ThunderCloud

We know that A sees B=334 and c=334 - A CANNOT see B=335 and C=334. (Not to confound with: "As this information in not available to him, he ASSUMES he has 334 or 335.") So my first answer was correct. In the 2nd try, I got lost and could not correct quickly enough.

I think you are on the right track, and very close.

No. The logicians cannot agree upon a strategy in advance. However, you may assume each of them to be "perfect" in that they will deduce all that they logically can. You may further assume that each logician will assume the others to behave this way as well (i.e., that it is generally known that all three logicians are "perfect").

I know a better and harder version, if I find it, I'll post it. Does it matter if it is in French? I am not sure I follow your reasoning here -- care to elaborate further?

Still unanswered... Can you find the complete sequence of responses? It is considerably shorter than one might initially think...

I might have missed the gist of this question the first time around... Another way to express 1/13 is 0.076923076923076923... It is clumsy compared with the bar method in this example (unlike .999...), but still understandable.

______ 1/13 = 0.076923 The bar covers the part that repeats. This can be done, but would be rather tedious (particularly as the decimal expansion for 45/89 has a relatively long repeating digit string). The sum will also have a repeating pattern, of some size less than or equal to the LCM of the repeating digit string lengths of the two operands.

The puzzle states: "There is a spy who informs the PO of the strategy. So in placing the hats, the PO does his best to foil it." I interpreted this to mean the PO placed the hats himself, as diabolically as he could. I follow you. And the OP, again, to be fair to you, does not preclude your comment. The difference I drew was between knowing the strategy in general terms and being privy to the details of its implementation. And that is a finer distinction than the OP makes. BTW, This may distinguish between Phil's and Rainman's (post 11) approaches. I don't think your PO tactic can defeat Rainman's solution. Or can it? The Evil PO would be out of luck with Rainman's approach, so far as I can tell.

The puzzle states: "There is a spy who informs the PO of the strategy. So in placing the hats, the PO does his best to foil it." I interpreted this to mean the PO placed the hats himself, as diabolically as he could.

I think the environmental officials would be alright with that approach. Well done!

Actually, that conniving PO could foil this plan too, saving only 48...

With a tip of the hat to wolfgang... Lake Champlain has long been surrounded by rumors of a mysterious creature akin to the Loch Ness Monster. The locals call him "Champ," and he is officially a protected species in both Vermont and New York State. And after so many decades of stories and speculation, Champ has been spotted!! Scientists, eager to learn as much as possible about the potentially prehistoric lake monster before he disappears again, have quickly assembled all of the maritime and scientific equipment that they might possibly need. However, owing to Champ's thoroughly protected status, officials have mandated that no experiment shall be conducted which might pose an immediate risk to Champ's life or health; in particular, they stipulate that Champ can not be removed from the lake at any time, even momentarily (nor can the ecosystem of the lake be impacted such as by draining the lake). Can you find a method to determine Champ's inherent buoyancy? Can you find a method to determine how much Champ would weigh if he were allowed to be hoisted out of the water and placed on a scale?

I may be missing something obvious, but why is the expected time to see "willywilly" not 26^10?

Yes. Both sides of the relations specify real numbers, which are well-ordered. So, exactly one of the relations must be true.