TITLE: Covering a random graph with spanning trees.
QUESTION [10 upvotes]: Let $G=(V,E)$ be a connected graph, say $V=\{1,\ldots,n\}$. Let $F=(V,E')$ be a uniformly random forest in $G$. (In other words, $E'$ is a subset of edges $E$ not containing a cycle, and it is uniformly chosen over all such sets.) 
Associated to the random forest $F$ are marginals $\{p_e:e \in E\}$, where $p_e = \mathbb{P}(e \in E')$. Now let $E^*$ be a random subset of $E$, chosen by independently including each edge $e \in E$ with probability $p_e$. 
Finally, let $N$ be the (random) smallest number of spanning trees of $G$ whose union contains $E^*$. 

What is known about the distribution of $N$? How does $\sup_{G} \mathbb{E}(N)$, the largest expected value of $N$ over all $n$-vertex graphs, grow? Is it $O(\log n)$? Is it $O(1)$?

Edit: is it $O(\sqrt{\log n})$? Fedor has a nice example showing that it is not $O(1)$. I believe optimizing Fedor's example yields a lower bound of order $(\log n/\log\log n)^{1/2}$.
Note: the question also makes sense if $E'$ is the edge set of a uniformly random spanning tree, and Fedor's example applies in either case.

REPLY [2 votes]: Consider the path P of length n. Now replace every edge by n parallel edges. A random tree has $p_e=1/n$ and $p_e=1/(n+1)$ in a random forest for each edge $e$. Now, consider the set E^*. There is a constant probability that $\Omega(\frac{\log n}{\log \log n})$ copies of some parallel edge will show up. Therefore, the expectation of $N$ will be $\Omega(\frac{\log n}{\log \log n})$. Am I missing anything here? 
I believe the upperbound will be the same as well.