chinese emperor Has anyone got an answer to this question? The link is http://www.techinterview.org/Puzzles/fog0000000031.html I've been trying to list down all the possibilities but it seems like the last one to guess will never be sure of the answer if the first and second ones are both having white balls. cyw Wednesday, April 7, 2004 If [1] and [2] have white hats and it's [3]'s turn, he will shout "My hat is white." Why? Because if [3]'s hat were black, [2] would have known that his ([2]'s) hat is white. And why that? Because if [3]'s hat were black, and [2]'s hat were black, [1] would have said "My hat is white." Why? Because there are only two black hats total. The trick in these questions is that you have to deduce something from the fact that somebody else couldn't deduce anything. Vigor Thursday, April 8, 2004 I just noticed that they are "wearing" balls, not hats. Doesn't change the logic, though. Vigor Thursday, April 8, 2004 well Here Is the Complete Solution there are Three Cases B - Black Ball W - White Ball 1.  BBW 2. WBW 3. WWW now the trick is to solve the problem in 3 rounds Case I BBW (Answered in 1 Round) in round 1 person who sees two BB on other 2 person head will shout that i have a white ball on my head since its only the round1 and 1 person has said that he has a white ball will make other 2 persons sure that they have black balls on thier head CASE II BWW(Answered in 2nd round) in round 1 all person will pass there turn now since all have passed there turn so the CASE BBW is not there now in 2nd round when a person sees BW balls on other two person he will say that i have a White Ball on my head and then the next person who sees BW ball will say W and the last person left will say B CASE III WWW(answered in 3rd round) in round 1 and 2 all will pass there turn  because they will see WW on others head and wont be able to conclude anything since nobody concluded anything in 2 round that means all of them have white ball BondOfUk Sunday, April 11, 2004 If Sage 1 sees 2W,3W Then he can tell nothing If Sage 1 sees 2W,3B Then he can tell nothing If Sage 1 sees 2B,3W Then he can tell nothing If Sage 1 sees 2B,3B Then he can tell 1 must have W If Sage 2 sees 1W,3W Then he can tell nothing If Sage 2 sees 1B,3W Then he can tell nothing If Sage 2 sees 1B,3B Then he can tell 2 must have W If Sage 2 sees 1W,3B Then he can tell 2 must have W (since the only case where 3 could have black from person 1's analysis is where 2 has W or where 1 would know his color) Sage 3 doesn't even need to look, he knows the only case 1 or 2 wouldn't know what color they had is where he has White. Manfred Tuesday, April 20, 2004   Fog Creek Home