6. Choose 3 different numbers from $1,2, \cdots, 20$, then the probability that these 3 numbers form an arithmetic sequence is $\qquad$
The first term of 3 numbers being 1 has $1,2,3 ; 1,3,5 ; \cdots ; 1,10,19$ for a total of 9; The first term of 3 numbers being 2 has $2,3,4 ; 2,4,6 ; \cdots$; $2,11,20$ for a total of 9; The first term of 3 numbers being 3 has $3,4,5 ; 3,5,7 ; \cdots ; 3,11,19$ for a total of 8; The first term of 3 numbers being 4 has $4,5,6 ; 4,6,8 ; \cdots ; 4,12,20$ for a total of 8; $\cdots$; The first term of 3 numbers being 17 has $17,18,19$ for a total of 1; The first term of 3 numbers being 18 has $18,19,20$ for a total of 1. Therefore, the number of ways 3 numbers can form an arithmetic sequence is $2(1+2+\cdots+9)=90$, let $A=$ “3 numbers form an arithmetic sequence”, so $P(A)=\frac{90}{C_{20}^{3}}=\frac{3}{38}$.