Let $S$ be a finite ordered set of non-intersecting finite line segments in the plane. Let's randomly shuffle the elements of $S$ such that each possible permutation of those elements has equal probability of appearance.
Suppose we choose the first element $s_1$ of $S$, split the plane by the infinite extension $\ell$ (called a splitting line) of $s_1$ into a half-plane $\ell^+$ above and $\ell^-$ below $\ell$ and recursively continue this process for both of the subsets of $S$ induced by $\ell^+$ and $\ell^-$.
The splitting lines may cut segments into fragments. The process described above continues until there is only one fragment in the interior of each region induced by the surrounding splitting lines. I hope the following figure will illustrate the result of this process:
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What's the expected number of fragments generated by this process?
Let's fix some $s\in S$, let $\ell$ be the splitting line induced by $s$ and define $$\operatorname{dist}t:=\begin{cases}\text{number of segments }u\in S\text{ between }s\text{ and }t\text{ with }\ell\cap u\ne\emptyset&\text{, if }\ell\cap t\ne\emptyset\\\infty&\text{, otherwise}\end{cases}$$ for $t\in S\setminus\left\{s\right\}$.
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Let $$C_n:=\left\{t\in S\setminus\left\{s\right\}:\operatorname{dist}t=n\right\}\;\;\;\text{for }n\in\mathbb N_0$$ (and notice that $|C_n|\le 2$ for all $n\in\mathbb N_0$, which might be useful).
Now, let
- $t\in C_n$ for some $n\in\mathbb N_0$
- $u_1,\ldots,u_n\in S\setminus\left\{s\right\}$ be the segments between $s$ and $t$
- $E$ be the event that $t$ is cut when $\ell$ is added as the next splitting line
I want to show that $$\Pr\left[E\right]\le\frac 1{n+2}\tag 1\;.$$
In order for $E$ to occur, $s$ must come before $t$ and $u_1,\ldots,u_n$ in the random ordering (since $u_1,\ldots,u_n$ shield $t$ from $s$).
I guess I'm thinking way too complicated; my ideas to proceed are as follows:
- There is $1$ possibility to place $s$ at his position $i\in\left\{1,\ldots,N\right\}$ where $N:=|S|$
- There are $N-i$ possibilities to place $t$ at some position $j\in\left\{i+1,\ldots,N\right\}$
- There are $N-i-1$ possibilities to place $u_1$ at some position $k_1\in\left\{i+1,\ldots,N\right\}\setminus\left\{j\right\}$
- $\cdots$
- There are $N-i-n$ possibilities to place $u_n$ at some position $k_n\in\left\{i+1,\ldots,N\right\}\setminus\left\{j,k_1,\ldots,k_{n-1}\right\}$
- There are $\left(\left(i-1\right)+\left(N-\left(\left(i-1\right)+1+1+n\right)\right)\right)!=(N-2-n)!$ possibilities to place the remaining segments at some position
So, there are $$1\cdot\prod_{a=0}^n(N-i-a)\cdot(N-2-n)!=\frac{(N-i)!}{(N-i-n-1)!}(N-2-n)!=:R$$ possible orderings for which $E$ would occur. Since there are $N!$ possible orderings at all, we should have $$\Pr\left[E\right]=\frac{R}{N!}\;.\tag 2$$
Unless the right-hand side of $(2)$ equals the right-hand side of $(1)$, it seems like I made some mistake. Please note that in either case $(2)$ would be an inequality. This is due to the fact that there might be segments that aren't cut by $\ell$ but whose extension shields $t$ (thus we are concerned with the "worst case").
So, how can we show $(1)$?
Your first question, "What's the expected number of fragments generated by this process?", is not well-defined unless you specify a distribution for the segments.
In the second part, your inequality ($1$) is right. By symmetry, all of the $n+2$ segments involved have the same probability of being the first among them, so this probability is $\frac1{n+2}$.
In the third part, you have a free variable $i$ which, if I understand correctly what you're trying to do, shouldn't be free but should be summed over from $1$ to $N$. (I haven't given any thought to whether and how this would yield the correct result $\frac1{n+2}$.)