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I think the solution should be something like:

If the first prisoner sees an even number of black hats, he calls black. If he sees an odd number of black hats he sayas white. The second prisoner uses the info that the first prisoner saw an even/odd number of black hats to deduce his hat color and so forth.

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Pretty sure this is correct. 99.5/100 on average go free unless I'm missing something.

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thats only if the hats are distributed 50/50.

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No.. it should work no matter what. It's only the first guy who may or may not go free.