11. Intersecting Diagonals 2 (from 9th grade. 3 points). 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.
Solution. In a $(2n+1)$-gon,

$$
C_{2 n+1}^{2}-(2 n+1)=\frac{(2 n+1) \cdot 2 n}{2}-(2 n+1)=2 n^{2}-n-1
$$

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 random diagonals intersect is

$$
\frac{C_{2 n+1}^{4}}{C_{2 n^{2}-n-1}^{2}}=\frac{(2 n+1) n(2 n-1)(n-1) \cdot 2}{6 \cdot\left(2 n^{2}-n-1\right)\left(2 n^{2}-n-2\right)}=\frac{n(2 n-1)}{3\left(2 n^{2}-n-2\right)}
$$

Answer: $\frac{2 n^{2}-n}{3\left(2 n^{2}-n-2\right)}$.

Comment. As $n$ grows, the obtained probability approaches $1 / 3$. This leads to the conjecture that if we take an ellipse or a circle and ask about the probability of two random chords intersecting, the answer should be $1 / 3$. Unfortunately, it's not that simple. It matters how exactly the random chords are chosen, since there are infinitely many chords in a circle, unlike the diagonals of a polygon. For example, we could do the following: choose four points $A, B, C$, and $D$ on the circle randomly and independently. What is the probability that the chords $A B$ and $C D$ intersect? Here, the answer is indeed $1 / 3$. You can try to prove this.