I have the following problem. Suppose we have a hypercube of dimension $L$. Then the number of nodes is given by $n=2^L$ and the number of edges is given by $e=L*2^{L-1}$. Now each node is deleted with a probability given by $(1−p)$. The question is, what is the average number of edges per node $\bar{d}(L,p)$ of the resulting graph for given $L$ and $p$. I know that the mean number of edges is given by $\bar{e}=p^2*L*2^{L-1}$ and the mean number of nodes is given by $\bar{n}=p*2^L$. The number of edges per node is given by $d=2*e/n$. But since $\bar{e}$ and $\bar{n}$ are depentend variables $\bar{d}=2*\bar{e}/\bar{n}=p*L$ is wrong.
Edit 1: I was able to calculate $\bar{d}(L,n)$ that is the mean number of edges per node for a random hypercube with dimension $L$ and a given number $n$ of "survived" nodes. Maybe this helps?
$$\bar{d}(L,n)=\frac{(n-1)*L}{2^L-1}$$
With $\bar{d}=2*\bar{e}/n$ in this case this also leads to
$$\bar{e}=\frac{L*n(n-1)}{2*(2^L-1)}$$
Edit 2: With the following equation given by gt6989b I get:
$$\bar{d}(L,p)=\sum_{v=0}^{2^L}\overline{n}(L,v) \Pr (v; p, 2^L)=\sum_{v=0}^{2^L}\frac{(v-1)L}{2^L-1} \binom{2^L}{v}p^v(1-p)^{2^L-v}=\frac{(2^L p-1)L}{2^L-1}$$
So I get from $\bar{d}(L,n)$ to $\bar{d}(L,p)$ by replacing $n$ with $p*2^L$ which somehow makes sense. So far everything sounds right, but I noticed that
$$\bar{d}(L,p)=\frac{(2^L p-1)L}{2^L-1}$$
becomes negative for sufficiently small values of $p$. I am now wondering if something is still wrong.
Given you computed $d(L,n)$ correctly, you have $$ \mathbb{E}[d(L)] = \sum_{k=0}^{2^L} d(L,k) \mathbb{P}[n=k] $$ where the probability in question is easy since $n \sim\mathcal{B}(2^L,p)$.