Can two dice be weighted so that the probability of each of the numbers 2, 3, …, 12 is the same?

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I think this problem was first proposed by J.B. Kelly in the American Mathematical Monthly in 1950. The answer is no. Let a_{j} and b_{j} be the probabilities that the number j turns up on each of the two dice, and let P_{j} be the probability of getting the sum j on the pair of dice. So
P_{2} = a_{1}b_{1} = a_{6}b_{6} = P_{12}.
Now, to maintain this balance, if the first die favors 1 over 6, then the second must favor 6 over 1, and vice versa, so that
(a_{1} – a_{6})(b_{1} – b_{6}) ≤ 0.
But this means that
P_{2} + P_{12} = a_{1}b_{1} + a_{6}b_{6} ≤ a_{1}b_{6} + a_{6}b_{1} ≤ P_{7}.
The probabilities of rolling a 2, a 7, and a 12 cannot be equal.
