Consider a minimally connected graph (i.e., a spanning tree) on $n$ nodes, $\mathcal{T}=(\mathcal{V},\mathcal{E}_{\tau})$,

and its complement $\overline{\mathcal{T}}=(\mathcal{V},\overline{\mathcal{E}}_{\tau})$.

That is, $\mathcal{T} \cup \overline{\mathcal{T}} = K_n$, the complete graph on $n$ nodes.

Consider an edge $e_i \in \overline{\mathcal{E}}_{\tau}$. Then when this edge is added to the spanning tree $\mathcal{T}$, it forms a cycle. Denote the length of this cycle as $l(e_i)$, and let $c(e_i)\subseteq \mathcal{E}_{\tau}$ be the set of edges in $\mathcal{T}$ that the cycle uses.

I would like to determine the probability that two edges $e_i,e_j \in \overline{\mathcal{E}}_{\tau}$ when added to $\mathcal{T}$ form cycles that share $k$ edges;i.e., $|c(e_i) \cap c(e_j)|=k$.

More generally, what is the probability that $|\cap_{i=1}^p c(e_i)|=k$ for $e_i \in \overline{\mathcal{E}}_{\tau}$.

I hope the statement of this problem is clear.

I have begun thinking about this, but have been stuck. Here is what I have.

Let $E(\mathcal{T})$ and $E(\overline{\mathcal{T}})$ be the incidence matrix of $\mathcal{T}$ and $\overline{\mathcal{T}}$ respectively. Then $$T = \left(E(\mathcal{T})^TE(\mathcal{T}) \right)^{-1} E(\mathcal{T})^TE(\overline{\mathcal{T}})$$ is a matrix such that its $i$th column describes which edges in $\mathcal{T}$ are used to create a cycle with the edge $e_i \in \overline{\mathcal{E}}_{\tau}$.

The matrix $TT^T$ then gives information about how many times edges $i$ and $j$ are used in the same cycle. That is $[TT^T]_{ij}$ is an integer number that says how many times $e_i,e_j \in \mathcal{E}_{\tau}$ are used in the same cycle.

Also, the element $[T^TT]_{ii}$ is the length of the cycle formed from

$e_i \in \overline{\mathcal{E}}_{\tau}$,

and $[T^TT]_{ij}$ is the number of edges two cycles share (with a plus/minus sign in there).

Furthermore, there are a total of $(n/2)(n-1)-(n-1)$ possible cycles to form.

I am not very good at combinatorics and am having problems putting the pieces together. Thanks in advance!