# Unusual pair of dice

Suppose we roll two ordinary, 6-sided dice, $f_i=s_i=\{1,2,3,4,5,6\}$. All $36$ possible rolls with a pair of dice (first dice - $f_i$, second dice $s_i$):

$$\begin{array}{ccccccc} f_i/s_i & 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ 6 & 7 & 8 & 9 & 10 & 11 & 12 \end{array}$$

So we have $1$ variant to roll the sum $2, 2$ variants to roll $3, 3$ variants to roll $4, ..., 2$ variants to roll $11, 1$ variant to roll $12.$

Is there some another pair $(f_i, s_i)$ of $6$-sided dice with different numbers on its sides, but with the same set of variants to roll the same sums $(2,\ldots,12)$? And all $f_i>0, s_i>0$. And how much?

$f_i$ and $s_i$ in such pair can be different.

Special upd $f_i \in \mathbb{N}$, $s_i \in \mathbb{N}$.

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Yes. If we factor the generating function $(x+x^2+x^3+x^4+x^5+x^6)^2$, we obtain $$(x+x^2+x^3+x^4+x^5+x^6)^2= x^2(x+1)^2(x^2-x+1)^2(x^2+x+1)^2$$ and have the option of distributing the factors unevenly to the dice. But (apart from the standard dice) the only possibility to have nonnegative coefficients, no constant term and coefficient sum six is given by $$(x^2 -x+1)^2(x^2+x+1)(x+1)=x^8 + x^6 + x^5 + x^4 + x^3 + x$$ and $$(x^2+x+1)(x+1)x=x^4 + 2x^3 + 2x^2 + x$$ corresponding to dice with labels $(1, 3, 4, 5, 6, 8)$ and $(1, 2, 2, 3, 3, 4)$.
There is no other way because each die must obtain one factor of $x$ (to have positive labels), one factor of $(x+1)$ to have even face count and one factor of $1+x+x^2$ to have face count divisible by three. Only the factors $x^2-x+1$ are "expendible".