Let $G$ be a connected simple graph with two distinct edges $e,f \in E(G)$. Choose a random spanning tree $T\subseteq G$, my question is whether there are any known upper bound for the following 
\begin{equation} \frac{\mathbb{P}(e\in E(T))\mathbb{P}(f\in E(T))}{\mathbb{P}(e,f\in E(T)) } \end{equation}
for all possible choices of $e$ and $f$? 



Conditional probability that a random spanning tree contains the edge e
There is no constant upper bound, as shown by the following example. Take two vertices $v, u$ and connect them with $n \geq 2$ edge-disjoint paths of two edges. This graph has $n 2^{n - 1}$ spanning trees. The number of spanning trees containing any fixed edge $e$ is $(n + 1)2^{n - 2}$. However, for any $v-u$ path $e, f$ the number of spanning trees containing both $e$ and $f$ is only $2^{n - 1}$. The ratio is thus $\frac{(n + 1)^2}{4n} = \Omega(n)$.


It may still be interesting to determine the largest possible ratio among graphs on $n$ vertices as a function of $n$.


