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The coprime dice relabeling conjecture

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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.

References

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
Posted

Progress summary

Refreshed
Claimed solved

A posted construction claims the conjecture is false for infinitely many coprime dice pairs, but no independent verification has appeared.

The conjecture, extending the prime-sized case, says that coprime-sized dice cannot be relabeled while preserving every sum frequency. The broader relabeling problem was posed by Gallian and Rusin; Chao, Gabel, Larson, and Nasr studied it in 2024.

Known results

  • Distinct prime side counts admit no nontrivial relabeling (Chao, Gabel, Larson, and Nasr, 2024).
  • The 2024 paper reports only preliminary results for unequal side counts in general.
  • For side counts pp and pkp^k, there are kk sum-preserving relabelings, but these sizes are not coprime (except trivially).

Posted attempt

A construction claims a counterexample for every s≥5s\ge5 with gcd⁡(s,6)=1\gcd(s,6)=1: relabel the six-sided die as (1,2,2,3,3,4)(1,2,2,3,3,4) and the ss-sided die as (1,3,4,…,s,s+2)(1,3,4,\ldots,s,s+2). It claims identical sum frequencies, giving the smallest instance (r,s)=(6,5)(r,s)=(6,5) and infinitely many counterexamples. This complete counterexample claim has not been independently verified.

Current status (as of August 2026): The distinct-prime case is settled, while the conjecture for arbitrary relatively prime side counts has a claimed but unverified infinite family of counterexamples.

Sources

Solutions 1

CounterexampleThis solution needs a summarySee full solutionHide full solution

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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Shivam Patel Posted