Assume an undirected simple graph $G=(V,E)$ is created randomly: an edge $e=(u,v)$ is in $E$ with probability $p$, independent of other edges.
Assume we select a random cut $(S,T = V\setminus S)$ in the graph. What is the expected value of its size? (The size of the cut is the number of edges $(u,v)\in E, u\in S, v\in T$)
I have made an attempt with the following approach: the maximal possible size of such a cut is $|S|\cdot|T|$ (every vertex in $S$ is connected to every vertex in $T$). So the mean cut can be calculated as follows:
$$\sum_{i=0}^{|S|\cdot|T|}i \cdot P(\text{size of the cut is } i)$$
However, I am having difficulty with $P(\text{size of the cut is } i)$. Somehow I must refer to the total number of edges. It is the probability that an edge $(u,v)$ exists ($p$), and that $u \in S$ and $v\in T$ (the probability that $u\in S, v\in T$ are not given...) and that all other edges do not satisfy any of those three conditions. Therefore, this sounds like an infeasible approach.
This sort of depends on how you select the random cut. We might do it uniformly: for every vertex $v$, we put $v \in S$ with probability $\frac12$, $v \in T$ otherwise, and make this decision independently for all vertices.
If we fix the cut $(S,T)$, then the formula $$\mathbb E[\text{size of $(S,T)$ cut}] = \sum_{i=0}^{|S||T|} i \cdot \Pr[\text{size is }i]$$ is valid, but it's much easier to approach this by linearity of expectation: each of the $|S||T|$ edges is present with probability $p$, so the expected number of edges present is $p|S||T|$.
This is a formula that depends on $S$ and $T$ (or rather, on their sizes). To get the overall formula, we can substitute the previous result into the summation
$$\mathbb E[\text{size of cut}] = \sum_{S \subseteq [n]} \mathbb E[\text{size of $(S,[n]\setminus S)$ cut}] \cdot \Pr[\text{the cut is }(S,[n]\setminus S)].$$
In the case where $S$ is chosen uniformly from all subsets of $[n]$, this is $$\sum_{S \subseteq [n]} p |S|(n-|S|) \cdot 2^{-n} = \sum_{k=0}^n pk(n-k) \binom nk 2^{-n}$$ where we get the second sum by grouping the sets $S$ together by size. This simplifies to $\frac14 n(n-1)p$.
We can also get this result much more directly by observing that each edge of the graph $G$ is present with probability $p$, and its endpoints end up on different sides of the cut with probability $\frac12$, so it ends up contributing $1$ to the size of the cut with probability $\frac p2$. By linearity of expectation, the size of the cut is $\frac p2 \binom n2 = \frac14 n(n-1)p$.