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# Two Problems Concerning Dice

The following are two of my favorite recreational math problems.

The first time that I encountered this problem was in Béla Bollobás’s book, The Art of Mathematics: Coffee Time in Memphis, which is an excellent read.  There are several ways to solve this problem.

We have two loaded dice (not identical) with 1, . . . , 6 coming up with various probabilities when each one is rolled. Is it possible that when we roll them all of the sums 2, . . . , 12 come up with equal probability 1/11?

The following is a cute problem that I first saw several years ago in Matt Baker’s class Problem Solving for the Putnam at Georgia Tech.

Is it possible create a nonstandard pair of six-sided dice with sides labeled by positive integers so that the odds of rolling a $2,3, \ldots, 12$ are the same as normal dice?

Comment: Numbers can be repeated.
Solutions below the fold…

This solution is from József Balogh’s Probabilistic Methods course and is very elegant.  Bollobás interviewed candidates for scholarships at Trinity College, Cambridge and used this problem as a way to weed out candidates. However, in all of his years of interviewing, no one came up with this solution. Let $P(x) = p_1 + p_2 x + p_3 x^2 + p_4 x^3 + p_5 x^4 + p_6 x^5$ and $Q(x) = q_1 + q_2 x + q_3 x^2 + q_4 x^3 + q_5 x^4 + q_6 x^5$,
where $p_i$ is the probability that the first die shows $i$ and $q_j$ is the probability that the second die shows $j$. As $P$ and $Q$ are real polynomials with degree 5, each must have at least one real root. The coeffficients of the product $P(x) \cdot Q(X)$ correspond to probabilities of rolling various sums.
Suppose for a contradiction that all coefficients are $1/11$, then
$P(x) \cdot Q(X) = \frac{1}{11}(1 + x + \cdots + x^{10}) = \frac{(x^{11}-1)}{11(x-1)}.$
However, the cyclotomic polynomial on the right has zero real roots, the desired contradiction.

George Sicherman discovered this pair of dice, which are known as the Sicherman dice. One die has the numbers 1,2,2,3,3,4 and the other has 1,3,4,5,6,8. From probability theory, the generating function for throwing a die is $x+x^2+x^3+x^4+x^5+x^6$. Therefore the generating function for throwing a pair of dice is $(x+x^2+x^3+x^4+x^5+x^6)(x+x^2+x^3+x^4+x^5+x^6)$
$= x^2 + 2x^3 + x^4+4x^5+5x^6+6x^7+5x^8+4x^9+3x^{10}+2x^{11}+x^{12}$. Using cyclotomic polynomials, we may factor this generating function $x(x+1)(x^2+x+1)(x^2-x+1)x(x+1)(x^2+x+1)(x^2-x+1)$. We must partition the factors to form two legal dice. The only possible partition is $x(x+1)(x^2+x+1) = x+2x^2 +2x^3 +x^4$ and $x(x+1)(x^2+x+1)(x^2-x+1) = x+x^3+x^4+x^5+x^6+x^8$.