We know that the sum of two standard 6-sided dice follows: $$P(X_1+X_2=2)=\frac{1}{36},\dots, P(X_1+X_2=6)=\frac{5}{36}$$ $$P(X_1+X_2=7)=\frac{6}{36}, P(X_1+X_2=8)=\frac{5}{36},\dots, P(X_1+X_2=12)=\frac{1}{36}$$ How to design a 4-sided die and an 8-sided die and the numbers on the dice and the probability of getting each face is chosen by the designer such that the sum of the dice has the same distribution as the sum of the above 2 standard dice?

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    $\begingroup$ I expect you need 4 and 9 sided dice, to give 36 outcomes $\endgroup$ – Empy2 Nov 21 at 22:50
  • $\begingroup$ But then it adds up to 13. $\endgroup$ – marty cohen Nov 21 at 22:53

Mimic the generating function approach used for Sicherman dice. You want a product $$p(x)q(x)=(x+x^2+x^3+x^4+x^5+x^6)^2=[x(x+1)(x^2+x+1)(x^2-x+1)]^2$$ with $p(0)=q(0)=0$, $p(1)=4$, and $q(1)=8$. But $p(1)q(1)=36$, so $p(1)=4$ implies $q(1)=9 \not= 8$. You can do it with a 4-sided die and a 9-sided die: \begin{align} p(x) &= x(x+1)^2 = x+2x^2+x^3 \\ q(x) &= x(x^2+x+1)^2(x^2-x+1)^2 = x+2x^3+3x^5+2x^7+x^9 \end{align} That is, $(1,2,2,3)$ and $(1,3,3,5,5,5,7,7,9)$.

There are exactly two other possibilities, based on where you put the two copies of factor $x^2-x+1$.

I had missed the part about also deciding the probability of each face. Here is such a solution for a 4-sided die and an 8-sided die: \begin{matrix} \text{face} &1 &2 &3 &4 &5 &6 &7 &8 \\ \hline \text{die 1} &1/6 &1/3 &1/3 &1/6 \\ \text{die 2} &1/6 &0 &1/6 &1/6 &1/6 &1/6 &0 &1/6 \\ \end{matrix}

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