---
title: "Trefoil Probabilities and Polyhedra"
authors:
  - "Joseph Geisz"
  - "Chris Peterson"
  - "Clayton Shonkwiler"
year: 2026
venue: "arXiv:math.GT"
arxiv: "2609.24860"
doi: null
source: "https://arxiv.org/abs/2609.24860"
topics:
  - "curiosity"
  - "knot-theory"
  - "geometric-probability"
  - "polyhedra"
seed_rank: 1849
seed_batch: "curiosity-2026-10-07"
reviewed: "2026-10-07"
pool: "maths-foundations"
relevance_score: 9
lineage: random-knots
cites:
  - title: "Trefoil Probabilities and Polyhedra"
    url: "https://arxiv.org/abs/2609.24860"
    year: 2026
    arxiv: "2609.24860"
    doi: null
---

# Trefoil Probabilities and Polyhedra

## One-sentence takeaway

Join six uniformly random points on the unit sphere into a closed hexagon: it is knotted (necessarily a trefoil) with probability exactly 1/(16 pi^2).

## Why it's lovely

Why you might love this: it is a Jane Street puzzle with a real answer. The proof turns a knot question into polyhedral combinatorics: the convex hull of six random points is almost surely one of two octahedra; the combinatorially regular one admits exactly one trefoil Hamiltonian cycle, the irregular one admits none. Stereographic projection then turns "probability of the regular hull" into a Sylvester-type random-triangle problem in the plane, solved with Stokes' theorem and Blaschke–Petkantschin. Answer chain: regular octahedron with probability 15/(4 pi^2), trefoil with probability 1/(16 pi^2).

## Key ideas

- Random closed hexagon from six i.i.d. uniform points on the sphere; the only possible nontrivial knot is the trefoil.
- Hull types: combinatorially regular vs non-regular octahedron (almost surely one of the two).
- Regular type: exactly one unoriented trefoil Hamiltonian cycle on the straight-line complete graph; irregular type: none.
- Stereographic projection to a Sylvester-type problem for the planar beta-prime measure dx dy / (pi (1 + x^2 + y^2)^2).
- Stokes plus Blaschke–Petkantschin give the expected squared mu-content of a random triangle, hence 15/(4 pi^2) and 1/(16 pi^2).

## Caveats

Specific to hexagons with vertices on the sphere; other models (e.g. uniform random equilateral polygons) and longer polygons are not covered by the exact formula. Twelve-page preprint, recent.

## Links

- arXiv abs: https://arxiv.org/abs/2609.24860
- PDF: https://arxiv.org/pdf/2609.24860