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Let $X$ count copies of the complete graph $K_4$ in $G$. Its expected value is

Write $X$ as a sum of indicator random variable. Indicators for two distinct copies are independent when the copies share at most one vertex, since their edge sets are then disjoint. Pairs sharing two vertices have eleven edges in their union, while pairs sharing three have nine. It follows thatThus $\mathrm{var}(X)/{\mu }^2\to 0$. By the second moment method, $X/\mu \to 1$ in probability, so $\mathbb{P}(X\ge 100)\to 1$. This is the sparse clique-count concentration estimate.

For $p=n^{-2/3}\log n$, the expected number of pairs of $K_4$ copies sharing a vertex isThus, with probability tending to one, all the $K_4$ copies are vertex-disjoint. Together with sparse clique-count concentration, this gives arbitrarily many vertex-disjoint copies with probability tending to one.