Past exam of the graph Theory and Probabilistic Combinatorics course of the University of Oxford / 2008 / B / 8 / b / iv

For $p=1/4$, use $k={\log }_{4}n+\omega (n)$. Then $\mathbb{E}X=4^{-\omega (n)}$, so its limits are exactly those needed for the same two arguments.

For $p=1/{\log }_{e}n$, a convenient definition that also accounts for the varying probability isIt makes $\mathbb{E}X=e^{-\omega (n)}$. Thus $\omega \to +\infty$ gives no runs with high probability, and $\omega \to -\infty$ gives runs and relative concentration. For sufficiently large $n$, $p\le 1/2$, so the same uniform variance bound applies. Definitions are made from the actual integer $k$; no continuous rounding assumption is required. These conclusions retain the stipulated range $1\le k\le n/2$.