> Its interesting to notice that a larger number of hat colors poses no problem here. For any set of hat colors , the prisoners can pick an abelian group struc

> Its interesting to notice that a larger number of hat colors poses no problem here. For any set of hat colors , the prisoners can pick an abelian group structure on . Then, the first prisoner guesses the ’sum’ of all the hat colors he can see. The next guy can then subtract the sum of the hat colors he sees from the hat color the first guy said to find his own hat color. Again, this argument repeats, and so everyone except the first guy gets out. For the case of black and white, the previous argument used black = 0 (mod 2) and white = 1 (mod 2). So the first guy says the sum, not a color? If he's allowed to say arbitrary things he can as well tell everyone their color "next guy is white, next green, etc.".