Number of lonely edges in a graph Suppose we have a graph with $n$ vertices and $n$ edges, where the edges draw a circle (so vertex $1$ connects to vertex $2$, vertex $2$ to $3$, and so on).
Now, randomly select $t$ edges from this graph, without reposition. We're interested in the count and length of the selected segments of the graph. For example, an edge that does not have any adjacent element also selected will have length 1; two adjacent edges will have length 2 and will count as one segment, and so on.
Now my question is: what is the expected count of length-1, length-2, etc, segments of the graph, and, instead of the expected count, can this amount be lower-bounded with a reasonable probability?
Finally, is this a well-studied problem under some other guise? I couldn't find a match for it. 
 A: Let $E$ be the set of edges in the graph and $T$ the set of size $t$ subsets of $E$. For $X\in T$ let $c_k(X)$ be the number of length-$k$ segments in $X$. We are interested in the expected number of length-$k$ segements,
$$
  E(c_k)=\frac{\sum_{X\in T}c_k(X)}{|T|}.
$$
It is easy to see that for $k=n$ or $n-1$, $E(c_n)$ is $1$ if $t=k$ and $0$ otherwise. Suppose $k<n-1$.
For any $X\in T$ and $e\notin X$, let $s(X,e)$ denote the length of the segment immediately clockwise of $e$ in $X$ (this can be zero if the next edge clockwise from $e$ is also not in $X$). Then
$$
  c_k(X)=\#\{e\in E\mid e\notin X\text{ and }s(X,e)=k\}.
$$
Thus
$$\begin{eqnarray*}
  E(c_k)|T|
    &=&\sum_{X\in T}c_k(X)\\
    &=&\sum_{X\in T}\#\{e\in E\mid e\notin X\text{ and }s(X,e)=k\}\\
    &=&\sum_{e\in E}\#\{X\in T\mid e\notin X\text{ and }s(X,e)=k\}.
\end{eqnarray*}$$
For any fixed $e\in E$,
$$
  \#\{X\in T\mid e\notin X\text{ and }s(X,e)=k\}=\binom{n-k-2}{t-k}.
$$
Indeed the conditions imply that $X$ contains the $k$ edges clockwise of $e$, and $X$ does not contain the two edges either side of that segment, leaving $n-k-2$ edges from which to choose the remaining $t-k$ edges.
Thus
$$
  E(c_k)=\frac{n\binom{n-k-2}{t-k}}{\binom{n}{t}}.
$$
