11. When seven dice are rolled simultaneously, the probability that the sum of the numbers on the seven faces is 10 is equal to the probability that the sum of the numbers on the seven faces is $a(a \neq 10)$. Then, $a=$ $\qquad$
11.39.
$$
\text { Given }\left(x_{1}, x_{2}, \cdots, x_{7}\right) \rightarrow\left(7-x_{1}, 7-x_{2}, \cdots, 7-x_{7}\right)
$$

we know that the probability of the sum of the points being 10 is the same as the probability of the sum of the points being $49-10$ $=39$.