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.
 A: Another way to count the number of intersection points created by two diagonals strictly inside the heptagon is to note that if you choose any $4$ vertices $A,B,C,D$ (names in clockwise order, say), then these $4$ points give you exactly one intersection point inside the heptagon, namely $AC\cap BD$, so each of the diagonal-pair-intersections inside the heptagon corresponds to a choice of $4$ vertices of the heptagon, so the number of such intersections is $\binom74=35$. With $14$ available diagonals, you have $\binom{14}2=91$ options for choosing a pair of diagonals, so the required probability is $\dfrac{35}{91}=\dfrac{5}{13}$.
Note: However, this calculation won't be sufficient if the regular polygon has $n$ sides, with $n$ even and $n\gt 4$, because in those cases there are intersection points where more than two diagonals are concurrent. Here the geometry of angles becomes important.

The number of diagonals in any $n$-gon is $\dfrac{n(n-3)}2$. 
Moreover, if $n>4$ is odd, then for a regular $n$-gon, no three diagonals are concurrent inside the polygon, so the number of such intersections being $\binom{n}4$, this probability in an $n$-gon is $$\dfrac{\binom{n}4}{\binom{\frac{n(n-3)}2}2}=\dfrac{\frac{n(n-1)(n-2)(n-3)}{24}}{\frac{\frac{n^2-3n}2\left(\frac{n^2-3n}2-1\right)}2}=\dfrac{(n-1)(n-2)}{3(n^2-3n-2)}\\=\dfrac{n^2-3n+2}{3(n^2-3n-2)}$$
A: The diagonals come in two lengths you've called short and long, and there are $7$ of each, so there are $(7\times6+7\times4)/2=35$ intersections. There are $\binom{14}{2}=91$ diagonal pairs, each with at most one intersection. The probability they intersect is $\tfrac{35}{91}=\tfrac{5}{13}$.
A: 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 $5\over13$.
