Chinese Emperor
Actually the game is fair if the balls are randomly distributed. Each sage has a 1 in 3 chance of "winning", assuming they don't guess.
w = white
b = black
x = don't care
There are three possibilities for the first round:
Sage 1's turn:
x b b Sage 1 knows he is w
x w w Sage 1 passes
x w b Sage 1 passes
Sage 2's turn:
b x w Sage 2 passes
w x w Sage 2 passes
x x b Sage 2 wins (he knows that if he were "b" Sage 1 would have guessed his color)
Sage 3's turn:
The only way for Sage 3 to get a turn is if he is white.
Lars
Friday, September 19, 2003
I think you have over simplified the problem.
If we take all probabilities into account, we find that
P(sage 1 guessing)=0.1
P(sage 2 guessing)=0.3
P(sage 3 guessing)=0.6
This is because the occurences have different probabilities.
Vineet
Saturday, November 8, 2003
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