6. The probability that the product of the points obtained from rolling a die three times is divisible by 6 is $\qquad$
6. $\frac{133}{216}$.

Let $P_{3}$ and $P_{6}$ represent the probabilities that the product is divisible by 3 and by 6, respectively.
Then $P_{3}=1-\left(\frac{4}{6}\right)^{3}=\frac{152}{216}$.
Let $P_{3}^{\prime}$ denote the probability that the product is divisible by 3 and is also odd.
Then $P_{3}^{\prime}=\left(\frac{3}{6}\right)^{3}-\left(\frac{2}{6}\right)^{3}=\frac{19}{216}$.
Therefore, $P_{6}=P_{3}-P_{3}^{\prime}=\frac{133}{216}$.