jmc

On each die the probability of rolling $k$, for $1\le k\le 6$, is $$\frac{k}{1+2+3+4+5+6}=\frac{k}{21}.$$ There are six ways of rolling a total of 7 on the two dice, represented by the ordered pairs $(1,6)$, $(2,5)$, $(3,4)$, $(4,3)$, $(5,2)$, and $(6,1)$. Thus the probability of rolling a total of 7 is $$\frac{1\cdot 6+2\cdot 5+3\cdot 4+4\cdot 3+5\cdot 2+6\cdot 1}{21^2}=\frac{56}{21^2}=\boxed{\frac{8}{63}}.$$