Probability of k-clique in a random graph
The value is $2\log_2 n - 2\log_2\log_2 n + O(1)$.

Let $k_0$ be the maximal value such that the expected number of cliques of size $k_0$ is at least 1. A boring calculation shows that $k_0 = 2\log_2 n - 2\log_2\log_2 n + O(1)$. It is a classical result that with high probability, $G(n,1/2)$ contains a $(k_0-1)$-clique. This implies that $k &gt; k_0-1$. On the other hand, it is known that the expected number of cliques of size $k_0+1+C$ is $O((\log n/n)^C)$ for any constant $C$ (this is because the expected number of cliques drops by $\Theta(\log n/n)$ near $k_0$). This shows that $k \leq k_0+3$. We conclude that $k_0 \leq k \leq k_0+3$.

You can probably decrease the length of this interval by 1, following the proof of the classical result which shows concentration of the clique number of two values rather than three.