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(a) For this problem, define a nonstandard die as a 6-sided die that is equally likely to come up on each side, but has a different set of numbers than the usual 1,2,3,4,5,6 on its sides. A standard die will be the usual fair die bearing the numbers 1,2,3,4,5,6.

Is it possible to design a pair of nonstandard dice, with positive integers on the faces (not necessarily all distinct), so that the probability of rolling any given total on those two dice is the same as the probability of rolling that total on two standard dice? The two dice do not need to be identical. If it is possible, do it. If it is not possible, prove that it is not possible.

Some hints I got were to try to construct a pair of nonstandard dice such that the generating function for their sum matches the corresponding g.f. for a pair of standard dice and to note that the g.f. for a pair of standard dice can be factored quite a bit (you can reduce it to factors of degree 2 or less).

I still can't figure it out after rereading the textbook and looking at a few examples. Can anyone help me?

Sam
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1 Answers1

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Use a generating function for the dice. Normal dice have faces with 1 to 6 dots, so they are represented by the polynomial:

$\begin{align} D(z) &= z + z^2 + z^3 + z^4 + z^5 + z^6 \\ &= z \frac{1 - z^6}{1 - z} \end{align}$

Throwing two dice gives the distribution:

$\begin{align} D^2(z) &= z^2 + 2 z^3 + 3 z^4 + 4 z^5 + 5 z^6 + 6 z^7 + 5 z^8 + 4 z^9 + 3 z^{10} + 2 z^{11} + z^{12} \end{align}$

We want polynomials $D_1(z)$ and $D_2(z)$ that give:

$\begin{align} D_1(z) D_2(z) = D^2(z) \end{align}$

It is clear that $D_i(1)$ is the number of faces of the die, so we want:

$\begin{align} D_1(1) = D_2(1) = 6 \end{align}$

If we want no blank faces, it means $D_i(0) = 0$ (no constant term).

Let's factor:

$\begin{align} D(z) &= z (z + 1) (z^2 - z + 1) (z^2 + z + 1) \end{align}$

If we evaluate the factors at $z = 1$:

$\begin{align} D(1) &= 1 \cdot 2 \cdot 1 \cdot 3 \end{align}$

Thus, in order to have $D_i(1) = 6$, each has to get a factor $z + 1$ and a factor $z^2 + z + 1$. As we want $D_i(0) = 0$, each one has to get a factor $z$ too. This leaves two factors $z^2 - z + 1$ (one from each normal die) to distribute, giving the dice:

$\begin{align} D_1(z) &= z (z + 1) (z^2 - z + 1)^2 (z^2 + z + 1) \\ &= z + 2 z^2 + 2 z^3 + z^4 \\ D_2(z) &= z (z + 1) (z^2 + z + 1) \\ &= z + z^3 + z^4 + z^5 + z^6 + z^8 \end{align}$

These is dice with faces $\{1, 2, 2, 3, 3, 4\}$ and $\{1, 3, 4, 5, 6, 8\}$ The only other solution is normal dice (and exchanging them, obviously).

They are called Sicherman dice.

vonbrand
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