## Archive for May, 2017

### Intransitive dice V: we want a local central limit theorem

May 30, 2017

It has become clear that what we need in order to finish off one of the problems about intransitive dice is a suitable version of the local central limit theorem. Roughly speaking, we need a version that is two-dimensional — that is, concerning a random walk on $\mathbb Z^2$ — and completely explicit — that is, giving precise bounds for error terms so that we can be sure that they get small fast enough.

There is a recent paper that does this in the one-dimensional case, though it used an elementary argument, whereas I would prefer to use Fourier analysis. Here I’d like to begin the process of proving a two-dimensional result that is designed with our particular application in mind. If we are successful in doing that, then it would be natural to try to extract from the proof a more general statement, but that is not a priority just yet.

As people often do, I’ll begin with a heuristic argument, and then I’ll discuss how we might try to sharpen it up to the point where it gives us good bounds for the probabilities of individual points of $\mathbb Z^2$. Much of this post is cut and pasted from comments on the previous post, since it should be more convenient to have it in one place.
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### Intransitive dice IV: first problem more or less solved?

May 27, 2017

I hope, but am not yet sure, that this post is a counterexample to Betteridge’s law of headlines. To back up that hope, let me sketch an argument that has arisen from the discussion so far, which appears to get us close to showing that if $A,B$ and $C$ are three $n$-sided dice chosen independently at random from the sequences model (I will recap some of these definitions in a moment), then the probability that $A$ beats $C$ given that $A$ beats $B$ and $B$ beats $C$ is $1/2+o(1)$. Loosely speaking, the information that $A$ beats $B$ and $B$ beats $C$ tells you nothing about how likely $A$ is to beat $C$. Having given the argument, I will try to isolate a statement that looks plausible, not impossible to prove, and sufficient to finish off the argument.

### Basic definitions

An $n$sided die in the sequence model is a sequence $(a_1,\dots,a_n)$ of elements of $[n]=\{1,2,\dots,n\}$ such that $\sum_ia_i=n(n+1)/2$, or equivalently such that the average value of $a_i$ is $(n+1)/2$, which is of course the average value of a random element of $[n]$. A random $n$-sided die in this model is simply an $n$-sided die chosen uniformly at random from the set of all such dice.

Given $n$-sided dice $A=(a_1,\dots,a_n)$ and $B=(b_1,\dots,b_n)$, we say that $A$ beats $B$ if

$|\{(i,j):a_i>b_j\}|>|\{(i,j):a_i

If the two sets above have equal size, then we say that $A$ ties with $B$.
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### Intransitive dice III

May 19, 2017

I now feel more optimistic about the prospects for this project. I don’t know whether we’ll solve the problem, but I think there’s a chance. But it seems that there is after all enough appetite to make it an “official” Polymath project. Perhaps we could also have an understanding that the pace of the project will be a little slower than it has been for most other projects. I myself have various other mathematical projects on the boil, so can’t spend too much time on this one, but quite like the idea of giving it an occasional go when the mood takes me, and trying to make slow but steady progress. So I’ve created a polymath13 category, into which this post now fits. I’ve also retrospectively changed the category for the previous two posts. I don’t think we’ve got to the stage where a wiki will be particularly useful, but I don’t rule that out at some point in the future.

In this post I want to expand on part of the previous one, to try to understand better what would need to be true for the quasirandomness assertion to be true. I’ll repeat a few simple definitions and simple facts needed to make the post more self-contained.
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### 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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