11. Let the set $T=\{0,1, \cdots, 6\}$,
$$
M=\left\{\left.\frac{a_{1}}{7}+\frac{a_{2}}{7^{2}}+\frac{a_{3}}{7^{3}}+\frac{a_{4}}{7^{4}} \right\rvert\, a_{i} \in T, i=1,2,3,4\right\} \text {. }
$$

Arrange the elements of set $M$ in descending order, then the 2015th number is $\qquad$ .
11. $\frac{386}{2401}$.

Multiplying each number in set $M$ by $7^{4}$ gives
$$
\begin{aligned}
M^{\prime} & =\left\{7^{3} a_{1}+7^{2} a_{2}+7 a_{3}+a_{4} \mid a_{i} \in T, i=1,2,3,4\right\} \\
& =\left\{\left(a_{1} a_{2} a_{3} a_{4}\right) \mid a_{i} \in T, i=1,2,3,4\right\} .
\end{aligned}
$$

The largest number in set $M^{\prime}$ is
$$
(6666)_{7}=(2400)_{10} \text {. }
$$

In decimal numbers, the 2015th number from the largest to the smallest starting from 2400 is $2400-2014=386$, which is the 2015th number in set $M^{\prime}$ when arranged from largest to smallest. Dividing this number by $7^{4}$ gives $\frac{386}{2401}$, which is the 2015th number in set $M$ when arranged from largest to smallest.