8. Roll a die three times, the obtained points are $m$, $n$, $p$. Then the function $y=\frac{2}{3} m x^{3}-\frac{n}{2} x^{2}-p x+1$ is an increasing function on $[1,+\infty)$ with a probability of $\qquad$
8. $\frac{11}{24}$.

Notice,
$$
f(x)=\frac{2}{3} m x^{3}-\frac{n}{2} x^{2}-p x+1
$$

being an increasing function on $[1,+\infty)$ is equivalent to
$$
f^{\prime}(x)=2 m x^{2}-n x-p>0
$$

holding for all $x$ in $[1,+\infty)$, which is equivalent to
$f^{\prime}(1)>0$, i.e., $2 m>n+p$.
When $m=2$, $n+p \leqslant 3$, there are 3 cases;
When $m=3$, $n+p \leqslant 5$, there are 10 cases;
When $m=4$, $n+p \leqslant 7$, there are 21 cases;
When $m=5$, $n+p \leqslant 9$, there are 30 cases;
When $m=6$, $n+p \leqslant 11$, there are 35 cases.

Therefore, the required probability is $\frac{3+10+21+30+35}{6^{3}}=\frac{11}{24}$.