jmc

The probability of rolling a number other than 1 on one die is $\frac{5}{6}$, so the probability of rolling zero 1's on two dice is ${\left(\frac{5}{6}\right)}^2=\frac{25}{36}$. The probability of rolling a 1 on one die is $\frac{1}{6}$, so the probability of rolling 1's on two dice (for a total of two 1's) is ${\left(\frac{1}{6}\right)}^2=\frac{1}{36}$. Since the only other possibility is rolling exactly one 1, the probability of rolling one 1 is $1-\frac{25}{36}-\frac{1}{36}=\frac{10}{36}$. We find the expected number of 1's to be $E=\frac{1}{36}\cdot 2+\frac{10}{36}\cdot 1+\frac{25}{36}\cdot 0=\boxed{\frac{1}{3}}$.