Sum Seven With a Weighted Die

The total number of dots is $1+2+\dots+6=21$, so $\mathbb{P}(\text{face } n)=\frac{n}{21}$ for $1\le n\le 6$.

The sum is $7$ exactly when the first roll is some $i\in\{1,\dots,6\}$ and the second roll is $7-i$, which is also in $\{1,\dots,6\}$. These six cases are disjoint. The rolls are independent, so

Evaluating the sum term by term gives

As a check, $7\sum i-\sum i^2=7\cdot 21-91=56$. Therefore

This is slightly less than the fair-die value $\frac16\approx 0.167$. The weighting favours high faces, and pairing a high face with a low one to make $7$ is penalised.

Key idea. The answer is the coefficient of $x^7$ in $\bigl(\tfrac{1}{21}\sum_{n=1}^6 n x^n\bigr)^2$. Computing that coefficient is the same as summing products over complementary pairs $(i,7-i)$.