opened if Door A has been selected and also hides the gold is 1/2, since our host could have just as likely opened Door C. And the unconditioned probability that the gold lies behind Door A, namely P[A] is clearly 1/3. Thus N_{1} = P[A_{s}] / 6.
**Turning to N**_{2}

N_{2} = P[A_{s}B_{o}] = P[B_{o}|A_{s}] P[A_{s}]
= {P[B_{o}|A_{s}A] P[A] + P[B_{o}|A_{s}B] P[B]
+ P[B_{o}|A_{s}C] P[C]} P[A_{s}] =

{(1/2)(1/3) + (0)(1/3) + (1)(1/3)}P[A_{s}] = P[A_{s}] / 2

Thus the probability that the gold lies behind Door A, given that Door A was selected by the player and Door B was opened by the host is N_{1} / N_{2}, or 1/3, as was to be shown.