6. From $1,2, \cdots, 20$, choose any five different numbers, the probability that at least two of them are consecutive is $\qquad$
6. $\frac{232}{323}$.

Let $a_{1}<a_{2}<\cdots<a_{5}$ be chosen from $1,2, \cdots, 20$.
If $a_{1}, a_{2}, \cdots, a_{5}$ are not adjacent, then
$$
1 \leqslant a_{1}<a_{2}-1<a_{3}-2<a_{4}-3<a_{5}-4 \leqslant 16 \text {. }
$$

This implies that the number of ways to choose five non-adjacent numbers from $1,2, \cdots, 20$ is the same as the number of ways to choose five different numbers from $1,2, \cdots, 16$, which is $\mathrm{C}_{16}^{5}$. Therefore, the required probability is
$$
1-\frac{C_{16}^{5}}{C_{20}^{5}}=\frac{232}{323} .
$$