Number of house subgraphs in a Erdős-Rényi graph

Question

Considering we have a "house subgraph", which is a graph composed of 5 nodes, $v_1, \ldots v_5$ in which the edges
$\{v_1,v_2\}, \{v_1,v_3\}, \{v_2,v_3\}, \{v_2,v_4\}, \{v_3,v_5\}, \{v_4,v_5\}$ exist, and no other edge between them exists
(see Figure below). What is the expected number of house subgraphs in a random graph according
to $G(n, p)$ (Erdős-Rényi graph with $n$ vertices and edge probability $p$)?

House subgraph figure:

Answer 1

Let's solve a more general problem:

Suppose $\Gamma$ is a graph with $m$ vertices and $k$ edges. What is the expected number of induced subgraphs isomorphic to $\Gamma$ in $G(n, p)$ ($n$-vertex Erdos-Renyi random graph with edge probability $p$)?

Let's solve it first for the case, where $n = m$. Then there is only one induced subgraph of size $m$ in our random graph. What is the probability of it being isomorphic to $\Gamma$?

There are $m!$ bijections from its vertices to the vertices of $\Gamma$, but only $\frac{m!}{|Aut(\Gamma)|}$ are distinct in respect to automorphisms of $\Gamma$. For all those distinct bijections the events of them being isomorphisms are disjoint and probability of each of them equals $p^k (1 - p)^{\frac{m(m-1)}{2} - k}$. Thus the total probability of $G(m, p)$ being isomorphic to $\Gamma$ is $\frac{m! p^k (1 - p)^{\frac{m(m-1)}{2} - k}}{|Aut(\Gamma)|}$.

Now assume $n$ is arbitrary. For any $m$-element set $A$ of vertices of our random graph assume $I_A$ to be the Bernoulli random variable that equals $1$ if induced subgraph on those vertices is isomorphic to $\Gamma$ and $0$ otherwise. Then, due to linearity of expectation, the expected number of induced subgraphs isomorphic to $\Gamma$ is the sum of expected values of all $I_A$. We have already found, that the expectation of each of them (equal to their probability to be $1$) is $\frac{m! p^k (1 - p)^{\frac{m(m-1)}{2} - k}}{|Aut(\Gamma)|}$. And the number of $m$-element subsets of vertices to which they correspond is $C_n^m = \frac{n!}{m!(n-m)!}$.

Thus, the answer is

$$\frac{n!p^k (1 - p)^{\frac{m(m-1)}{2} - k}}{(n-m)!|Aut(\Gamma)|}$$

Now you can calculate the corresponding numbers for your specific case (i.e. the number of vertices, edges and automorphisms of the "house graph") and put them into this formula.