[ Combinations and Permutations ]  $[$ Polygons (other) ]

In a convex polygon with an odd number of vertices, equal to $2 n+1$, two random diagonals are chosen independently of each other.

Find the probability that these diagonals intersect inside the polygon.
In a $(2 n+1)$-gon, there are $\quad C_{2 n+1}^{2}-(2 n+1)=\frac{(2 n+1) \cdot 2 n}{2}-(2 n+1)=2 n^{2}-n-1 \quad$ diagonals, and, consequently, the number of pairs of diagonals is $C_{2 n^{2}-n-1}^{2}=\frac{\left(2 n^{2}-n-1\right)\left(2 n^{2}-n-2\right)}{2}$.

Two intersecting diagonals are uniquely determined by the choice of four vertices, so the number of ways to choose two intersecting diagonals is $C_{2 n+1}^{4}=\frac{(2 n+1) 2 n(2 n-1)(2 n-2)}{24}=\frac{(2 n+1) n(2 n-1)(n-1)}{6}$.

Diagonals are chosen randomly, so all combinations are equally likely. Therefore, the probability that two randomly chosen diagonals intersect is $\frac{C_{2 n+1}^{4}}{C_{2 n^{2}-n-1}^{2}}=\frac{(2 n+1) n(2 n-1)(n-1) \cdot 2}{6 \cdot\left(2 n^{2}-n-1\right)\left(2 n^{2}-n-2\right)}=\frac{n(2 n-1)}{3\left(2 n^{2}-n-2\right)}$.