chinese emporer There was no posted solution, so here goes: Solution 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 David Woodruff Tuesday, February 19, 2002   Fog Creek Home