smc

The third toss cannot be $1$, since the minimal sum on the other two tosses is $2$. If the third roll is $2$, then the first two rolls must be $(1,1)$. If the third roll is $3$, then the first two rolls must be $(1,2)$ or $(2,1)$. If the third roll is $4$, then the first two rolls must be $(1,3)$, $(2,2)$, or $(3,1)$. If the third roll is $5$, then the first two rolls must be $(1,4)$, $(2,3)$, $(3,2)$, or $(4,1)$. If the third roll is $6$, then the first two rolls must be $(1,5)$, $(2,4)$, $(3,3)$, $(4,2)$, or $(5,1)$ There are $15$ possibilities for the three tosses. Of those possibilities, $7$ have a $2$ in the first two rolls, and $1$ has a $2$ in the third. Therefore, the answer is $\boxed{\frac{8}{15}}$.