I have been investigating the behavior of graphs that can have a vertex at every point $(x,y,z)$ in an $N\times N\times N$ cube. Each vertex has an edge connecting it with adjacent points (not including diagonals).
I have derived that the maximum number of edges is $3(N-1)N^2$.
The vertex at each point exists with probability, $p$, and otherwise does not exist.
By counting edges of randomly generated cubes using a java program I am fairly certain that the expected number of edges is $p^2 3(N-1)N^2$.
However, I'm not able to work out why this would be the case.
Each of the $3(N-1)N^2$ edges of the original cube graph is present in the random cube graph whenever both endpoints of the edge "survive" to the random graph, which happens with probability $p^2$.
So the expected number of surviving edges is the total number multiplied by the probability of survival, or $p^2 \cdot 3(N-1)N^2$.
Formally, for each edge $e$ in the original graph, we can define a random variable $X_e$ which is $1$ if the edge survives, and $0$ if it does not. We have $\mathbb E[X_e] = \Pr[X_e = 1] = p^2$. The total number of edges is in the random graph is $X = \sum_{e \in E} X_e$, and by linearity of expectation, $$\mathbb E[X] = \sum_{e \in E} \mathbb E[X_e] = \sum_{e \in E}p^2 = p^2|E| = p^2 \cdot 3(N-1)N^2.$$