Triangle variance in a binomial random graph ($\mathrm{Var}X=\left(\genfrac{}{}{0ex}{}{n}{3}\right)p^3(1-p^3)+12\left(\genfrac{}{}{0ex}{}{n}{4}\right)p^5(1-p)$)

For the triangle count in a binomial random graph, distinct triangle indicators are independent unless they share an edge. A shared-edge pair has covariance $p^5(1-p)$, and there are $12\left(\genfrac{}{}{0ex}{}{n}{4}\right)$ ordered pairs of that type. The individual indicator variances give the first term. Equivalently,by counting ordered pairs according to the number of shared vertices.

If $np\to 0$, the triangle count in a binomial random graph has expectation at most $(np{)}^3/6\to 0$, so the Markov inequality shows there are no triangles in a graph with probability tending to one. If $np\to \infty$, the triangle variance in a binomial random graph divided by the squared mean is $O((np{)}^{-3})+O((n^{2}p{)}^{-1})\to 0$. The second moment method then gives at least one triangle in a graph with probability tending to one. At the intermediate scale $np\to c\in (0,\infty )$, a nontrivial Poisson limit for triangles in a binomial random graph describes the count.