Three players enter a room and a red or blue hat is placed on each
person's head. The color of each hat is determined by a coin toss,
with the outcome of one coin toss having no effect on the others.
Each person can see the other players' hats but not his own.
No communication of any sort is allowed, except for an initial
strategy session before the game begins. Once they have had a
chance to look at the other hats, the players must simultaneously
guess the color of their own hats or pass. The group shares a
hypothetical $3 million prize if at least one player guesses
correctly and no players guess incorrectly. What strategy should they use to maximize their chances of success?
(From  http://www.princeton.edu/~sjmiller/riddles/riddles.html)
The possible hat configurations areas follows (each occurs with equal probability 1/8)
rrr
rrb
rbr
rbb
brr
brb
bbr
bbb
In six of these 8 cases, 1 player will see two hats of color opposite to his. In these six cases, the other 2 players will see two different colored hats. If the players agree to the rule that they will pass if they see different hats, and guess the opposite color if they see the same color hats, then they will win in these 6 cases. In the other 2 cases they lose.
Overall, they win 6 out of 8 times with this strategy. (I offer no proof, just this example strategy  proof by jawboning?)

Posted by jim
on 20031029 18:14:24 