jmc

We instead find the probability that the sum of the numbers showing is not between 3 and 11. Since each die's face contains the numbers 1-6, we see that this can only be the case if we roll two 1s or two 6s. Thus the chance that the sum is not between 3 and 11 is $\frac{1}{6}\cdot \frac{1}{6}+\frac{1}{6}\cdot \frac{1}{6}$, or $\frac{1}{18}$. So our desired probability is $1-\frac{1}{18}=\boxed{\frac{17}{18}}$.