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For fixed $0<p<1$, the missing common-neighbour count in a binomial random graph has expectation tending to zero. The Markov inequality therefore shows that every pair has a common neighbour with probability tending to one. On that event every graph distance is at most two, so the graph is connected and its graph diameter is at most two.

Let $I_{ij}$ be the indicator random variable that the unordered vertex pair $\{i,j\}$ has no common neighbour. For each of the other $n-2$ vertices, the two required edges are both present with probability $p^2$. The edge pairs for different candidate neighbours are disjoint sets of independent random choices. Thus $\mathbb{E}{I}_{ij}=(1-p^2{)}^{n-2}$. Linearity of expectation yieldsthe missing common-neighbour count in a binomial random graph.