First of all, which are the possible cases?
Now, let’s take all the cases, one by one. The mistake would be to start from case WWW, which is the most confusing. One other thing, each sage sees the two other sages’ hats, so he can assume two possible combinations, except for the case that the two others are wearing black hats. And a last one, the first sage will speak only in the obvious case of WBB.
• BBW: the first sage can’t figure out anything so he passes. The second one sees B-W, so it’s either BBW or BWW, so he passes. The third sage sees BB-, so it’s obvious for him that he’s wearing a white hat.
• BWB: the first sage passes. The second one sees B-B so it’s not magical that he knows he’s wearing a white hat.
• BWW: the first sage again passes. The second one sees B-W, giving him two possible solutions, BWW or BBW, no solution for him. The third one though sees BW- giving him the alternatives BWW and BWB, but the second one passed, so it wasn’t BWB. So it was BWW.
• WBB: the first sage is lucky and knows he is wearing a white hat.
• WBW: the first sage can’t guess. The second sees W-W and again can’t guess. The third one sees WB- so it’s either WBW or WBB, in which case though the first one would have already guessed, so it’s not it, it’s WBW.
• WWB: the first sage can’t guess again. The second one sees W-B, giving him the alternatives WWB or WBB, in which case the first one would have spoken, so it’s WWB.
• WWW: the first one passes. The second one also passes. The third one sees WW-, so it’s WWW or WWB, in which case, as we saw the second sage would have spoken. So he assumes he’s wearing a white hat.
The trick here is not to start with the case of WWW, but to leave it last and better right after case WWB, so that it will be obvious.
Monday, April 18, 2005
So the sage's "unfair" pleas were not quite precise. better said: "The first man to guess who can see that everyone who has not guessed yet is wearing a black ball, will know he is wearing a white ball".*
In fact, their complaint (which seemed to suggest that the last one to guess would have an unfair advantage) was even less accurate: suppose the king, having heard their reasoning, decided to make them play anyway, and picked his favourite, Wisey, to go third. Assuming all the sages are intelligent and play to the odds, what are the chances of Wisey winning? I've just worked it out and it's less than you think. Let me know if you want the answer.
*To see this, look through the previous answer and check it, or simply follow this reasoning:
The first man will only guess if both the others are wearing black, in which case he must be wearing white.
The second man, if it gets to him, will know that he and the third are not both wearing black. So he will guess only if the third man is wearing black, in which case he must be wearing white.
The third man, if it gets to him, will therefore know he is not wearing black, and thus must be wearing white.
Wednesday, April 27, 2005
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