jmc

We compute the probability that the sum is three or less and subtract from 1. A sum of 2 or 3 can be obtained only with the following tosses: $(1,1),(2,1),(1,2)$. There are 36 total toss possibilities, so the probability of getting a sum of 2 or 3 is $\frac{3}{36}=\frac{1}{12}$. Therefore, the probability of getting a sum greater than 3 is $1-\frac{1}{12}=\boxed{\frac{11}{12}}$.