Sum of seven with a dot-weighted die

A six-sided die is loaded so that each face comes up with probability proportional to its number of pips (face $n$ has weight $n$). If it is rolled twice, what is the probability that the two results add up to $7$?

$8/63$. Face $i$ has probability $i/21$. Summing over the pairs $(i,7-i)$ gives $\tfrac{1}{441}\sum_{i=1}^6 i(7-i)=\tfrac{147-91}{441}=\tfrac{56}{441}=\tfrac{8}{63}$.