6. From $1,2, \cdots, 20$ choose any 5 different numbers, the probability that at least two of them are consecutive is
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From $1,2, \cdots, 20$ choosing 5 different numbers has $C_{20}^{5}$ ways. Considering the scenario where the 5 numbers are pairwise non-adjacent, let $a<b<c<d<e$ be chosen from $1,2, \cdots, 16$, then $a, b+1, c+2, d+3, e+4$ are 5 numbers chosen from $1,2, \cdots, 20$ that are pairwise non-adjacent, and vice versa. Therefore, choosing 5 pairwise non-adjacent numbers from $1,2, \cdots, 20$ has $C_{16}^{5}$ ways.
Let $A=$ “choosing 5 different numbers from $1,2, \cdots, 20$, where at least two are adjacent”, then $P(A)=1-P(\bar{A})=1-\frac{C_{16}^{5}}{C_{20}^{5}}=1-\frac{91}{323}=\frac{232}{323}$.