10. (20 points) The rules of a card game are as follows: Arrange nine cards labeled $1,2, \cdots, 9$ randomly in a row. If the number on the first card (from the left) is $k$, then reverse the order of the first $k$ cards, which is considered one operation. The game stops when no operation can be performed (i.e., the number on the first card is “1”). If a permutation cannot be operated on and is obtained from exactly one other permutation by one operation, it is called a “second-terminal permutation”. Among all possible permutations, find the probability of a second-terminal permutation occurring.
10. Obviously, for a secondary terminating permutation, the first card must be 1. Let the number marked on the card at the $i$-th position be $a_{i}$, then $\left(a_{1}, a_{2}, \cdots, a_{9}\right)$ is a permutation of $1,2, \cdots, 9$, and $a_{1}=1$.

Since there is a unique permutation $\left(a_{1}, a_{2}, \cdots, a_{9}\right)$ that can be obtained after one operation, there is exactly one $k$ in $\{2,3, \cdots, 9\}$ such that $a_{k}=k$, and this $k$ can have eight different choices. The other seven numbers must satisfy $a_{j} \neq j(j \notin\{k, 1\})$ (otherwise, if there exists $j_{0} \notin\{k, 1\}$ such that $a_{j_{0}}=j_{0}$, consider the permutation $A$ obtained by reversing the first $k$ cards and the permutation $B$ obtained by reversing the first $j_{0}$ cards. Since 1 is in the $k$-th and $j_{0}$-th positions in permutations $A$ and $B$ respectively, they cannot be the same, but they both result in the same secondary terminating permutation after one operation, which is a contradiction). Therefore, the number of derangements for the seven positions should be
$$
\begin{array}{l}
D_{7}=7!\left(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\frac{1}{7!}\right) \\
=1854 .
\end{array}
$$

The number of secondary terminating permutations is
$$
8 \times 1854=14832 \text {, }
$$

Thus, the probability of its occurrence is $\frac{14832}{9!}=\frac{103}{2520}$.