## Two infinities that are surprisingly equal

September 19, 2017

It has been in the news recently — or rather, the small corner of the news that is of particular interest to mathematicians — that Maryanthe Malliaris and Saharon Shelah recently had an unexpected breakthrough when they stumbled on a proof that two infinities were equal that had been conjectured, and widely believed, to be distinct. Or rather, since both were strictly between the cardinality of the natural numbers and the cardinality of the reals, they were widely believed to be distinct in some models of set theory where the continuum hypothesis fails.

A couple of days ago, John Baez was sufficiently irritated by a Quanta article on this development that he wrote a post on Google Plus in which he did a much better job of explaining what was going on. As a result of reading that, and following and participating in the ensuing discussion, I have got interested in the problem. In particular, as a complete non-expert, I am struck that a problem that looks purely combinatorial (though infinitary) should, according to Quanta, have a solution that involves highly non-trivial arguments in proof theory and model theory. It makes me wonder, again as a complete non-expert so probably very naively, whether there is a simpler purely combinatorial argument that the set theorists missed because they believed too strongly that the two infinities were different.

I certainly haven’t found such an argument, but I thought it might be worth at least setting out the problem, in case it appeals to anyone, and giving a few preliminary thoughts about it. I’m not expecting much from this, but if there’s a small chance that it leads to a fruitful mathematical discussion, then it’s worth doing. As I said above, I am indebted to John Baez and to several commenters on his post for being able to write much of what I write in this post, as can easily be checked if you read that discussion as well.
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## Intransitive dice VII — aiming for further results

August 12, 2017

While Polymath13 has (barring a mistake that we have not noticed) led to an interesting and clearly publishable result, there are some obvious follow-up questions that we would be wrong not to try to answer before finishing the project, especially as some of them seem to be either essentially solved or promisingly close to a solution. The ones I myself have focused on are the following.

1. Is it true that if two random elements $A$ and $B$ of $[n]^n$ are chosen, then $A$ beats $B$ with very high probability if it has a sum that is significantly larger? (Here “significantly larger” should mean larger by $f(n)$ for some function $f(n)=o(n^{3/2})$ — note that the standard deviation of the sum has order $n^{3/2}$, so the idea is that this condition should be satisfied one way or the other with probability $1-o(1)$).
2. Is it true that the stronger conjecture, which is equivalent (given what we now know) to the statement that for almost all pairs $(A,B)$ of random dice, the event that $A$ beats a random die $C$ has almost no correlation with the event that $B$ beats $C$, is false?
3. Can the proof of the result obtained so far be modified to show a similar result for the multisets model?

The status of these three questions, as I see it, is that the first is basically solved — I shall try to justify this claim later in the post, for the second there is a promising approach that will I think lead to a solution — again I shall try to back up this assertion, and while the third feels as though it shouldn’t be impossibly difficult, we have so far made very little progress on it, apart from experimental evidence that suggests that all the results should be similar to those for the balanced sequences model. [Added after finishing the post: I may possibly have made significant progress on the third question as a result of writing this post, but I haven’t checked carefully.]
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## Another journal flips

July 27, 2017

There is widespread (even if not universal) agreement that something is deeply wrong with the current system of academic publishing. The basic point, which has been made innumerable times by innumerable people, is that the really hard parts — the writing of papers, and the peer review and selection of the ones to publish — are done voluntarily by academics, and modern technology makes things like typesetting and dissemination extremely cheap. And yet publishers are making more money than ever before. They do this by insisting that we give them ownership of the content we produce (though many journals will publish papers even if you strike out the part of the contract that hands them this ownership — these days I never agree to give copyright to a publisher, and I urge you not to either), and by bundling their journals together so that libraries are forced into an all-or-nothing decision. (As another aside, I also urge libraries to look closely at what is happening in Germany, where they have gone for the “nothing” option with Elsevier and the world has not come to an end.)

What can be done about this? There are many actions, none of which are likely to be sufficient to bring about major change on their own, but which in combination will help to get us to a tipping point. In no particular order, here are some of them.

1. Create new journals that operate much more cheaply and wait for them to become established.
2. Persuade libraries not to agree to Big Deals with the big publishers.
3. Refuse to publish with, write for, or edit for, the big publishers.
4. Make sure all your work is freely available online.
5. Encourage journals that are supporting the big publishers to leave those publishers and set up in a cheaper and fairer way.

Not all of these are easy things to do, but I’m delighted to report that a small group I belong to, set up by Mark Wilson, has, after approaching a large number of maths journals, found one that was ready to “flip”: the Journal of Algebraic Combinatorics has just announced that it will be leaving Springer. Or if you want to be more pedantic about it, a new journal will be starting, called Algebraic Combinatorics and published by The Mersenne Centre for Open Scientific Publishing, and almost all the editors of the Journal of Algebraic Combinatorics will resign from that journal and become editors of the new one, which will adhere to Fair Open Access Principles.

If you want to see change, then you should from now on regard Algebraic Combinatorics as the true continuation of the Journal of Algebraic Combinatorics, and the Journal of Algebraic Combinatorics as a zombie journal that happens to have a name that coincides with a former real journal. And of course, that means that if you are an algebraic combinatorialist with a paper that would have been suitable for the Journal of Algebraic Combinatorics, you should understand that the reputation of the Journal of Algebraic Combinatorics is being transferred, along with the editorial board, to Algebraic Combinatorics, and you should therefore submit it to Algebraic Combinatorics. This has worked with previous flips: the zombie journal rarely thrives afterwards and in some notable cases has ceased to publish after a couple of years or so.
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## Intransitive dice VI: sketch proof of the main conjecture for the balanced-sequences model

July 25, 2017

I have now completed a draft of a write-up of a proof of the following statement. Recall that a random $n$-sided die (in the balanced-sequences model) is a sequence of length $n$ of integers between 1 and $n$ that add up to $n(n+1)/2$, chosen uniformly from all such sequences. A die $(a_1,\dots,a_n)$ beats a die $(b_1,\dots,b_n)$ if the number of pairs $(i,j)$ such that $a_i>b_j$ exceeds the number of pairs $(i,j)$ such that $a_i. If the two numbers are the same, we say that $A$ ties with $B$.

Theorem. Let $A,B$ and $C$ be random $n$-sided dice. Then the probability that $A$ beats $C$ given that $A$ beats $B$ and $B$ beats $C$ is $\frac 12+o(1)$.

In this post I want to give a fairly detailed sketch of the proof, which will I hope make it clearer what is going on in the write-up.
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## 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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## A potential new Polymath project: intransitive dice

April 28, 2017

A few days ago I received an email from Brian Conrey, who has a suggestion for a possible Polymath project. The problem he wants to see solved is a little different from the problems in most previous projects, in that it is not a well known and long-standing question of acknowledged importance. However, that is not an argument against trying it out, since it is still far from clear what kinds of problems are best suited to the polymathematical approach, and it would be good to get more data. And this problem has other qualities that could make it very suitable indeed. First of all, it is quite new — it arises from a paper published last year, though it appeared on arXiv in 2013 — so we do not yet have a clear idea of how difficult it is, which should give us hope that it may turn out to be doable. Secondly, and more importantly, it is a very attractive question.

Suppose you have a pair of dice $D_1,D_2$ with different numbers painted on their sides. Let us say that $D_1$ beats $D_2$ if, thinking of them as random variables, the probability that $D_1>D_2$ is greater than the probability that $D_2>D_1$. (Here, the rolls are of course independent, and each face on each die comes up with equal probability.) It is a famous fact in elementary probability that this relation is not transitive. That is, you can have three dice $D_1,D_2,D_3$ such that $D_1$ beats $D_2$, $D_2$ beats $D_3$, and $D_3$ beats $D_1$.

Brian Conrey, James Gabbard, Katie Grant, Andrew Liu and Kent E. Morrison became curious about this phenomenon and asked the kind of question that comes naturally to an experienced mathematician: to what extent is intransitivity “abnormal”? The way they made the question precise is also one that comes naturally to an experienced mathematician: they looked at $n$-sided dice for large $n$ and asked about limiting probabilities. (To give another example where one might do something like this, suppose one asked “How hard is Sudoku?” Well, any Sudoku puzzle can be solved in constant time by brute force, but if one generalizes the question to arbitrarily large Sudoku boards, then one can prove that the puzzle is NP-hard to solve, which gives a genuine insight into the usual situation with a $9\times 9$ board.)
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## Timothy Chow starts Polymath12

February 24, 2017

This is a quick post to draw attention to the fact that a new and very interesting looking polymath project has just started, led by Timothy Chow. He is running it over at the Polymath blog.

The problem it will tackle is Rota’s basis conjecture, which is the following statement.

Conjecture. For each $i$ let $B_i=\{e_{i1},\dots,e_{in}\}$ be a basis of an $n$-dimensional vector space $V$. Then there are $n$ disjoint bases of $V$, each containing one element from each $B_i$.

Equivalently, if you have an $n\times n$ matrix where each row is a basis, then you can permute the entries of the rows so that each column is also a basis.

This is one of those annoying problems that comes into the how-can-that-not-be-known category. Timothy Chow has a lot of interesting thoughts to get the project going, as well as explanations of why he thinks the time might be ripe for a solution.