Japan Mathematical Olympiad

Let $N_i$ be the number written on the card placed on the $i$-th position from the left. If $N_i\ge i$ is satisfied for all $i$ ($1\le i\le 5$), then $N_5$ has 2 possibilities as it can either be 5 or 6. When $N_5$ is determined, $N_4$ has 2 possibilities as it can be one of the numbers 4, 5, 6 different from $N_5$. Furthermore, when $N_4,N_5$ are determined, $N_3$ can be one of the numbers 3, 4, 5, 6 different from $N_4,N_5$ so there are 2 possibilities for $N_3$. Finally, $N_2\ge 2,N_1\ge 1$ are always satisfied. Therefore, the probability in question is $$\frac{2\times 2\times 2\times 2\times 1}{5\times 4\times 3\times 2\times 1}=\frac{2}{15}.$$