4. From the numbers $1,2, \cdots, 20$, choose any three different numbers. Then the probability that these three numbers form an arithmetic sequence is ( ).
(A) $\frac{3}{19}$
(B) $\frac{1}{19}$
(C) $\frac{3}{38}$
(D) $\frac{1}{38}$
4. D.

From these 20 numbers, any three numbers can form $\mathrm{A}_{20}^{3}$ sequences.
If the three numbers $a, b, c$ form an arithmetic sequence, then $a+c=2b$.
Therefore, $a$ and $c$ have the same parity, and once $a, c$ are determined, $b$ is also determined.
Thus, the required probability is $p=\frac{2 \mathrm{~A}_{10}^{2}}{\mathrm{~A}_{20}^{3}}=\frac{1}{38}$.