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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.

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.

Saturday, November 8, 2003

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