Archive for May 12th, 2017

Intransitive dice II

May 12, 2017

I’m not getting the feeling that this intransitive-dice problem is taking off as a Polymath project. However, I myself like the problem enough to want to think about it some more. So here’s a post with some observations and with a few suggested subproblems that shouldn’t be hard to solve and that should shed light on the main problem. If the rate of comments by people other than me doesn’t pick up, then I think I’ll simply conclude that there wasn’t sufficient interest to run the project. However, if I do that, I have a back-up plan, which is to switch to a more traditional collaboration — that is, done privately with a small number of people. The one non-traditional aspect of it will be that the people who join the collaboration will select themselves by emailing me and asking to be part of it. And if the problem gets solved, it will be a normal multi-author paper. (There’s potentially a small problem if someone asks to join in with the collaboration and then contributes very little to it, but we can try to work out some sort of “deal” in advance.)

But I haven’t got to that point yet: let me see whether a second public post generates any more reaction.

I’ll start by collecting a few thoughts that have already been made in comments. And I’ll start that with some definitions. First of all, I’m going to change the definition of a die. This is because it probably makes sense to try to prove rigorous results for the simplest model for which they are true, and random multisets are a little bit frightening. But I am told that experiments suggest that the conjectured phenomenon occurs for the following model as well. We define an $n$-sided die to be a sequence $A=(a_1,\dots,a_n)$ of integers between 1 and $n$ such that $\sum_ia_i=n(n+1)/2$. A random $n$-sided die is just one of those chosen uniformly from the set of all of them. We say that $A$ beats $B$ if $\sum_{i,j}\mathbf 1_{[a_i>b_j]}>\sum_{i,j}\mathbf 1_{[a_i
That is, $A$ beats $B$ if the probability, when you roll the two dice, that $A$ shows a higher number than $B$ is greater than the probability that $B$ shows a higher number than $A$. If the two probabilities are equal then we say that $A$ ties with $B$.
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