Is it possible to build two dice such that the sum of the two dice follows an uniform distribution on $\{2,\dots,12\}$ ?
EDIT : Both dice must have sides 1-6, but the probability that the dice falls on a face can be chosen.
I am looking for a solution that gives the answer and a somewhat detailed reasoning, even though it might seem really easy for some people. I have some difficulties understanding how to solve that kind of problems (what method to follow), any comment towards that issue would be welcomed.
Let $a_i, b_i\in[0,1]$ be the probabilities that die $A, B$ shows $i+1$. (According to the edit, $i$ can only range over $\{0,1,2,3,4,5\}$ for both dice). Let $f(z)=\sum a_iz^i$ and $g(z)=\sum b_i z^i$. Then the desired result is that $$f(z)g(z)=\sum_{j=0}^{10}\frac1{11}z^j,$$ a polynomial without real roots. Hence both $f$ and $g$ have no real roots, i.e. they must both have even degree. This implies $a_5=b_5=0$ and the the coefficient of $z^{10}$ in their product is also $0$, contradiction.
Remark: Why does $h(z)=\sum_{j=0}^{10}z^j$ not have real roots? We have $h(z)=\frac{1-z^{11}}{1-z}$ and the numerator has only one real root at $z=1$. But that is no root of $h$.