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After that last game, the warden didn't like that they escaped. So he made them all play again. Like before, they know there are 6 prisoners, 3 red hats and 3 white hats. They know one prisoner is completely hidden behind the wall. They know they should not speak except to correctly answer which color hat he is wearing or they will be executed. And they know that one of the prisoners is mute. But this time the witless warden decides to give them a clue. He tells them the mute is wearing a red hat. This is how the arrangement landed:

Hats2

After a while, one of the prisoners shouts out his hat color with confidence! Which one?

Assumptions: All prisoners are totally infallible and quick and precise in their logic and they are given as much time as they need to make a guess.

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C announces he is white.

Reasoning:

C Reasons as follows:

"Suppose my hat is red. Then A and B cannot both be red, since there are only three red hats and D has one."

"Suppose B is the third red. Then A could see three red hats: B, C, and D. Therefore he could deduce he must be white. Since whites are not mute, he would be able to announce this."

"This has not happened, so B's and my hats are not both red."

"If my hat really is red, then B could also follow this same reasoning."

"By knowing that my hat is red but both of ours aren't, B would deduce he had a white hat, and announce it."

"B has not announced, so B is not white, therefore he is red."

"But B's and my hats are not both red."

"Therefore mine must be white."

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  • $\begingroup$ Nice. I didn’t even know this solution existed. @Oray’s answers was what I was thinking, but this is good. $\endgroup$ – tyobrien Sep 5 '18 at 20:49
  • $\begingroup$ I believe I have a solution where B speaks, but his logic would probably take longer. See my answer below. $\endgroup$ – tyobrien Sep 7 '18 at 4:42
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C would make the call

Reasoning:

C understands that C and D has got to have different hats by this logic:
Red hat = r, white hat = w, (I left out the one behind the wall)

Option 1: w r r r w -> A would call his color (three red hats infront)
Option 2: w w r r w -> B would know that he has white if A doesn't call his color
Option 3: r w r r w -> Same as option 2

C calls white.

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  • $\begingroup$ No, sorry: Your option 3 is wrong: A could be the mute... then he would not call despite of knowing $\endgroup$ – Torsten Link Sep 6 '18 at 13:01
  • $\begingroup$ @Torsten Link It doesn't matter whether A is mute or not in option 3. He couldn't have said anything even if he wasn't mute. There are 2 red hats infront of B, which means that either he himself has a red hat (and A with a white hat would call) or he has a white hat (where A would not say anything in any case). $\endgroup$ – Unerean Sep 6 '18 at 20:07
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A

Because

A can see 2 white hats infront of them and he knows that he is not mute thus he can only have a white hat on.

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    $\begingroup$ Only one of the red hats is mute. There are two red hats who can speak, so being able to speak does not imply that you have a white hat. $\endgroup$ – EightAndAHalfTails Sep 6 '18 at 9:19
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I’ve accepted @EightAndAHalfTails’ answer because it is a valid solution.

However my original solution was where B speaks and it is dependent on C’s silence.

But the accepted solution is dependent on B’s silence so what is the correct answer?

I suppose it is whoever’s logic is quicker... Anyway, here’s the alternative solution assuming C doesn’t speak before B.

B knows that A isn’t speaking - either because A sees two of each color in front of him or A is mute.

B sees one red hat in front of him on D. So B’s only hope to be the one to answer is to deduce that D is actually the mute.

C doesn’t speak at this point either (and remember we’re assuming he doesn’t speak before B).

B reasons the following:

“Suppose D is not mute.

D realizes that C and D cannot both be white otherwise A or B would have certainly spoken. So C and D must be red and red, red and white, or white and red.

D also realizes the only case where he is white (C red and D white) is the exact same scenario as the last game. In which case B would speak if C is mute, and C would speak if C is not mute.

Since this doesn’t happen D would realize that this case (C red and D white) must not be true. Therefore one of the other two cases must be true (red and red, or white and red). In either case D is red and he would announce this as his color.

Since D has not announced this, he must be mute and therefore A is not mute.

And therefore A sees two of each color in front of him.

Therefore I am red!”

Since this answer appears to have more steps than EightAndAHalfTails’ answer, then B would actually not speak before C.

However if we choose to believe that these prisoners have instantaneous god-like logic, then we come to a loop where no one can answer. How interesting!

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  • $\begingroup$ I think the crux of the apparent paradox is this: In my answer, C's reasoning is predicated on B not announcing that he is white, not on him not saying anything at all. C's reasoning is something like "If B were white, he would be able to deduce and announce that fact." In that case, B saying nothing, or B saying "red", both contradict the supposition and allow C's reasoning to continue. Of course, in the latter case, the game is already over. I imagine the reverse is true as well, for your reasoning. $\endgroup$ – EightAndAHalfTails Sep 7 '18 at 12:24
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C is correct. C can see one red and one white hat in front of him. If C and B both had red hats then A could see three red hats on B, C and D so he would announce his color, C can understand that since A has not announced his hat color therefore only one of them (B or C) has a red hat. If C had a red hat then B could see two red hats in front of him and B would think that if he had a red hat then A would speak but A does not so C got the fact that his hat can't be red hence C knew that he had a white hat.

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    $\begingroup$ You are misunderstanding about A. A prisoner can be red and not mute. $\endgroup$ – tyobrien Sep 6 '18 at 12:59
  • $\begingroup$ @tyobrien oh yes that was my mistake. $\endgroup$ – MEDLEY Sep 6 '18 at 14:37

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