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

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 $ACcap BD$, 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 $ngt 4$, because in those cases there are intersection points where more than two diagonals are concurrent. Here the geometry of angles becomes important.

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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 $$dfrac{binom{n}4}{binom{frac{n(n-3)}2}2}=dfrac{frac{n(n-1)(n-2)(n-3)}{24}}{frac{frac{n^2-3n}2left(frac{n^2-3n}2-1right)}2}=dfrac{(n-1)(n-2)}{3(n^2-3n-2)}\=dfrac{n^2-3n+2}{3(n^2-3n-2)}$$