Suppose I have a connected graph with $n$ vertices and $n−1$ edges, that is in form of a tree. Now, I will add the number of vertices in the tree and uniformly randomly select a vertex. I break the graph at this selected vertex, deleting the vertex and the edges connected to it. Now, I again repeat the same thing at all of the left components (sub-trees) until I am left with one vertex for which the answer is inherently $1$. What will be the expectation of the sum obtained by adding the number of vertices recursively in such a problem?
For example, I have 3 vertices, A, B and C in which A is connected to B as well as C, but B and C are not connected. Like, B--A--C. Now, I could select any one vertex out of three with probability 1/3, so I add 3. Now, if I had selected A, I would repeat this for B and C separately adding one for each case. Otherwise, selecting B or C, I would have repeated it on A--C or B--A respectively, adding 2 and selecting each node again with probability 1/2.
I agree the problem is weakly written, here is a formulation suggested by Karolis Juodelė: I'm looking for the expected value of $f(G) = \left|V_G\right| + \sum f(G_i)$ where $G_i$ are components remaining after removing a random vertex of $G$.