6. Among the triangles formed by the vertices of a convex 18-sided polygon, if any triangle is chosen, the probability that the chosen triangle has no common sides with the 18-sided polygon is $\qquad$
6. $\frac{91}{136}$.

The number of triangles formed by the vertices of a convex 18-gon is $\mathrm{C}_{18}^{3}$.

For any vertex $A$ of a convex 18-gon, to select a vertex of a triangle that does not share any side with the convex 18-gon, the other two vertices $B$ and $C$ cannot be the other two vertices of the two sides of vertex $A$ in the convex 18-gon, but must be two non-adjacent vertices among the other 15 vertices, with a total of $\mathrm{C}_{15}^{2}-14$ different selection methods. Therefore, the number of triangles that do not share any side with the original convex 18-gon is $\frac{1}{3} \times 18\left(\mathrm{C}_{15}^{2}-14\right)$.
Thus, the required probability is
$$
\frac{\frac{1}{3} \times 18\left(\mathrm{C}_{15}^{2}-14\right)}{\mathrm{C}_{18}^{3}}=\frac{91}{136} .
$$