Correlated Hats

Well, my thesis is essentially done, now its just time for the eternal editing of the damn thing (also, its only like 65 pages, somehow I had hoped for more). Anyway, time to post the solution the the hat problem from last time.

For the first part of the puzzle, consider that the logicians have some sort of strategy, which must be based on the hats of their fellow logicians and possibly some constant which varies for each individual logician and possibly some random variable. You get into the room ("you" being one of the logicians) and you check your strategy, and it says to say that your hat was white. OK, great, but since your strategy is independent of your own hat colour, your hat colour could randomize as we do this and your answer won't change. Thus you are 50% to be correct, no matter what. This is true for each and every logician. This means that no strategy can save more than 50% on average. In fact, no strategy can save less than 50% either. All strategies have an expectation value of saving 50 of the logicians.

You might think that is weird, since in so many other hat puzzles it has been that case that the logicians pull off something amazing with no information about their own hat colours. If you check carefully though, in all of those, when they were going to guess they were 50% chance to be correct, the most they ever pulled off was correlation with the other logicians being right.

Alright, now, is there a strategy that guarantees saving 50? Note that if the logicians knew if the total number of white hats was even, they could always guess their own hat, and similarly if they knew the total number of white hats was odd. So, the strategy is to have half of the logicians assume the total number of white hats is even, the other half assume that it is odd. Exactly one of those two groups will be right, and you will save 50 of them.

Finally, a strategy that saves all or none of them. This is simple given the last paragraph, simply have everybody assume the total number of white hats is even (or odd, if you prefer). They will either all be correct or all be wrong, with 50-50 chance.

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