8. The probability of selecting four vertices from a regular 12-sided polygon such that no two are adjacent is $\qquad$ .
8. $\frac{10}{99}$.

Notice that, the total number of ways to choose four vertices from 12 vertices is $\mathrm{C}_{12}^{4}$.

Label the vertices of the regular 12-gon as $1 \sim 12$, and let the numbers of the four chosen vertices, in ascending order, be $a, b, c, d$.
If they are pairwise non-adjacent, then
$$
\begin{array}{l}
x=b-a-2 \geqslant 0, \\
y=c-b-2 \geqslant 0, \\
z=d-c-2 \geqslant 0 .
\end{array}
$$

When $a=1$, it satisfies $d \leqslant 11$, i.e., $x+y+z \leqslant 4$. The number of $(x, y, z)$ is $\mathrm{C}_{6}^{2}=15$;

When $a \geqslant 2$, it satisfies $d \leqslant 12$, i.e., $x+y+z \leqslant 6-a$. The number of $(x, y, z)$ is $\mathrm{C}_{8-a}^{2}$.
Therefore, the probability that they are pairwise non-adjacent is
$$
\frac{\mathrm{C}_{6}^{2}+\mathrm{C}_{6}^{2}+\mathrm{C}_{5}^{2}+\mathrm{C}_{4}^{2}+\mathrm{C}_{3}^{2}+\mathrm{C}_{2}^{2}}{\mathrm{C}_{12}^{4}}=\frac{10}{99} .
$$