My solution relies on a known result about this hat problem for finitely
many prisoners. See http://en.wikipedia.org/wiki/Hat_puzzle for example
(Ebert's version). It is known that there is a strategy, using Hamming
codes, whereby n = 2^k - 1 players can with probability n/(n+1) arrange
that at least one of them correctly guesses the color of his own hat and no
one guesses incorrectly (although of course many players pass). My solution
for infinitely many players, then, is for them to split into groups of 2^k
- 1 players for k = 10, 11, 12, .... Anyone not in a group passes. Players
within each group play according to the aforementioned strategy. With high
probability each group will produce at least one correct guess and no wrong
guesses. Since there are infinitely many groups, if all succeed, then the
infinite set of prisoners will win. To calculate the probability we just
need the infinite product of (2^k - 1)/2^k for k from 10 to infinity, which
maple tells me is about 99.8%.