Expected number of random diagonals before intersection Background Motivation: There are various questions asked about randomly drawn chords and their number of intersections in a circle; for example, MSE 73033 and MO 284124. I am interested here as to a discretized version where the circle is replaced by an $n$-gon and, if all goes well, then one can consider what happens as $n \rightarrow \infty$. (I originally asked this question on twitter, and, although there are some proposed small case computations, I cannot vouch for their correctness. See the thread here.)

Question: Consider an $n$-gon ($n \geq 4$) and a list of all non-adjacent vertex pairs. Pick an unchosen pair from the list at random and connect the vertices. If you continue this process without replacement, then what is the expected number of diagonals drawn before two intersect in the interior of the $n$-gon?

(I am, in particular, looking for a formula that is a function of $n$.)
Note 1: If there is an alternative formulation of the problem with a different method of choosing the diagonals "at random" that yields a different result, then I would welcome such answers in that direction, too.
Note 2: If this result is already known, then a pointer to its answer would be appreciated! I did not locate an answer in my exploration of MSE, but it seems a natural enough question for me to have missed it here (or elsewhere).
 A: Unless I'm mistaken, the asympotic expected number (for large $n$) of non-inserting chords seems quite simple to compute.
In that range, the restriction that we can't select neighbours vertexes should be negligible. Then, we can consider $2n$ points over a circle and consider the event that, for a random pairing, the resulting $n$ chords do not intersect.
In this setup, the total number of pairings is $$T_n=\frac{(2n)!}{2^n n!}=  2^{-n} \binom{2n}{n}n!\,$$
And the number of pairings that do not cross is given by the Catalan number
$$C_n = \binom{2n}{n}\frac{1}{n+1}$$
Hence the probability that $n$ chords do not intersect is
$$P_n = \frac{C_n}{T_n}=\frac{2^n}{(n+1)!}$$
This formula is valid also for the initial trivial values $P_0=P_1=1$.
Let $X$ be the number of chords drawn when the first crossing occurs.
Then $P(X>n)=P_n$ and
$$E[X] = \sum_{n=0}^\infty P_n
=\frac12 \sum_{m=1}^\infty \frac{2^m}{m!}
=\frac12 \left(\sum_{m=0}^\infty \frac{2^m}{m!}-1 \right)
=\frac{1}{2}(e^2-1) = 3.1945\cdots
$$
A: Let $X_n$ be the  number of diagonals drawn until two intersect in the interior of the $n$-gon.
We can write $\mathbb{E}X_n$ as
$$\mathbb{E}X_n = \sum_{k=1}^{\infty}P(X_n\geq k)=\sum_{k=1}^{n-2}P(X_n\geq k).$$
Note that event $\{X_n\geq k\}$ occurs iff the first $k-1$ sampled diagonals don't intersect.
The probability $P(X_n\geq k)$ can be written as
$$P(X_n\geq k)= \frac{s_{n,k-1}}{d_{n,k-1}},$$
where $d_{n,k-1}$ is the number of ways to choose $k-1$ diagonals, and $s_{n,k-1}$ is the number of ways to dissect the $n$-gon with $k-1$ non-crossing
diagonals.
The first term is given by:
$$d_{n,k-1} =\binom{\binom{n}{2}-n}{k-1}=\binom{\frac{n(n-3)}{2}}{k-1}.$$
The second term is given by a generalization of the Catalan numbers, called the Kirkman-Cayley dissection numbers [1]:
$$s_{n,k-1}=\frac{1}{k}\binom{n+k-2}{k-1}\binom{n-3}{k-1}.$$
Putting this together gives
$$\mathbb{E}X_n = \sum_{k=1}^{n-2}P(X_n\geq k)=\sum_{k=1}^{n-2}\frac{\binom{n+k-2}{k-1}\binom{n-3}{k-1}}{k\binom{\frac{1}{2}n(n-3)}{k-1}}.$$
I'm not sure if there is a way to simplify this further.
Update (1/1): Sil found that WA gives a closed form for this sum in terms of the hypergeometric function:
$$\frac{1}{2}(_{2}F_{1}(−n+2,n−1;−\frac{(n−1)(n−2)}{2};1)−1).$$
The first few values:




$n$
$\mathbb{E}X_n$
$\text{approx.}$




4
$2$
2.0


5
$\frac{5}{2}$
2.5


6
$\frac{11}{4}$
2.75


7
$\frac{413}{143}$
2.888


8
$\frac{3839}{1292}$
2.971


9
$\frac{1809}{598}$
3.025


10
$\frac{374329}{122264}$
3.061


10000
-
3.193




Which agrees with leonbloy's answer for large $n$.
