The first prisoner that is able to announce the color of his hat correctly will be released. No communication between the prisoners is allowed. In this case A, B and C would remain silent for some time, until A finally deduces that he must have a white hat because C and B have remained silent for some time. As mentioned, there are three white hats and two black hats in total, and the three prisoners know this. In this riddle, you can assume that all three prisoners are very clever and very smart.

If C could not guess the color of his own hat that is because he saw either two white hats or one of each color. If he saw two black hats, he could have deduced that he was wearing a white hat. In this variant there are 3 prisoners and 3 hats. Each prisoner is assigned a random hat, either red or blue. In all, there are three red hats and two blue. Each person can see the hats of two others, but not their own. On a cue, they each have to guess their own hat color or pass.

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They win release if at least one person guessed correctly and none guessed incorrectly passing is neither correct nor incorrect. Which strategy has the highest probability of winning?

If you think of colors of hats as bits, this problem has some important applications in coding theory. The solution and the discussion of this puzzle can be found here also a solution to the analogous 7-hat puzzle and other 3 variants are available on this Logic Puzzles page they are called Masters of Logic I-IV.

In this variant there are 10 prisoners and 10 hats. Each prisoner is assigned a random hat, either red or blue, but the number of each color hat is not known to the prisoners. The prisoners will be lined up single file where each can see the hats in front of him but not behind. Starting with the prisoner in the back of the line and moving forward, they must each, in turn, say only one word which must be "red" or "blue".

If the word matches their hat color they are released, if not, they are killed on the spot. What is the plan to achieve the goal? The prisoners agree that if the first prisoner sees an odd number of red hats, he will say "red". This way, the nine other prisoners will know their own hat color after the prisoner behind them responds. As before, there are 10 prisoners and 10 hats. The prisoners are distributed in the room such that they can see the hats of the others but not their own.

Now, they must each, simultaneously, say only one word which must be "red" or "blue". If the word matches their hat color they are released, and if enough prisoners resume their liberty they can rescue the others. A sympathetic guard warns them of this test one hour beforehand. If they can formulate a plan following the stated rules, 5 of the 10 prisoners will definitely be released and be able to rescue the others. The prisoners pair off. In a pair A, B of the prisoners A says the color he can see on the head of B, who says the opposite color he sees on the head of A.

Then, if both wear hats with the same color, A is released and B is not , if the colors are different, B is released and A is not.

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In total, 5 prisoners answer correctly and 5 do not. This assumes the pair can communicate who is A and who is B, which may not be allowed. Alternatively, the prisoners build two groups of 5. One group assumes that the number of red hats is even, the other assumes that there is an odd number of red hats.

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Similar to the variant with hearing, they can deduce their hat color out of this assumption. Exactly one group will be right, so 5 prisoners answer correctly and 5 do not. Note that the prisoners cannot find a strategy guaranteeing the release of more than 5 prisoners. Indeed, for a single prisoner, there are as many distributions of hat colors where he says the correct answer than there are where he does not.

Hence, there are as many distributions of hat colors where 6 or more prisoners say the correct answer than there are where 4 or fewer do so. In this variant, a countably infinite number of prisoners, each with an unknown and randomly assigned red or blue hat line up single file line. Each prisoner faces away from the beginning of the line, and each prisoner can see all the hats in front of him, and none of the hats behind. Starting from the beginning of the line, each prisoner must correctly identify the color of his hat or he is killed on the spot.

As before, the prisoners have a chance to meet beforehand, but unlike before, once in line, no prisoner can hear what the other prisoners say. The question is, is there a way to ensure that only finitely many prisoners are killed? If one accepts the axiom of choice , and assumes the prisoners each have the unrealistic ability to memorize an uncountably infinite amount of information and perform computations with uncountably infinite computational complexity , the answer is yes.

In fact, even if we allow an uncountable number of different colors for the hats and an uncountable number of prisoners, the axiom of choice provides a solution that guarantees that only finitely many prisoners must die provided that each prisoner can see the hats of every other prisoner not just those ahead of them in a line , or at least that each prisoner can see all but finitely many of the other hats. The solution for the two color case is as follows, and the solution for the uncountably infinite color case is essentially the same:.

The prisoners standing in line form a sequence of 0s and 1s, where 0 is taken to represent blue, and 1 is taken to represent red.

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Before they are put into the line, the prisoners define the following equivalence relation over all possible sequences that they might be put into: Two sequences are equivalent if they are identical after a finite number of entries. From this equivalence relation, the prisoners get a collection of equivalence classes.

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Assuming the axiom of choice, there exists a set of representative sequences—one from each equivalence class. Almost every specific value is impossible to compute, but the axiom of choice implies that some set of values exists, so we assume that the prisoners have access to an oracle. When they are put into their line, each prisoner can see all but a finite number of hats, and can therefore see which equivalence class the actual sequence of hats belongs to.

This assumes that each prisoner can perform an uncountably infinite number of comparisons to find a match, with each class comparison requiring a countably infinite number of individual hat-comparisons. They then proceed guessing their hat color as if they were in the representative sequence from the appropriate equivalence class. Because the actual sequence and the representative sequence are in the same equivalence class, their entries are the same after some finite number N of prisoners. All prisoners after these first N prisoners are saved.

It may seem paradoxical that an infinite number of prisoners each have an even chance of being killed and yet it is certain that only a finite number are killed. However, there is no contradiction here, because this finite number can be arbitrarily large, and no probability can be assigned to any particular number being killed. This is easiest to see for the case of zero prisoners being killed.

This happens if and only if the actual sequence is one of the selected representative sequences. If the sequences of 0s and 1s are viewed as binary representations of a real number between 0 and 1, the representative sequences form a non-measurable set. This set is similar to a Vitali set , the only difference being that equivalence classes are formed with respect to numbers with finite binary representations rather than all rational numbers. Hence no probability can be assigned to the event of zero prisoners being killed. The argument is similar for other finite numbers of prisoners being killed, corresponding to a finite number of variations of each representative.

This variant is the same as the last one except that prisoners can hear the colors called out by other prisoners. The question is, what is the optimal strategy for the prisoners such that the fewest of them die in the worst case? To do this, we define the same equivalence relation as above and again select a representative sequence from each equivalence class. Now, we label every sequence in each class with either a 0 or a 1.

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First, we label the representative sequence with a 0. Then, we label any sequence which differs from the representative sequence in an even number of places with a 0, and any sequence which differs from the representative sequence in an odd number of places with a 1. In this manner, we have labeled every possible infinite sequence with a 0 or a 1 with the important property that any two sequences which differ by only one digit have opposite labels. Now, when the warden asks the first person to say a color, or in our new interpretation, a 0 or a 1, he simply calls out the label of the sequence he sees.

Given this information, everyone after him can determine exactly what his own hat color is. The second person sees all but the first digit of the sequence that the first person sees. Thus, as far as he knows, there are two possible sequences the first person could have been labeling: one starting with a 0, and one starting with a 1. Because of our labeling scheme, these two sequences would receive opposite labels, so based on what the first person says, the second person can determine which of the two possible strings the first person saw, and thus he can determine his own hat color.

Similarly, every later person in the line knows every digit of the sequence except the one corresponding to his own hat color. He knows those before him because they were called out, and those after him because he can see them. With this information, he can use the label called out by the first person to determine his own hat color. Thus, everyone except the first person always guesses correctly. Ebert's version of the problem states that all players who guess must guess at the same predetermined time, but that not all players are required to guess.