On each die the probability of rolling $ k $, for $ 1\leq k\leq 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 \[ \frac{1\cdot6+2\cdot5+3\cdot4+4\cdot3+5\cdot2+6\cdot1}{21^2}=\frac{56}{21^2}=\boxed{\frac{8}{63}}.\]