Luke Pebody

The first person to work out the hat thing has the biggest number.
The other two immediately know he has the biggest number and that
theirs is simply the difference between the other two hats.
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Someone asked whether the other two players can always determine their numbers right after one does. Another person noted that the player with the largest number always finds it first. If this observation is true, then the other players know they have smaller numbers and know to subtract the other numbers to find their own.
 
Bryce Herdt
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Dear Ed,

It occurs to me that if the logicians always identify the greatest
number first, then the other two logicians will identify their
numbers immediately, because they know that the first number is
the greatest.

So it suffices to prove that the greatest number is always first
to be identified.  There's probably an induction proof. . . .

						--George Sicherman

Dear Ed,

I see now that the logicians can always eventually identify their
numbers.  It's a simple induction on the max number.  If enough
rounds have passed to identify any smaller max number, and they're
still saying "I don't know," then the logician with the greatest
number knows that his number is the sum of the other two.

I suspect that there's a formula for how many rounds it takes.
For example, if A has the max number M and C has the min number m,
it looks as if he needs 1 + [(M - 1) / 2m] rounds.  But I'm still
experimenting.

The case 8-3-5 is deeper than 5-2-3!

	A: could be 8 or 2. "I don't know."
	B: could be 13 or 3.  Either way A would say "I don't know."
	   "I don't know."
	C: could be 11 or 5.  Either way A and B would say "I don't
	   know." "I don't know."
	A: could be 8 or 2.  If it were 2, C would say "could be 5
	   or 1.  If it were 1, B would say `could be 3 or 1.  If it
	   were 1, A would say <could be 2 or 0, but can't be 0, so
	   must be 2.> But he didn't, so must be 3.' But he didn't,
	   so must be 5." But he didn't, so must be 8.

But this wouldn't work for the original puzzle.  A also answers in
the second round with 8-6-2.

Have you looked at the case of more than 3 logicians?

-- 
Col. G. L. Sicherman