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The coprime dice relabeling conjecture

From papers

Let rr and ss be relatively prime positive integers, and consider one die with rr sides and one die with ss sides. A relabeling is a reassignment of the labels on the two dice; it is frequency-preserving when it leaves unchanged the frequencies of all possible sums of the two dice.

Coprime dice relabeling conjecture. There are no ways to relabel a die of size rr and a die of size ss without changing the frequencies of their sums.

This extends the preceding proposition from prime-sized dice to relatively prime sizes. The supplied text reports it as a conjecture, and gives no evidence of a resolution.

Progress summary

Open

The extension from prime-sized dice to relatively prime sizes remains open, with no verified proof or counterexample found.

The conjecture asserts that two dice whose side counts share no common factor cannot be relabeled while preserving every sum frequency. A 2024 treatment records the broader relabeling question and leaves this relatively-prime extension unresolved.

Known results

  • Distinct prime side counts admit no such relabeling (Proposition 7.1, Chao, Gabel, Larson, and Nasr, 2024).

August 2024 paper

Chao, Gabel, Larson, and Nasr formulate the two-die relabeling problem and establish the distinct-prime case, but provide no proof or counterexample for all relatively prime side counts.

Current status (as of August 2026): The distinct-prime case is settled negatively, while the conjecture for arbitrary relatively prime side counts remains open and no verified resolution was found.

Sources
Sources & referencesView supporting material

Primary source

Yikai Chao, Josh Gabel, Carlye Larson and George David Nasr, “Revisiting Dice Relabeling using Cyclotomic Polynomials”, arXiv:2408.10331 (2024).

combinatoricscyclotomic-polynomialsdice-relabeling

Solutions 1

Counterexample

Write

[m]x=1+x+⋯+xm−1.[m]_x=1+x+\cdots+x^{m-1}.

For every integer s≥2s\ge2, consider the six-sided die with faces

(1,2,2,3,3,4)(1,2,2,3,3,4)

and the ss-sided die with faces

(1,3,4,…,s,s+2).(1,3,4,\ldots,s,s+2).

Their generating polynomials are

A(x)=x(1+2x+2x2+x3)=x(1+x)(1+x+x2)=x[6]xx2−x+1A(x) =x(1+2x+2x^2+x^3) =x(1+x)(1+x+x^2) =\frac{x[6]_x}{x^2-x+1}

and

Bs(x)=x(1+x2+x3+⋯+xs−1+xs+1)=x[s]x(x2−x+1).\begin{aligned} B_s(x) &=x(1+x^2+x^3+\cdots+x^{s-1}+x^{s+1})\\ &=x[s]_x(x^2-x+1). \end{aligned}

Indeed,

[s]x(1−x+x2)=1+∑j=2s−1xj+xs+1,[s]_x(1-x+x^2) =1+\sum_{j=2}^{s-1}x^j+x^{s+1},

with an empty middle sum when s=2s=2. In particular, all coefficients are nonnegative and

A(1)=6,Bs(1)=s.A(1)=6,\qquad B_s(1)=s.

Their product is

A(x)Bs(x)=x2[6]x[s]x,A(x)B_s(x)=x^2[6]_x[s]_x,

exactly the generating polynomial for the sums of standard six- and ss-sided dice.

Whenever s≥5s\ge5 and gcd⁡(s,6)=1\gcd(s,6)=1, these are nonstandard relabelings of two coprime-sized dice. Thus the conjecture fails for infinitely many pairs (6,s)(6,s).

For the smallest instance, s=5s=5, the relabeled dice are

(1,2,2,3,3,4)and(1,3,4,5,7).(1,2,2,3,3,4) \quad\text{and}\quad (1,3,4,5,7).

Their sum frequencies for totals 2,…,112,\ldots,11 are

(1,2,3,4,5,5,4,3,2,1),(1,2,3,4,5,5,4,3,2,1),

identical to those of standard six- and five-sided dice.

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