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.

You can write a polynomial that encodes the probabilities for each die:
$$ P(x) = p_1 x^1 + p_2 x^2 + p_3 x^3 + p_4 x^4 + p_5 x^5 + p_6 x^6 $$
and similarly
$$ Q(x) = q_1 x^1 + q_2 x^2 + q_3 x^3 + q_4 x^4 + q_5 x^5 + q_6 x^6. $$
Then the coefficient of $x^n$ in $P(x) Q(x)$ is exactly the probability that the sum of your two dice is $n$. As Robin Chapman points out, you want to know if it's possible to have
$$ P(x) Q(x) = (x^2 + \cdots + x^{12})/11 $$
where $P$ and $Q$ are both sixth-degree polynomials with positive coefficients and zero constant term.  
For simplicity, I'll let $p(x) = P(x)/x, q(x) = Q(x)/x$.  Then we want
$$ p(x) q(x) = (1 + \cdots + x^{10})/11 $$
where $p$ and $q$ are now fifth-degree polynomials. We can rewrite the right-hand side to get
$$ p(x) q(x) = {(x^{11}-1) \over 11(x-1)} $$
or 
$$ 11 (x-1) p(x) q(x) = x^{11} - 1. $$
The roots of the right-hand side are the eleventh roots of unity.  Therefore the roots of $p$ must be five of the eleventh roots of unity which aren't equal to one,  and the roots of $q$ must be the other five. 
But the coefficients of $p$ and $q$ are real, which means that their roots must occur in complex conjugate pairs.  So $p$ and $q$ must have even degree!  Since five is not even, this is impossible.
(This proof would work if you replace six-sided dice with any even-sided dice. I suspect that what you want is impossible for odd-sided dice, as well, but this particular proof doesn't work.)
