Instead of looking at a die in its sequence representation (a_1, a_2, … a_n), consider it in the multiset representation (m_1, m_2, .. m_n) where m_i is the number of dice face with value i.

Then the score function is just an antisymmetric bilinear function, and we can use linear algebra.

With the matrix:

S_{ij} = sign(j-i)

and denoting the column vector v(A) as the vector with components equal to the multiset representation of the die A,

the score function is then:

score(A,B) = v(A)^T . S . v(B)

In this representation, the standard die is a vector with all entries 1.

v(std) = [1,1,1,…,1]

Since the multiset representation already restricts the values of die face to the set of integers 1 to n, the only remaining constraint is the sum. This sum constraint is equivalent to saying:

score(std,A) = v(std)^T . S . v(A) = 0

As we are only considering multisets which meets that constraint, there is a more convenient representation, where we only consider the difference to the std die. Denoting this representation as w(A),

w(A) = v(A) – v(std)

the score function retains the same form:

score(A,B) = w(A)^T . S . w(A)

and it is obvious that the standard die ties with everything because it is just the zero vector in this representation.

If we constraint to the dice with multiplicity of values <= 2, now it becomes obvious what the "negative" relation is and why this restriction allows it.

w(-A) = – w(A)

And since the score function is bilinear, we immediately get:

score(-A,B) = – score(A,B)

The reason this doesn't work more generally, is that it doesn't make sense to have a negative number of faces with some number j.

The "one step" dice considered in the arxiv paper Gowers linked when starting this discussion, are a "basis" for the space of all dice in that these steps can be added as vectors in the w(A) representation to get to all other dice. Exploring what the score function on the space of all dice pairs looks like, using this basis as a guide, may be fruitful because the scores has already been discussed for this basis in the arxiv paper.

The "max2 multiset dice" is the maximal expansion about the origin (the standard die) with the "one step" basis, while still allowing all dice to have a natural "negative" die. I did some computational tests and this is enough to make the fraction of ties go to zero as n increases, and yet the larger "random tournament" conjecture still appears to fail. Strangely, it even appears to be failing with the same ratios for k=4 (I did not heavily test this, and at this point still likely to be a coincidence).

So this appears to be a nice small model that has all the same features as the full larger model.

]]>Consider sorted sequence proper dice, where no number is repeated more than twice. This model of dice have a nice concept of a “negative” die, an involution such that the score(A,B), which is the number of roll pairs where A beats B – pairs where B beats A, has the nice property:

score(A,B) = – score(A,-B)

So you automatically get that for all die A, it beats exactly as many dice and it loses to. The distribution is perfectly symmetric.

Now, this says nothing about the number of ties, but with your approach would this be enough to conclude the following little conjecture?

For dice (in this restricted model) A,B,C such that A > B > C, (where > is the “beats” relation), is it just as probable that A > C as C > A.

To what extent does this approach rely on that?

]]>For simplicity, let us talk about ties conjecture, for which it suffices to prove that converges to $0$.

Let , and let be an estimate of obtained by integrating density of standard normal distribution around point . Then converges to , so all we need is to control the ratio .

Now, substituting x=(0,0) into the formula (2.5) on page 25 in the book, we get , while (again according to the book) is of order $c/\sqrt{n}$. This seemed to clearly imply that converges to $1$, and we would be done. The proof that converges to 1/2 is similar.

However, while the constant in Berry–Esseen theorem was universal (0.4748 works), the constant c in the formula (2.5) in the book seems to depend on our initial random vector (X,Y). In our case, the vector is not fixed but different for every n, so c depends on n, which makes the inequality meaningless.

So, we need a version of Local Central Limit Theorem, as in the book, but with all constants universal, and the dependence of the initial random vector (X,Y), if any, should be explicit (in terms of moments, etc.) It seems not easy to extract this from the proof in the book.

]]>It seems to me that there will probably be some kind of technical lemma needed to argue that the probability that the entire random walk lives in a sublattice tends to zero. But probably this too follows reasonably easily from the fact that for every .

]]>However, in fact , where are frequencies for each value, and implies , hence . So, .

On the other hand, most of proof strategies we had for the dice problem, if worked, could also be used to prove that converges to 0, while converges to 1/2…

]]>The theorem in the paper measures closeness to a Gaussian as the maximum discrepancy between the Gaussian probability of being a convex set and the probability of the given distribution being in that set. However, the convex set that we want to restrict to will be a one-dimensional set, and therefore very small.

The kind of problem this could cause is illustrated by the following simple example. Let be a random variable that has mean zero and is supported on a finite subset of . Let and be independent copies of . Then a sum of independent copies of will approximate a normal distribution with mean and a certain variance.

Now let’s suppose that is non-zero. Then the normal distribution we converge to will assign non-zero probabilities to every point in .

But if instead we start with a different random variable with the same mean and variance as but supported on the even numbers, then the normal distribution we converge to will assign zero probability to numbers with an odd coordinate and will make up for it by assigning more weight to points in .

This is no problem when it comes to the probability of landing in a convex set, but it is more worrying when we condition on one of the coordinates being zero, which will lead to the probabilities being doubled, in a certain sense.

In our case, we should be OK, because the variable whose value we are conditioning on will not be supported in a residue class, but I think it means we may have to delve into the proof of the multidimensional Berry-Esseen theorem rather than just quoting it — unless of course someone else has proved a version that is better adapted to our purpose.

]]>This mathoverflow question points us to a multidimensional Berry-Esseen theorem. In particular, it looks as though Theorem 1.1 in this paper is probably what we want. Does it look strong enough to you? (I could probably work this out, but I think you’ll be able to do so a lot more quickly.)

]]>One thing it gives us quite easily is a way of proving statements about the function . For convenience I’ll take a set of size as my basic object. Then if we list the elements of as , the elements of the multiset are (with multiplicity equal to the number of times they appear in this list). I’ll call this multiset and I’ll write for the number of elements of , with multiplicity, that are at most .

It doesn’t seem to be wholly pleasant to define directly in terms of the set : it is the maximal such that . However, we do at least have that the “inverse” is a nice function. That is, we know that the th largest face of the die is . And that is quite helpful. For instance, if instead of choosing a random that adds up to 1 we instead choose a random subset of where each element is chosen independently with probability , then we would normally expect to be around , and not to differ from by much more than . So we get that for this completely random model that the th largest face of the die typically has value close to , and is well concentrated. If we now condition on the size of the set being and the sum of the faces of the die being , this will not change too much. And now we can just invert and say the same about .

We also have that beats if the number of such that is greater than the number of such that . This may turn out to be reasonably easy to work with — I’m not sure.

]]>Small conjecture: if beats and beats then the probability that beats is approximately 1/2.

Big conjecture: the tournament obtained by the relation “beats” is quasirandom.

We now know that the small conjecture is equivalent to the statement that almost every die beats approximately half the other dice, and the big conjecture is equivalent to the statement that for almost every *pair* of dice, the remaining dice are split into four approximately equal classes according to the four possibilities for which of the two they beat.

I’ve only really been thinking about the sequences model, but I think the experimental evidence is suggesting that the same should be true for the multisets model as well.

Also, I think we may already essentially have a proof of the small conjecture for the sequences model, but I’m waiting to see whether Bogdan Grechuk agrees with me about this.

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