Two diagonals of a regular heptagon are chosen. What is the probability that they intersect inside the heptagon? 
Two diagonals of a regular heptagon are chosen. What is the probability that they intersect inside the heptagon?

I know there are 14 diagonals and the longer diagonals have 6 total intersections each and the shorter diagonals have 4 intersections each but now I'm stuck.

Consider the first of the two diagonals chosen.
If it is a “short” diagonal, a quick sketch shows that $4$ of the other $13$ diagonals intersects it within the heptagon, and $9$ of the other diagonals do not. So for a short diagonal, there is a $4\over13$ chance that the second diagonal intersects it within the heptagon.
If it is a “long” diagonal, $6$ of the other $13$ diagonals intersects it within the heptagon, for a $6\over13$ chance of inside intersection.
There are the same number of short and long diagonals, so the probability that the second diagonal intersects the first one within the heptagon is the average of the probabilities for the short and long diagonals, which is $5\over13$.
