On the Forged Dice

It is a well{known exercise in elementary probability that it is impossible to forge two ordinary dice with numbers 1;. . .; 6 in such a way that when one throws both and takes the sum one gets the numbers 2; 3;. . .; 12 all with the same probability. preceding assertion is just the statement that for m = 10, k 1 = k 2 = 5 there are no such probabilities. It will be shown how the k 1 ;. . .; k s for which P 1 ;. . .; P s exist can be described explicitly (and how the P look like in this case) in terms of the divisors of m + 1. In the case s = 2 the description reads as follows: Write m + 1 = g 1 g t and deene inductively k 0 1 := g t ? 1, k 00 1 := 0, k 0 := g t?+1 k 00 ?1 + g t?+1 ? 1, k 00 = g t?+1 k 0 g with k 1 + k 2 = m, then the tuple (k 1 ; k 2) can be obtained as (k 0 t ; k 00 t) (or (k 00 t ; k 0 t)) as above. Many devisors give rise to many families k 1 ;. . .; k s so that our description can be considered as an extension of an old result of Raikov's (which states that no k 1 ;. . .; k s exist if m + 1 is prime). Our original problem is equivalent with representing 1+x+ +x m as a product of polynomials A 1 ;. . .; A s with nonnegative coeecients. Hence we get a catalogue of all such factorizations, and the explicit form of the measures P (they are uniform on subsets of f0;. . .; k g) gives rise to an explicit description of the partition of the (m + 1){th roots of unity diierent from 1 which correspond to the factorization 1 + x + + x m = A 1 (x) A s (x).