This is a logic puzzle. This is the page that has the ANSWER. If you want to see the puzzle without the answer, go **HERE**.

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There were three prisoners in certain jail, one of whom had normal sight, one of whom had only one eye, and the third of whom was completely blind. The jailer liked to play tricks on his prisoners, so one day he brought them all three out and had them stand in a line. He then showed them five hats, three white and two red, which he hid in a bag. From these five hats, he selected three and put one on each of the prisoners’ heads. None of the men could see what color hat he himself wore. The jailor offered freedom to the sighted man, if he could tell him what color his hat was (and to prevent guessing, he threatened execution for a wrong answer). The sighted man could not tell what color hat he wore, and declined to answer. The jailor next offered the same terms to the one-eyed man, who also could not tell what color hat he wore, and also declined to answer. At this point the jailor thought his game was done, but was delighted when the blind prisoner spoke up and asked if he didn’t get a turn? Laughing, the jailer offered the blind man the same terms as the others. The blind prisoner smiled and replied,

I do not need my sight;

From what my friends have said,

I clearly see my hat is _____!

And so he won his freedom. How did the blind prisoner know what color his hat was? And what color was it?

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ANSWER: The blind prisoner’s hat is WHITE.

PROOF: There are seven possible permutations of hats:

1 W R R

2 R W R

3 W W R

4 W R W

5 R W W

6 R R W

7 W W W

(1) If the permutation 1 were correct, then the 1st prisoner would see two RED hats and *know* his hat was WHITE. But the 1st prisoner doesn’t *know* what color his hat is, so permutation 1 is ruled out.

(2) The 2nd prisoner looks at the 3rd prisoner. If he sees a RED hat on the 3rd prisoner, then he will *know* that his hat must be WHITE. That is, if the 3rd prisoner has a RED hat, the 2nd prisoner would *know* that either permutation 2 or permutation 3 is correct, and either way, his hat must be WHITE. The only permutation where the 3rd prisoner could have a RED hat and the 2nd prisoner NOT have a WHITE hat is permutation 1, which is already ruled out already in (1). So if the 2nd prisoner sees a RED had on the 3rd prisoner, we would *know* his hat is WHITE. But he does not *know* what color his hat is. Therefore, he does not see a RED hat on the 3rd prisoner. Therefore, the 3rd prisoner’s hat must be WHITE.

(3) To sum up, the 1st prisoner’s ignorance rules out permutation 1. The 2nd prisoner’s ignorance rules out permutation 2 and permutation 3. Since permutations 1-3 are ruled out, the correct answer must be 4, 5, 6 or 7. And it doesn’t matter which one is correct, because in all of these, the 3rd prisoner’s hat is WHITE, which is all he needs to *know*.

[…] To see the solution, go HERE. […]

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Thanks, this was fun!

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This class of problem has the general issue that it assumes hyperrationality, which is completely alien to actual human behavior. Were I the blind prisoner, I certainly wouldn’t stake my life on the proposition that two other prisoners had both thought through a mildly tricky logic problem and determined they didn’t have enough information, as opposed to missing the clues and playing it safe.

Of course, the dire half-dragon version of this problem is the “blue-eyed natives on the island” puzzle, which is even less plausible.

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Yes, obviously, it presumes the perfect rationality of all three prisoners, but it isn’t meant to be an illustration of human nature, so much as an exercise in pure logic.

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Indeed.

My favorite recent (well, “recent”) puzzle along these lines was probably the “Cheryl’s birthday” problem, e.g. here. Pretty similar line of logic.

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These sorts of problems are always a little under-specified; the brilliance of the one-eyed prisoner was not mentioned (or, alternatively, that the one-eyed prisoner was able to take into account what the two-eyed prisoner said). Granted, it’s the only way that the blind prisoner can be right, but it wasn’t obvious we were guaranteed that.

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It doesn’t say the one-eyed prisoner can only train his eye on one of the other prisoners or did I miss that? If he can look at the 3rd, then the 1st, he knows his hat must be white.

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He can look at both. The fact he has one eye is merely a colorful detail/red herring. He can see and is not color-blind.

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