6. If 3 dice are thrown randomly, then the probability that the sum of the points on 2 of the dice is 7 is


6. $\frac{5}{12}$.

Rolling 3 dice has a total of $6^{3}=216$ possibilities. Consider $7=1+6=2+5=3+4$.
When rolling three dice, the possibility of two dice showing 1 and 6 is $6 \times 6-6=30$ (divided into $(1,6, *)$, $(1, *, 6),(6,1, *),(6, *, 1),(*, 1,6),(*, 6,1)$, these 6 possibilities, each with 6 cases. Among them, $(1,6,1)$, $(1,6,6),(1,1,6),(6,1,1),(6,1,6),(6,6,1)$ are repeated).

Similarly, the possibility of two dice showing 2 and 5 when rolling three dice, and the possibility of two dice showing 3 and 4, are both 30.

Therefore, the number of possibilities where 2 out of 3 dice show a sum of 7 is $3 \times 30=90$. The required probability is $\frac{90}{216}=\frac{5}{12}$.