Loaded dice with a uniformly distributed sum

Two dice with integer labels are loaded, perhaps unevenly, and rolled together. If every possible total turns out to be equally likely, must each die be fair on the faces it can show? In algebraic form: if two polynomials with nonnegative coefficients multiply to a polynomial whose coefficients are all 0 or 1, must their own coefficients be 0 or 1 (after normalising)? The answer is yes when the totals form an interval, and every computer search so far agrees, but no proof is known in general.

Problem

Let $X$ and $Y$ be independent random variables, each taking finitely many integer values. Suppose that $X + Y$ is uniformly distributed on its support $S$ , that is, $\mathbb{P}(X + Y = s) = 1/|S|$ for every $s \in S$ . Must $X$ be uniformly distributed on its own support, and likewise $Y$ ?

Polynomial form (the unfair 0–1 polynomial conjecture). Let $P, Q \in \mathbb{R}[x]$ be monic polynomials with nonnegative coefficients. If every coefficient of $PQ$ lies in $\{0, 1\}$ , must every coefficient of $P$ and of $Q$ lie in $\{0, 1\}$ ?

The two forms are equivalent: after shifting $X$ and $Y$ to take nonnegative values, their generating functions $G_X(x) = \sum_n \mathbb{P}(X = n)\, x^n$ and $G_Y$ satisfy $G_X G_Y = G_{X+Y} = |S|^{-1} \sum_{s \in S} x^s$ , and dividing $G_X$ and $G_Y$ by their leading coefficients gives monic $P, Q$ with $PQ = \sum_{s \in S} x^s$ . Uniformity of $X$ means exactly that $P$ has 0–1 coefficients.

Background

Green records the question as asked by Emmanuel Amiot on Mathematics Stack Exchange in 2019, after which it became a popular MathOverflow question. Amiot came to it from tilings of the integers, where linear programming over the reals kept returning 0–1 factorisations. Ghidelli traces the general question back to G. Letac around 1969. A classical special case is the puzzle that two six-sided dice cannot be loaded so that all totals from 2 to 12 are equally likely.

What is known

Krasner and Ranulac (1937) proved the statement when the support of $X + Y$ is an interval $\{0, 1, \dots, N-1\}$ , and their argument extends to palindromic (self-reciprocal) 0–1 polynomials (as surveyed by Ghidelli, 2022).
Ghidelli (2022) settled the first family not covered by classical methods: one die with faces 0, 2 and 5, that is, a factor $x^5 + a x^2 + 1$ .
Dvorsky (2026) showed that no counterexample has a factor $1 + a x^2 + x^k$ with $0 < a < 1$ and odd $k \ge 341$ ; smaller odd $k$ are treated in a companion paper.
Hare (2023, Experimental Mathematics 2025) gave an algorithm that ruled out all but 975 of 7,141,686 candidate factor patterns of degree at most 15.
Exhaustive verification for products of degree at most 26 (Alekseyev), 32 (Mueller, with Gröbner bases) and, in 2026, 66 (Zhang, with Positivstellensatz certificates) has been reported on MathOverflow; no counterexample has been found.

What would count as progress

A counterexample: explicit monic $P$ , $Q$ with nonnegative (for instance algebraic) coefficients, not all in $\{0, 1\}$ , whose product has 0–1 coefficients. This can be checked exactly with computer algebra.
Verification to higher degrees with certificates that others can recheck independently.
Proofs for new structured families, such as factors with few nonzero terms or supports $S$ with special structure.
A formal proof of the Lean statement green_28 in GreensOpenProblems/28.lean (google-deepmind/formal-conjectures).

References

L. Ghidelli, Progress on the unfair 0-1-polynomials conjecture using linear recurrences and numerical analysis, arXiv:2209.09843 (2022).
K. G. Hare, Computational progress on the unfair 0–1 polynomial conjecture, Experimental Mathematics (2025); arXiv:2307.07363.
A. Dvorsky, The unfair 0–1 polynomial problem and high-degree trinomials, arXiv:2608.19173 (2026).
E. Amiot, question 3325163, Mathematics Stack Exchange (2019), and MathOverflow question 339137 (2019, with answers to 2026).

Source

Imported from Ben Green's 100 open problems 28, curated by Ben Green. Open according to the source as of 2026-10-20. The statement above is written in our own words; see the source for the original wording, history and discussion.