<s>[INST] Any sum of 2 dice with equal probability The question is the following: Can one create two nonidentical loaded 6-sided dice such that when one throws with both dice and sums their values the probability of any sum (from 2 to 12) is the same. I said nonidentical because its easy to verify that with identical loaded dice its not possible.
Formally: Let's say that $q_{i}$ is the probability that we throw $i$ on the first die and $p_{i}$ is the same for the second die. $p_{i},q_{i} \in [0,1]$ for all $i \in 1\ldots 6$. The question is that with these constraints are there $q_{i}$s and $p_{i}$s that satisfy the following equations:
$ q_{1} \cdot p_{1} = \frac{1}{11}$
$ q_{1} \cdot p_{2} + q_{2} \cdot p_{1} = \frac{1}{11}$
$ q_{1} \cdot p_{3} + q_{2} \cdot p_{2} + q_{3} \cdot p_{1} = \frac{1}{11}$
$ q_{1} \cdot p_{4} + q_{2} \cdot p_{3} + q_{3} \cdot p_{2} + q_{4} \cdot p_{1} = \frac{1}{11}$
$ q_{1} \cdot p_{5} + q_{2} \cdot p_{4} + q_{3} \cdot p_{3} + q_{4} \cdot p_{2} + q_{5} \cdot p_{1} = \frac{1}{11}$
$ q_{1} \cdot p_{6} + q_{2} \cdot p_{5} + q_{3} \cdot p_{4} + q_{4} \cdot p_{3} + q_{5} \cdot p_{2} + q_{6} \cdot p_{1} = \frac{1}{11}$
$ q_{2} \cdot p_{6} + q_{3} \cdot p_{5} + q_{4} \cdot p_{4} + q_{5} \cdot p_{3} + q_{6} \cdot p_{2} = \frac{1}{11}$
$ q_{3} \cdot p_{6} + q_{4} \cdot p_{5} + q_{5} \cdot p_{4} + q_{6} \cdot p_{3} = \frac{1}{11}$
$ q_{4} \cdot p_{6} + q_{5} \cdot p_{5} + q_{6} \cdot p_{4} = \frac{1}{11}$
$ q_{5} \cdot p_{6} + q_{6} \cdot p_{5} = \frac{1}{11}$
$ q_{6} \cdot p_{6} = \frac{1}{11}$

I don't really now how to start with this. Any suggestions are welcome.
 [/INST] I believe the following punchline to the generating function argument doesn’t depend on whether the number of sides on each die is even or odd. Plug $z = \zeta = \exp 2 \pi i /11$ into the purported equality $1+z+z^2+\dots+z^{10} = p(z) q(z)$ with $p,q$ quintic polynomials with nonnegative coefficients. The LHS vanishes but the RHS can’t since all terms contributing to $p(\zeta)$ lie in the upper half-plane and similarly for $q$.
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