2 curly brackets

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

$$p(x) q(x) = {(x^11-1) (x^{11}-1) \over 11(x-1)}$$

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.)

1

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.)