Let $G\in G(n,L)$ be an Erdoes-Renyi graph with $n$ vertices and $L$ edges and let $A=A(G)$ be the adjacency matrix of G. The vertices $i,j,k$ form a closed triplet iff $A_{ij}A_{jk}A_{ki}=1$ (without summation over repeated indices).
A given triplet $t=(i_1,i_2,i_3)$ may intersect another triplet $s=(j_1,j_2,j_3)$ in at most one edge, so $|t\cap s|=2$. Let's call them adjacent if they intersect in this way (rather than just in a point). How many adjacent triplets does a given triplet have on average in the ensemble $G(n,L)$?
Suppose one also knows that the number of triplets is $T$. What is then the number of adjacent triplets for a given triplet? The naive guess was \begin{equation} \sum_{j=0}^Tj\binom{T}{j}\left(\frac{9}{L}\right)^j\left(1-\frac{9}{L}\right)^{T-j}=\frac{9T}{L} \end{equation} where $\frac{9}{L}$ is the probability that one triplet intersects any one of three designated edges (namely those belonging to the other triplet). This is however naive: Suppose that there are on average $k>1$ triplets, then for every triplet that falls on the designated one, there are $k-1$ that must not intersect anymore, because they already do.