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Alejandro1 comments on Open Thread March 31 - April 7 2014 - Less Wrong Discussion
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Do the prisoners guess all at the same time, or in order (and thus after hearing finitely many other guesses)? If the second, is the guessing order known beforehand, or will it be at the whim of the guards?
All at the same time. Solving the case with guessing in order is a good intermediate step.
With guessing in order, I observe that for every finite subset there are either an even or an odd number of red hats, and prisoner 1 can indicate which it is by his guess; then everyone in that subset can count the red hats and figure out which colour his own must be to make the total number even or odd. Let the size of the finite subset go to infinity.
This is a good idea, and solves the similar problem with finitely many prisoners getting at most one guess incorrect. But...
I don't see how this immediately gives a solution; any finite set of hats has either an even or an odd number of red hats, but an infinite set of hats may have an infinite number of red hats and an infinite number of black hats, and infinities are neither odd nor even.
Guessing in order with the other prisoners hearing these guesses would violate the stricture that the prisoners cannot communicate with each other.