There was no posted solution, so here goes:
Sage 1 can only guess for one case:
- when ?BB can be seen, Sage 1 must have W. (Note: if this happened, Sage 2 would see W?B, and Sage 3 would see WB?)
- otherwise, Sage 1 must pass.
Sage 2 can guess for two cases:
- when B?B can be seen, Sage 2 must have W. (Note: if this happened, Sage 3 would see BW?)
- when W?B can be seen, Sage 2 must have W, otherwise Sage 1 would have seen ?BB and guessed. (Note: if this happened, Sage 3 would see WW?)
- otherwise, Sage 2 must pass.
Sage 3 will always know:
- when BB? can be seen, Sage 3 must have W.
- when BW? can be seen, Sage 3 must have W, otherwise Sage 2 would have seen B?B and guessed.
- when WB? or WW? can be seen, Sage 3 must have W, otherwise one of the other sages would have guessed.
So, in tabular form:
Result.. Sage sees... ...and guesses...
1 2 3 1 2 3
WBB ?BB W?B WB? W - -
WWB ?WB W?B WW? pass W -
BWB ?WB B?B BW? pass W -
BBW ?BW B?W BB? pass pass W
BWW ?WW B?W BW? pass pass W
WBW ?BW W?W WB? pass pass W
WWW ?WW W?W WW? pass pass W
Tuesday, February 19, 2002
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