6. Choose three different numbers from $1,2, \cdots, 20$, then the probability that these three numbers form an arithmetic sequence is $\qquad$
6. $\frac{3}{38}$.

Let the three different numbers drawn be $a, b, c$. The total number of different ways to choose these numbers is $\mathrm{C}_{20}^{3}$. If these three numbers form an arithmetic sequence, then $a+c=2b$.

Therefore, $a$ and $c$ must both be odd or both be even, and once $a$ and $c$ are determined, $b$ is also determined. Thus, the probability is
$$
P=\frac{\mathrm{C}_{10}^{2}+\mathrm{C}_{10}^{2}}{\mathrm{C}_{20}^{3}}=\frac{3}{38} \text {. }
$$