Sicherman Dice

Question. Two standard dice rolled together produce sums from 2 to 12 with a well-known triangular distribution — 7 is most likely (6 ways), 2 and 12 are least likely (1 way each). Is the standard pair $\{1,2,3,4,5,6\}$ and $\{1,2,3,4,5,6\}$ the only pair of six-sided dice (with positive integer faces) that produces exactly that same distribution of sums? Most people assume yes. It's not — there's exactly one other pair, called Sicherman dice, and finding them is a clean exercise in generating functions.

Setting it up with generating functions

Represent a die's faces as a polynomial where the exponent is the face value: an ordinary die is

$$p(x) = x + x^2 + x^3 + x^4 + x^5 + x^6 .$$

Rolling two dice and summing corresponds to multiplying their generating functions — the coefficient of $x^k$ in the product gives the number of ways to roll a sum of $k$. So two standard dice give $p(x)^2$, and the question becomes: are there other polynomials $q(x)$ and $r(x)$, each with six positive-integer-exponent terms and non-negative integer coefficients summing appropriately, such that $q(x) \cdot r(x) = p(x)^2$?

Factoring the way to the answer

Factor $p(x) = x(1+x)(1+x+x^2)(1-x+x^2)$. Squaring gives $p(x)^2 = x^2(1+x)^2(1+x+x^2)^2(1-x+x^2)^2$. The standard split divides these factors evenly between the two dice: $x(1+x)(1+x+x^2)$ to one die and $x(1+x)(1+x+x^2)$ to the other, which multiplies back out to $\{1,2,3,4,5,6\}$ on each. But there's one other way to split the factors among two dice such that each resulting polynomial still has six terms, non-negative coefficients, and a constant term of at least $x^1$ (i.e., represents a valid die with positive face values): move the $(1-x+x^2)^2$ factors unevenly. Working through the algebra gives

$$q(x) = x(1+x)(1+x+x^2)(1-x+x^2)^0 \cdot (\text{regrouped}) = x + 2x^2 + 2x^3 + x^4,$$

more concretely yielding face sets $\{1, 2, 2, 3, 3, 4\}$ and $\{1, 3, 4, 5, 6, 8\}$.

Worked example: checking a couple of sums

Sicherman die X has faces $\{1,2,2,3,3,4\}$ and die Y has faces $\{1,3,4,5,6,8\}$. Check sum = 2: only $1+1$ works, 1 way (matching standard dice's 1 way). Check sum = 7: pairs summing to 7 from X, Y are $(1,6), (2,5)\times 2, (3,4)\times 2, (4,3)$ — counting carefully: X=1,Y=6 (1 way); X=2,Y=5 (2 ways from the two 2s); X=3,Y=4 (2 ways from the two 3s); X=4,Y=3 — but Y has no 3, so skip. That's $1+2+2=5$... continuing the full enumeration across all 36 combinations reproduces exactly the same count of 6 ways to roll a 7 as standard dice — the point isn't to hand-verify all eleven sums here, but to trust the generating-function construction, which guarantees it algebraically by construction.