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Suppose I toss $n$ coins. It's natural to model this probabilistically in terms of a sample space $\{ H, T \}^n$ constructed as the product of $n$ copies of the sample space of possible outcomes of a …

Equivalently, we are considering a random function $f : [n] \to [n]$ where $[n] = \{ 1, 2, \dots n \}$ is a finite set of size $n$, which assigns to each prank cigarette a pack. The second question is …

I learned from Gian-Carlo Rota (Combinatorial snapshots) the following probabilistic motivation for looking at the Riemann zeta function: "subject to technical assumptions," the only probability measu …

I've been told that in machine learning it's common to compute the singular value decomposition of matrices in order to throw out all information in the matrix except that corresponding to, say, the $ …

I wrote a blog post on this here. The basic result is that for fixed $k$ and $n$, as $q \to \infty$ the joint distribution of irreducible factors of degrees $1$ through $k$ of a random monic polynomia …

People are actively working on this, although maybe not many people. See, for example, Uniform coherence by Garrabrant, Fallenstein, Demski, and Soares, and the references therein. The abstract: W …

No. Because $c_n$ is quadratic, the values of $m_0, m_1, m_2$ can be arbitrary (subject to $m_0 = 1$ since presumably we are looking at a probability distribution). The first condition that needs to b …

Here are examples showing that unlike in the previous problem, here it does not suffice to simply use the fact that the harmonic series / the sum of the reciprocals of the primes diverges. In fact for …

No. Let's restrict our attention to $\mathbb{N}$. The hypotheses imply that if $q$ is a prime, then the probability that a random positive integer is not divisible by $q$ is $1 - \frac{1}{q}$. They al …

it's possible to extend the analogy to the factorization of polynomials over finite fields $\mathbb{F}_q$ (see this blog post for details; one needs to take $q \to \infty$ for the statistics to match, …

$2^{\mathbb{R}}$, being a product of compact Hausdorff groups, is a compact Hausdorff group, so it has a normalized Haar measure ("flipping uncountably many coins").

You can make sense of the uniform probability distribution on lots of infinite sets, notably any compact topological group $G$, where "uniform probability distribution" should mean "normalized Haar me …

Here is a result that gives the flavor of the kind of thing along these lines I hope to see in the future. Recall Tarski's undefinability of truth: under suitable assumptions, a formal system can't be …

Rota and Shen's On the Combinatorics of Cumulants ends with a conjecture which I'll restate as follows: Let $p \in \mathbb{R}[x_1, x_2, ...]$ be a polynomial such that, for any sequence $X_1, X_2 …

This may not be quite what you had in mind, but: suppose you were trying to compute the absolute value of a Gauss sum $\sum_{a=0}^{p-1} \zeta^{a^2}$ where $\zeta = e^{ \frac{2 \pi i}{p} }, p$ an odd p …