# Weighted Dice Problem

### Problem

Is it possible to weight a pair of dice so that the probabilities of every possible outcome were the same?

### Solution 1

The feat is impossible. To see that introduce $a_i$ and $b_i$ to be the probabilities of $i$ turning up on each die, respectively, and let $P_i$ be the probability of getting the sum $i$ for a throw of the pair. Then,

$a_{1}b_{1}=P_{2}=P_{12}=a_{6}b_{6}.$

So that if $a_{1}\ge a_{6},$ then $b_{1}\le b_{6},$ and vice versa, implying $(a_{1}-a_{6})(b_{1}-b_{6})\le 0.$ This in turn leads to

$P_{2}+P_{12}=a_{1}b_{1}+a_{6}b_{6}\le a_{1}b_{6}+a_{6}b_{1}=P_{7}.$

Thus already the three probabilities $P_{2},P_{7},P_{12}$ could not be equal.

### Solution 2

$a_i$ and $b_a$ are as in Solution 1. Introduce the generating functions

$A(x)=a_{1}x+a_{2}x^{2}+\ldots +a_{6}x^{6}$ and
$B(x)=b_{1}x+b_{2}x^{2}+\ldots +b_{6}x^{6}.$

Form the product and note that for a pair of dice we expect eleven equal outcomes:

\begin{align} A(x)\cdot B(x) &=(a_{1}x+a_{2}x^{2}+\ldots +a_{6}x^{6})(b_{1}x+b_{2}x^{2}+\ldots +b_{6}x^{6})\\ &=a_{1}b_{1}x^{2}+(a_{1}b_{2}+a_{2}b_{1})x^{3}+\ldots +a_{6}b_{6}x^{12}\\ &=(x^{2}+x^{3}+\ldots +x^{12})/11. \end{align}

Factoring out $x^2$ we obtain

$\displaystyle (a_{1}+a_{2}x+\ldots +a_{6}x^{5})(b_{1}+b_{2}x+\ldots +b_{6}x^{5})=\frac{x^{11}-1}{11(x-1)}.$

But this is impossible because the right-hand side has ten complex (not real) roots, whereas in the left-hand side each of the fifth degree polynomials is bound to have at least one real root.

### Acknowledgment

The problem has been proposed in 1950 by J. B. Kelly (E925, Am Math Monthly 58 1951). I came across it in Murray Klamkin's 1994 paper complete with solutions and an extension of the problem.

### References

1. M. S. Klamkin, Mathematical Creativity in Problem Solving II, in In Eves' Circles, J. M. Anthony (ed.), MAA, 1994