jmc

There are $8\cdot 8=64$ ordered pairs that can represent the top numbers on the two dice. Let $m$ and $n$ represent the top numbers on the dice. Then $mn>m+n$ implies that $mn-m-n>0$, that is, $$1<mn-m-n+1=(m-1)(n-1).$$This inequality is satisfied except when $m=1$, $n=1$, or when $m=n=2$. There are 16 ordered pairs $(m,n)$ excluded by these conditions, so the probability that the product is greater than the sum is $$\frac{64-16}{64}=\frac{48}{64}=\boxed{\frac{3}{4}}.$$