# Distribution of the sum of points for non-standard dice

Question from my last exam:

A cubic die is called fair if its sides are labeled with integers, and each side has an equal probability of landing face up. When multiple fair dice are rolled, the events are independent. A standard die is defined as a cubic die with labels $$1+x, 2+x, 3+x, 4+x, 5+x, 6+x$$, for some $$x$$.

a) Prove that there are no two fair dice such that when rolled, the total result is always a number from 2 to 12, and each of these numbers has the same probability of occurring.

b) Prove that there exist two non-standard dice such that when rolled, the distribution of the total sum is the same as for two standard dice (with $$x = 0$$).

Hint: In part b), it may be helpful to express the generating function $$g_k$$ for a standard die and consider how to factorize it in two ways $$g_K = g_A \cdot g_B = g_C \cdot g_D$$, so that $$g_A \cdot g_C$$ and $$g_B \cdot g_D$$ correspond to non-standard dice.

My solution to a):
One die is of the form $$1+x,...,6+x$$, and the other is $$1-x,...,6-x$$.
Then, there is exactly one way to obtain a sum of $$12 = (6+x) + (6-x)$$, but there are two ways to get a sum of $$11 =(6+x) + (5-x)$$ or $$(5+x) + (6-x)$$, so $$11$$ and $$12$$ do not have the same probabilities.

Is that the right way to approach b)?
The generating function for the probabilities of a standard die is $$g_K = 1/36(t^1 + t^2 + t^3 + t^4 + t^5 + t^6)$$ We can rewrite it as $$g_k = \frac{1}{36} A\cdot B= \frac{1}{36}(1+t+t^2)(t+t^4)$$ or $$g_k = \frac{1}{36} C\cdot D= \frac{1}{36}(t+t^3+t^5)(1+t)$$
Then
$$A \cdot D = 1+2t+2t^2+t^3$$ $$B \cdot C = t^2+t^4+t^5+t^6+t^7+t^9$$ How to proceed?

Thank you.

Edit:
Thanks to the comment by @Jean-ClaudeArbaut, I see that we can denote the generating function for the sum of points as $$s_k = g_k \cdot g_k = t(t+1)(t^2+t+1)(t^2-t+1) \cdot t(t+1)(t^2+t+1)(t^2-t+1)$$ Now, if we want to achieve the same product using different dice, we can simply write: $$g_k = t(t+1)(t^2+t+1) \cdot t(t+1)(t^2+t+1)(t^2-t+1)^2,$$ resulting in two dice: $$t(t+1)(t^2+t+1)$$ and $$t(t+1)(t^2+t+1)(t^2-t+1)^2$$

1. Is there a general definition for a non-standard die? For example, can I take a die with faces $$t^5+t^4+t^3+t^2+t+1$$ (one face has $$0$$ points), or any other die that does not have six faces? In that case, my original solution would still be valid.

2. Is my solution for part a) correct?

• May help: Sicherman dice Jan 17 at 10:57
• Your AD stands for a die with faces $0,1,1,2,2,3$. Your BC, $2,4,5,6,7,9$. Can you see that these two dice give the same distribution as a pair of "normal" dice? Jan 17 at 12:10
• @GerryMyerson I drew a table, and each sum occurs as many times as in the table for a standard die, so I see that the distribution is the same. Therefore, is my solution correct? Jan 17 at 12:22
• I think the only difference between your dice and mine is that you've added $1$ to each face of AD, and subtracted $1$ from each face of BC. Michal, either way, it solves (b). Jan 17 at 20:41

In a), depending on which die the author had in mind, alternatively, one can say that the total number of possible outcomes from both dice is $$36$$, and the number of possible sums is $$11$$. Since $$11$$ does not divide $$36$$, one of the sums will occur more frequently than others.