4 ms·

I think a correct analysis of this problem requires you to know the guard's exact policy for what to say in every possible situation. The author has assumed tha

I think a correct analysis of this problem requires you to know the guard's exact policy for what to say in every possible situation. The author has assumed that in the case that A is pardoned, the guard randomly chooses whether to answer "B" or "C" with equal probability, but we have no evidence for this. Consider, for example, how you would analyze the situation where the guard says to A, completely unsolicited, "Hey, just so you know, B is still condemned to die." On the one hand, it seems that the author's exact same analysis could be used in this situation: the guard's policy here might be exactly the same as it was in the original situation (where A asks the guard to name one of the other prisoners). On the other hand, perhaps the guard's policy was to only offer this information to A if A were pardoned. Or maybe the guard would have only said this if C were pardoned. We just don't know, so we can't calculate the probability that A is pardoned.

vladf 9y ago
Yeah, looking at it with Bayes rule makes it sound more convincing. Here's what's assumed: * The guard can't express whether or not A was pardoned to A. * The guard is telling the truth. Then the named (named to prisoner A) executed person E must be either B or C if A is saved. Let the saved person be S. Then: P(S=A|E=B) = P(S=A) P(E=B|S=A) / sum_{S in {A, B, C}}(P(S) P(E=B|S)) = P(E=B|S=A) / (P(E=B|S=A) + P(E=B|S=B) + P(E=B|S=C)) = P(E=B|S=A) / (P(E=B|S=A) + 0 + 1 ) As mentioned in the OP, if S=B then E=C and if S=C then E=B necessarily: both these events happen with equal probability. Then, as you mention with your intuition, if, say, the guard prefers to name B in the case of S=A, then hearing that E=B really does give you some information, since there is more probability mass in situations where you're saved when looking at the subset of events where E=B. Put extremely, taking P(E=B|S=A) = 1, we find that indeed P(S=A|E=B) = 1/2!
FabHK 9y ago
> a correct analysis of this problem requires you to know the guard's exact policy for what to say in every possible situation. Just as with Monty Hall, where one ought to specify that the host knows exactly what's where, and always opens a remaining door such that it contains a goat (choosing uniformly if there's more than one such choice). The nice thing about the variant in the article is that it is fairly well and intuitively specified. (By the way, if the guard chooses B with probability x (rather than 1/2) in case A is pardoned, then the probability that A is pardoned is x/(1+x), which is between 0 and 1/2, but indeed is 1/3 only for x = 1/2.). A = A is pardoned etc. P(A) = P(B) = P(C) = 1/3 SA = guard says A still condemned etc. P(SB | A) = x P(SB | B) = 0 P(SB | C) = 1 P(SC | A) = 1-x P(SC | B) = 1 P(SC | C) = 0 P(SA) = 0 P(SB) = (1+x)/3 P(SC) = (2-x)/3 P(A | SB) = P(SB | A) P(A) / P(SB) = x * 1/3 / (1+x) * 3 = x/(1+x)

vladf 9y ago

Yeah, looking at it with Bayes rule makes it sound more convincing. Here's what's assumed: * The guard can't express whether or not A was pardoned to A. * The guard is telling the truth. Then the named (named to prisoner A) executed person E must be either B or C if A is saved. Let the saved person be S. Then: P(S=A|E=B) = P(S=A) P(E=B|S=A) / sum_{S in {A, B, C}}(P(S) P(E=B|S)) = P(E=B|S=A) / (P(E=B|S=A) + P(E=B|S=B) + P(E=B|S=C)) = P(E=B|S=A) / (P(E=B|S=A) + 0 + 1 ) As mentioned in the OP, if S=B then E=C and if S=C then E=B necessarily: both these events happen with equal probability. Then, as you mention with your intuition, if, say, the guard prefers to name B in the case of S=A, then hearing that E=B really does give you some information, since there is more probability mass in situations where you're saved when looking at the subset of events where E=B. Put extremely, taking P(E=B|S=A) = 1, we find that indeed P(S=A|E=B) = 1/2!

FabHK 9y ago

> a correct analysis of this problem requires you to know the guard's exact policy for what to say in every possible situation. Just as with Monty Hall, where one ought to specify that the host knows exactly what's where, and always opens a remaining door such that it contains a goat (choosing uniformly if there's more than one such choice). The nice thing about the variant in the article is that it is fairly well and intuitively specified. (By the way, if the guard chooses B with probability x (rather than 1/2) in case A is pardoned, then the probability that A is pardoned is x/(1+x), which is between 0 and 1/2, but indeed is 1/3 only for x = 1/2.). A = A is pardoned etc. P(A) = P(B) = P(C) = 1/3 SA = guard says A still condemned etc. P(SB | A) = x P(SB | B) = 0 P(SB | C) = 1 P(SC | A) = 1-x P(SC | B) = 1 P(SC | C) = 0 P(SA) = 0 P(SB) = (1+x)/3 P(SC) = (2-x)/3 P(A | SB) = P(SB | A) P(A) / P(SB) = x * 1/3 / (1+x) * 3 = x/(1+x)