## Two Dice

My friend Alex Ryba uses interesting math questions in the CUNY Math Challenge. For the 2016 challenge they had the following problem.

Problem.Eve owns two six-sided dice. They are not necessarily fair dice and not necessarily weighted in the same manner. Eve promises to give Alice and Bob each a fabulous prize if they each roll the same sum with her dice. Eve wishes to design her two dice to minimize the likelihood that she has to buy two fabulous prizes. Can she weight them so that the probability for Alice and Bob to get prizes is less than 1/10?

The best outcome for Eve would be if she can weight the dice so that the sum is uniform. In this case the probability that Alice and Bob get the prizes is 1/11. Unfortunately for Eve, such a distribution of weight for the dice is impossible. There are many ways to prove it.

I found a beautiful argument by Hagen von Eitzen on the stack exchange: Let *a _{i}* (correspondingly

*b*) be the probabilities that die A (correspondingly B) shows

_{i}*i*+ 1. It would be very useful later that that

*i*ranges over {0,1,2,3,4,5} for both dice. Let f(z) = ∑ a

_{i}z

^{i}and g(z) = ∑ b

_{i}z

^{i}. Then the desired result is that f(z)g(z) = ∑

_{j=0}

^{10}z

^{j}/11. The roots of the right side are the non-real roots of unity. Therefore both f and g have no real roots. So, they must both have even degree. This implies a

_{5}=b

_{5}=0 and the coefficient of z

^{10}in their product is also 0, contradiction.

Alex himself has a straightforward argument. The probabilities of 2 and 12 have to be equal to 1/11, therefore, a_{0}b_{0} = a_{5}b_{5} = 1/11. Then the probability of a total 7 is at least a_{0}b_{5} + a_{0}b_{5}. The geometric mean of a_{0}b_{5} and a_{0}b_{5}
is 1/11 (from above), so their arithmetic mean is at least 1/11 and
their sum is at least 2/11. Therefore, the uniform distribution for sums
is impossible.

So 1/11 is impossible, but how close to it can you get?

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