Strange Dice, Same Sums

Question

Relabel two six-sided dice with positive integers (repeats allowed, faces need not stay ≤ 6) so that the distribution of the sum of the two dice is exactly that of a standard pair, same probability for every total 2 through 12. Find such a labeling other than the standard one.

Accepted Answer

Die A: 1, 2, 2, 3, 3, 4. Die B: 1, 3, 4, 5, 6, 8. (Sicherman dice.) Generating functions. A die with faces f_1..f_6 is the polynomial sum x^f_i; rolling two dice multiplies polynomials. The requirement is A(x)B(x) = left(x + x^2 + cdots + x^6right)^2 = x^2(1+x)^2(1+x+x^2)^2(1-x+x^2)^2. Constraints on each factor: A(1) = B(1) = 6 (six faces each) and A(0) = B(0) = 0 with positive labels (a factor of x each). Evaluating the irreducible pieces at 1: (1+x) to 2, (1+x+x^2) to 3, (1-x+x^2) to 1. To make each die's count 6 = 2 times 3, every die must take exactly one (1+x) and one (1+x+x^2), plus its x, only the two "invisible" (1-x+x^2) factors (value 1 at x=1) are free to distribute. Splitting them 1–1 reproduces the standard dice; pushing both onto one die gives A = x(1+x)(1+x+x^2) = x + 2x^2 + 2x^3 + x^4, quad B = x(1+x)(1+x+x^2)(1-x+x^2)^2 = x + x^3 + x^4 + x^5 + x^6 + x^8, reading off the labelings above. Brute-force enumeration over all positive labelings confirms the sum distribution matches exactly, and that standard + Sicherman are the only two solutions. The takeaway: distributions of sums are products of generating functions, so "same sum law" is a factorization question, uniq

Strange Dice, Same Sums

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