The Three Hats

[Ghost Post]

Does anyone else find the statement of the puzzle and solution to The Three Hats to be overly simplified? By "overly simplified", I mean leaving out essential details. It's an old puzzle (and can be generalized to any number of hats, where any number >1 are blue), but as stated here it really has no solution.

First Edit:
Actually, on rereading it, I realize it is far more flawed than I had at first realized. Can anyone see why the "wisest sage" is really the dumbest? Yes, he got the answer right, but, despite the reasoning presented in the answer, he was taking a geuss, and was risking being beheaded.

[This message has been edited by extro... (edited 12-19-1999).]

[araya]

I suppose there are some ambiguities in the puzzle, but it makes pretty good sense overall. I'd say that anybody with half a brain, upon seeing a white hat and a blue hat on the other sage's heads, would figure out very quickly that his must be blue, assuming that the other sages are trying their best to win the contest. But when all three sages are faced with 2 blue hats, perhaps they wonder why the other men are not coming to the same conclusions as them, and start to doubt whether they are actually trying as hard as they can to figure out the puzzle.. maybe they just don't want to risk being beheaded. Indeed, a truly intelligent sage likely wouldn't put his life in the hands of a competitor.

[Ghost Post]

Here's the problem:

Sage A sees that sages B and C have blue hats, and considers whether he himself has a white hat or a blue hat, as follows:

If I (sage A) have a white hat, then B and C will each see one white hat, and one blue hat. Each of B and C will reason that if he himself has a white hat, the guy with the blue hat then sees two white hats, and knows then that he has a blue hat, and will say so. Each of B and C will deduce, from the fact that the other guy with the blue hat is remaining silent, that he himself does not have a white hat. So, if I (sage A) have a white hat, then after a short time - say T seconds - I'm not sure exactly how long - sages B and C will realize they have blue hats, and will say so.

So, all I need do is wait T seconds, and if B and C dont realize they have blue hats, then I must not have a white hat, but a blue hat.

But, if I (sage A) have a blue hat, then we all have blue hats, and sages B and C may be reasoning exactly as I have. Sage B, then, will be waiting T seconds, after which if neither I (sage A) nor sage C realize we have blue hats, he (sage B) will then know he has a blue hat. And similarly for sage C.

Now, if sage B and sage C have estimated T to be greater than I (sage A) have, then after T seconds (by my estimation of T) have elapsed, sages B and C will still be waiting, and will be silent whether I have a blue or white hat.

Therefore, to avoid having my head chopped off, I must make sure that my waiting time T is longer than that of sage B and C. So I'll increase it to T2.

But wait a minute. Sages B and C are not dummies. They too will want to set their waiting time to be the longer than the other two, so that they don't lose their heads.

How can I set my waiting time to be greater than that of sage B and sage C, when they are endeavoring to do the same?

[Andy]

Assume that all 3 sages are in fact loyal to the kingdom and will not wish to deprive the kingdom of the services of their fellows (at a lesser rank, of course).. Also, the puzzle states that the one sage who answered could see in his competitors' eyes that their thoughts were the same as his. A sage will speak up as soon as he has deduced the color of his own hat,rather than keep silent to induce his competitors to blunder.
A's waiting time T2 (for the 2 blue hats he sees) does not need to be longer than B's and C's T2 - just longer than their time T1 (if they saw only 1 blue hat). A doesn't need to wait until B and C conclude that there are 3 blue hats - only until they conclude that A and each other realize that there must be at least 2. In other words - if B sees one blue hat (on C) and one white hat (on A), he will not know immediately that his own hat is blue. He will, however, know fairly quickly that C doesn't see two white hats. Knowing this, he will speak up if he sees one white hat (on A) and one blue hat (on B). When he keeps silent, and C likewise keeps silent, and A keeps silent, each can conclude that neither of the others sees one blue hat and one white hat. The one who has the highest opinion of the others' intelligence will conclude soonest that his own hat must be blue. Note that this isn't necessarily the one who himself has the highest intelligence, demonstrating the fact that wisdom and intelligence are not identical.

[HyToFry]

I know this a little off the subject, and kinda stupid; BUT did anybody else notice that the king NEVER says the sages can't take their hats off and look at them? As for guessing goes, i thought about that to, but these are the three wisest sages in the kingdom, so there is NO WAY that the sages wouldn't have figured out the first "What if" part of the riddle, and so the third sage could be assured that he was not wearing a white hat, or one of the other sages would have jumped up and said "I must be wearing a blue hat or sage one would have noticed two white hats and jumped up"

Don't look at it as the sage discovering what hat he WAS wearing, but rather what hat he couldn't possibly be wearing.

[C.D.Wright]

Surely the most compelling "solution" is to assume
that the test is fair, therefore must be symmetrical,
and hence all hats must be the same colour?