Solutions to some Hat Problem AND some points of interest.

In my last blog here I asked three (known) hat problems since they may be new to you (one of them I just learned last week) and I had a point to make about them. I have WRITTEN UP the proofs  here since html is clumsy with math (or I'm clumsy with html-math), so this post is mostly about the points to make about these problems. I would urge you to read the writeup pointed to before reading the post.

1) N people 1,...,N, two colors R,B, Hats put on RANDOMLY (no adversary).

People are in a line and pe sees person j's hat iff i ≤ j .

There is a well known strategy where nobody passes which guarantees n-1 get it right (see here), but that strategy has EVERYONE get it right 1/2 of the time. We want MORE than that. LOTS more.

The following strategy works: For i=1,2,..., N person i does the following: if nobody has said RED yet AND ALL of the hats i sees are BLUE then i says RED. Otherwise Red passes

This fails on B^n. It works on everything else with the last R getting it right and everyone else passing. So the prob of getting it right is 1- 1/2^n.

POINT: I originally didn't have one to make, but a commenter misread the problem (or I miswrote it) in an interesting way. My problem was: Hats put on randomly, players are deterministic. They thought it was Hats put on by an adversary but players can use a randomized strategy. That problem (which frankly is more intersting) has a similar solution to the above: the players get a random string of R,B of length n and treat that like I treat B^n above.

2) omega people: 1,2,3,... and as above. We want to get all but a finite number of people get it right. See my writeup of it pointed to above. The proof I use uses the Axiom of choice and this is needed (see here).

POINT: some of my students didn't like that the players need uncountable memory. How much does this bother me: not even a little. A fellow blogger thought this result was so non-intuitive that he now thinks the axiom of choice is wrong (see here) Personally I am a lot more bothered by the Banach Tarski Paradox (see here), though that paradox has lead to what my wife calls either the best or the most obscure math joke ever: what is an anagram of Banach-Tarski? Answer: Banach-Tarski Banach-Tarski.

3) omega people: 1,2,3,... and as above but now we want to get at most ONE wrong. You CAN do this! see the writeup.

POINT: When I first learned problem (2) I assumed you could not get it down to a finite bound. And I was sure I could prove it, though I never got around to it, prob because I thought it was true and easy. Well, my turn to eat humble pie (an expression only said on TV and not in real live)--- you CAN do this with only one error.  The problem where you have an infinite number of people, they all see each others hats, and they all shout at the same time- that one I am sure you can't do with at most 1 error. I might need to eat humble pie once again.

4) n people, c colors, everyrone sees everyone else's hat, simul shouting, deterministic, and want to maximize how many get it right. OH- and adversarial.

Can do it with floor(n/c) but can't to better. See writeup.

POINT: The argument that you can't do better is a probabilistic argument! That's great! It may help bridge the gap between recreational and serious math (is there even a gap anymore?) that we use a Prob method on a fun hat problem!