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.
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?
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.