Probability that connecting two vertices in a dodecahedron will result in a segment inside the dodecahedron A regular dodecahedron is a convex polyhedron with 12 regular pentagonal faces and 20 vertices. If two distinct vertices are chosen at random, what is the probability that the line connecting them lies inside the dodecahedron?
I know that when you connect to points in the dodecahedron, it cannot result in a segment outside of the dodecahedron. So we have $1-P(\text{Segment On Dodecahedron})=$ desired result. So in order for the segment to be on the dodecahedron, the two vertices have to be on the same face. I'm thinking that the denominator, in this case, would be $\binom{20}2$. However, I'm not sure how to find the numerator.
Help is greatly appreciated.

There are two equivalent ways to find the probability that a segment does not lie in the dodecahedron:


*

*Select any vertex first; all of them are equivalent. As for the second vertex, the nine vertices that share a face with the first vertex (out of the 19 remaining vertices) will give a segment not in the dodecahedron.

*There are $\binom{20}2$ choices of vertex pairs, 90 of which give a segment not in the dodecahedron. These correspond one-to-one with the 30 edges and 60 face diagonals.


Both approaches give the auxiliary probability of $\frac9{19}$; the final answer is then $1-\frac9{19}=\frac{10}{19}$.
