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Brando's Dead (slight reprise) (Haiku) by ?-Dave_Mysterious-?

You come to me mhwrrr... Day of my daughter's wedding mhwrrr... to kill a man?

-=Dark_Angel=-, P.I. 12-Jul-04/4:22 AM
Yes, the correct answer is that you should switch. Opening the empty door cannot increase the probability that your first door was the right door, because you can always open an empty door from the other two, no matter what door you chose.

Your last point about having no information as to which door is correct isn't right. You are offered a choice between door A and door C, but the fact that the host opened door B actually teaches you something about door A because it rules out the case in which A is empty and B has the prize, leaving the cases in which both A and B are empty, or A has the prize and B is empty. This idea is formalised in the following analysis:

An element of the probability space is a 4-tuple (W, X, Y, Z) where W,X,Y are elements of {E, P} and Z is an element of {A, B}. The first three (W, X, Y) tell you the state of the three doors (either empty E, or with prize P). Z tells you which door the host chose - either A, or B. For example, the 4-tuple

(E, E, P, B)

Describes the situation in which door A is empty, door B is empty, door C has the prize, and door B was chosen by the host.

The probability space is given in the following table:

State--------------- Probability
(E, E, P, A)-------- (1/3)*(1/2) = (1/6)
(E, E, P, B)-------- (1/3)*(1/2) = (1/6)
(E, P, E, A)-------- (1/3)
(P, E, E, B)-------- (1/3)

The probability that the prize is in C given that the host opened B is obtained using Bayes' Theorem, and is:

(1/6)/(1/6 + 1/3) = 1/3

The probability that the prize is in A, given that the host opened B is:

(1/3)/(1/6 + 1/3) = 2/3

Had the host chosen his door at random, and then revealed that it happened to be empty, the choice would indeed be 50:50 so it would no difference.




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