3 white balls, 2 black balls, placed randomly on future advisers' heads.
If an adviser sees 2 black balls, he knows his is white. If an adviser passes, the advisers after him know he didn't see 2 black balls.
Adviser1, Adviser2, Adviser3 - guess1, guess2, guess3
1. b,b,w - pass, pass, white (sees 2 black)
2. b,w,b - pass, white (sees 2 black)
3. b,w,w - pass, pass, white (knows if he had had a black ball on his head, previous adviser wouldn't've passed, so knows his ball is white)
4. w,b,b - white (sees 2 black)
5. w,b,w - pass, pass, white (same reason as 3)
6. w,w,b - pass, white (same reason as 3)
7. w,w,w - pass, pass, white (Adviser3 knows that if he had had black on his head, Adviser2 would've known his own color was white)
so the probability of winning is 1/7 Adviser1, 2/7 Adviser2, 4/7 Adviser3. This is unfair in the favor of the last adviser to guess in round one.
Friday, January 03, 2003
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