8. If three dice are thrown at random, the probability that the numbers shown on the three dice can serve as the side lengths of a triangle is $\qquad$ .


8. $\frac{37}{72}$.

Let the required probability be $p$, and let the numbers shown on the three dice be $x, y, z$. It is clear that when $x \geqslant y+z, y \geqslant z+x$ or $z \geqslant x+y$, the numbers shown on the three dice cannot be the lengths of the sides of a triangle. Note that these three events are mutually exclusive, so we only need to calculate the probability of $x \geqslant y+z$. In fact, this probability is equal to $\frac{0+1+3+6+10+15}{216}=\frac{35}{216}$, hence the required probability is $1-3 \times \frac{35}{216}=\frac{37}{72}$.