I stumbled upon this interesting problem recently, which I thought deserved its own post.
The question is: "Given two n-faced dice numbered from 1 to n (with n > 1), is it possible that the sum of the two dice is equally probable?"

In other words, if I call the probability of getting i for the two dice \(p_i\) and \(q_i\) respectively, then the probability of getting n as their sum is:

\[P(n) = \sum_{i+j=n} p_i q_j\]

Now, we want all the sums to have equal probablity so:

\[P(2) = P(3) = ... = P(2n) = \frac{1}{2n-1}\]

Which means by replacing each probability with its components:

\[p_1 q_1 = p_1 q_2 + p_2 q_1 = ... = p_n q_n = \frac{1}{2n-1}\]

1st solution

The first way to approach this problem is to take the equations above and to see that:

(1) \(p_1 q_1 = \frac{1}{2n-1}\)

(2) \(p_n q_n = \frac{1}{2n-1}\)

(3) \(p_1 q_n + p_n q_1 \leq \sum_{i+j=n+1} p_i q_j = \frac{1}{2n-1}\)

So from (1) and (2) we know that \(p_1\), \(q_1\), \(p_n\) and \(q_n\) are > 0 and from (3) we conclude that:

\[p_1 q_n < \frac{1}{2n-1}\]

\[p_n q_1 < \frac{1}{2n-1}\]

But we know from (1) and (2) that \(p_1 q_1 = p_n q_n = \frac{1}{2n-1}\) so:

\(p_1 q_n < p_1 q_1\) and then \(q_n < q_1\)

\(p_n q_1 < p_n q_n\) and then \(q_1 < q_n\)

Which leads to a contradiction as \(q_n < q_1 < q_n\) so the answer to the problem is no, it's impossible.

2nd solution (for n even)

There is a beautiful solution in the case of n even which uses generating functions. Take the two functions (polynomials):

\(F(x) = \sum_{i=1}^n p_i x^i\) and \(G(x) = \sum_{i=1}^n q_i x^i\)

By multiplying the functions above, we get the following polynomials whose coefficients in \(x^i\) are the probabilities of getting a sum of dice equal to i:

\[F(x) G(x) = (\sum_{i=1}^n p_i x^i) (\sum_{i=1}^n q_i x^i)\]

So we can reformulate our problem by looking for two polynomials F and G such as:

\[F(x) G(x) = \sum_{i=2}^{2n} \frac{1}{2n-1} x^i\]

\[F(x) G(x) = \frac{x^2}{2n-1} \sum_{i=0}^{2n-2} x^i\]

But the right hand side is a geometric sum:

\[\sum_{i=0}^{2n-2} x^i = \frac{x^{2n-1}-1}{x-1}\]

So:

\[(2n-1) (x-1) F(x) G(x) = x^2 (x^{2n-1}-1)\]

\[(2n-1) (x-1) (\sum_{i=1}^n p_i x^i) (\sum_{i=1}^n q_i x^i) = x^2 (x^{2n-1}-1)\]

\[(2n-1) (x-1) (\sum_{i=1}^n p_i x^{i-1}) (\sum_{i=1}^n q_i x^{i-1}) = (x^{2n-1}-1)\]

\[(2n-1) (x-1) (\sum_{i=0}^{n-1} p_{i-1} x^i) (\sum_{i=1}^n q_{i-1} x^i) = (x^{2n-1}-1)\]

The zeros on the right hand side of the equation are the (2n-1)-th roots of unity. As 2n-1 is odd, that means we have 2n-2 complex roots and the last root is 1.
On the left hand side, (x-1) with one root equal to 1 and we have two polynomials of degree (n-1). The root equal to 1 is taken care of on both sides which means the two polynomials (with real coefficients) all have complex roots. But if a complex root is a root of a polynomial with real coefficients then so is its complex conjugate. So the two polynomials of degree n-1 must have an even number of roots, which is only possible if n is odd. So in the case of n even, it is impossible to find F and G and the problem has no solution.
This solution may be a out of place as it only covers n even. But this problem is usually asked for n = 6 and in this post I extended the result to any number n > 1.