English

Winning Probabilities of Balanced and Nontransitive n-tuples of Dice

Combinatorics 2025-05-29 v1

Abstract

For a positive integer nn, an nn-tuple of dice (A1,A2,,An)(A_1,A_2,\dots,A_n) is called balanced if P(A1<A2)=P(A2<A3)==P(An<A1)P(A_1<A_2) = P(A_2<A_3) = \cdots = P(A_n<A_1) and nontransitive if P(A1<A2),P(A2<A3),,P(An<A1)P(A_1<A_2), P(A_2<A_3), \dots, P(A_n<A_1) are each greater than 12\frac{1}{2}. For a balanced and nontransitive nn-tuple of dice (A1,A2,,An)(A_1,A_2,\dots,A_n), we define the winning probability w(A1,A2,,An):=P(A1<A2)w(A_1,A_2,\dots,A_n) := P(A_1 < A_2). The works of Trybula and Kim et al. together show that for a balanced and nontransitve triple of dice (A1,A2,A3)(A_1,A_2,A_3), the least upper bound on the winning probability is 1+52\frac{-1+\sqrt{5}}{2}. Kim et al. then asked what the least upper bound on the winning probability was for the n4n \geq 4 cases. Bogdanov and Komisarski independently have shown that for n3n\geq 3 and a balanced and nontransitive nn-tuple of dice (A1,A2,,An)(A_1,A_2,\dots,A_n), the winning probability is less than πn:=114cos2(πn+2)\pi_n := 1-\frac{1}{4\cos^2\left( \frac{\pi}{n+2} \right)}. In this paper, we will show that for n3n \geq 3 and every rational p(12,πn]p \in \left( \frac{1}{2}, \pi_n \right], there exists a balanced and nontransitive nn-tuple of dice with winning probability pp. Paired with Bogdanov and Komisarski's results, this fully answers the problem posed by Kim et al. and establishes a complete characterization of the winning probabilities for nontransitive and balanced nn-tuples of dice.

Keywords

Cite

@article{arxiv.2505.21950,
  title  = {Winning Probabilities of Balanced and Nontransitive n-tuples of Dice},
  author = {Joshua Rooney},
  journal= {arXiv preprint arXiv:2505.21950},
  year   = {2025}
}