7. From $S=\{0,1, \cdots, 7\}$, randomly select a subset $A$ $=\left\{a_{1}, a_{2}, a_{3}, a_{4}\right\}$, and the remaining elements form the complement set $B=$ $\left\{b_{1}, b_{2}, b_{3}, b_{4}\right\}$. Then the probability that the selected subset $A$ satisfies
$$
a_{1}+a_{2}+a_{3}+a_{4}>b_{1}+b_{2}+b_{3}+b_{4}
$$

is . $\qquad$
7. $\frac{31}{70}$.

The number of ways to choose a 4-element subset is $\mathrm{C}_{8}^{4}=70$, among which, the number of ways that satisfy
$$
a_{1}+a_{2}+a_{3}+a_{4}=b_{1}+b_{2}+b_{3}+b_{4}=14
$$

is 8:
$$
\begin{array}{l}
(7,6,1,0),(7,5,2,0),(7,4,3,0), \\
(7,4,2,1),(6,5,3,0),(6,5,2,1), \\
(6,4,3,1),(5,4,3,2) .
\end{array}
$$

By symmetry, in the remaining ways, the number of ways that satisfy
$$
a_{1}+a_{2}+a_{3}+a_{4}>b_{1}+b_{2}+b_{3}+b_{4}
$$

is half, which is $\frac{70-8}{2}=31$.
Therefore, the required probability is $\frac{31}{70}$.