6. Choose 5 different numbers from 1, 2, ..., 20, the probability that at least two of them are consecutive is $\qquad$ .
The answer is $\frac{232}{323}$.
Let $a_{1}<a_{2}<a_{3}<a_{4}<a_{5}$ be chosen from $1,2, \cdots, 20$. If $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are not adjacent, then
$$
1 \leq a_{1}<a_{2}-1<a_{3}-2<a_{4}-3<a_{5}-4 \leq 16,
$$

This means that the number of ways to choose 5 non-adjacent numbers from $1,2, \cdots, 20$ is the same as the number of ways to choose 5 different numbers from $1,2, \cdots, 16$, which is $C_{16}^{5}$. Therefore, the probability that at least two of the 5 different numbers chosen from $1,2, \cdots, 20$ are adjacent is
$$
\frac{C_{20}^{5}-C_{16}^{5}}{C_{20}^{5}}=1-\frac{C_{16}^{5}}{C_{20}^{5}}=\frac{232}{323}
$$