China Mathematical Competition

Suppose $a_1<a_2<a_3<a_4<a_5$ are taken from $1,2,\dots ,20$. If $a_1,a_2,a_3,a_4,a_5$ are not adjacent to each other, then we have $$1\le a_1<a_2-1<a_3-2<a_4-3<a_5-4\le 16,$$ from which we know that the number of ways to select five numbers not adjacent to each other from $1,2,\dots ,20$ is the same as selecting five different numbers from $1,2,\dots ,16$, i.e., $C_{16}^5$. Therefore, the required probability is $$\frac{C_{20}^5-C_{16}^5}{C_{20}^5}=1-\frac{C_{16}^5}{C_{20}^5}=\frac{232}{323}.$$ The answer is $\frac{232}{323}$.