Weighted Dice Problem

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.

$a_i$ 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.

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