Let $X$ be the total number of cycles,
$X_k$ the number of cycles of length $k$ in $G$, and
$Y_k$ the number of cycles of length $k$ in a random graph of size $k$.
Note that:
- ${\sf E}[X] = \sum_{k=2}^n {\sf E}[X_k]$.
- ${\sf E}[X_k] = {n \choose k} {\sf E}[Y_k]$.
- ${\sf E}[Y_k] = (k!/k)p^k = (k-1)!p^k$.
The last one is because a cycle of length $k$
is a permutation of numbers from $1$ to $k$
but the first vertex does not matter (a cycle does not have a first vertex).
The probability of each edge being present is $p$,
so the probability of $k$ edges being present is $p^k$.
We have:
\begin{align*}
{\sf E}[X] &= \sum_{k=2}^n E[X_k] \\
&= \sum_{k=2}^n {n \choose k} E[Y_k] \\
&= \sum_{k=2}^n {n \choose k} (k-1)! p^k \\
&= \sum_{k=2}^n \frac{n!}{k!(n-k)!} (k-1)! p^k \\
&= \sum_{k=2}^n \frac{n!}{(n-k)!k} p^k \\
&= \sum_{k=2}^n \frac{n\ldots (n-k+1)}{k} p^k \\
\end{align*}
For a simple lower bound
we can drop the first half of the sum:
\begin{align*}
&\geq \sum_{k=n/2}^{n} \frac{n\ldots (n-k+1)}{k} p^k \\
&\geq \sum_{k=n/2}^{n} \frac{(n/2)^{k}}{n} p^k \\
&\geq \frac{1}{n}\sum_{k=n/2}^{n} (np/2)^{k} \\
&= \frac{(\frac{np}{2})^{n+1}-(\frac{np}{2})^{\frac{n}{2}}}{\frac{np}{2}-1}
\end{align*}
If there are $3 \binom n3 p^3$ triangles in expectation, and $3 \binom n3 p^2$ connected triples, the global clustering coefficient should approach their ratio (which is $p$).
Of course, naively taking their ratios doesn't work: $\mathbb E[\frac {X}Y]$ is not the same thing as $\frac{\mathbb E[X]}{\mathbb E[Y]}$. This is one of the main challenges in dealing with the expected value of a ratio. Instead, we'll show that both quantities are concentrated around their mean, and proceed that way.
Let $X$ denote the number of triangles in $\mathcal G(n,p)$. It's easy to see if we properly define triangles that $\mathbb E[X] = 3\binom n3 p^3$, which for consistency with connected triplets I want to define as $3\binom n3$ choices of a potential path $P_3$, and a $p^3$ chance that both edges of the path and the edge that makes it a triangle are present.
Moreover, the number of triangles is $3n$-Lipschitz in the edges of the graph (changing one edge changes the number of triangles by at most $3n$) so by McDiarmid's inequality
$$
\Pr[|X - \mathbb E[X]| \ge n^{2.5}] \le 2 \exp \left(-\frac{2n^5}{\binom n2 (3n)^2}\right) \le 2 e^{-4n/9}.
$$
If we let $Y$ be the number of connected triplets, then the expected number of them is $3\binom n3 p^2$: there are $3\binom n3$ potential copies of $P_3$ in the graph, and each is realized with a probability of $p^2$. This is actually $2n$-Lipschitz in the edges of the graph (each edge is part of at most $2n$ paths on $3$ vertices) but we'll round that up to $3n$-Lipschitz so that the same bound applies.
As a result, with probability $1 - 4 e^{-4n/9}$, both values are within $n^{2.5}$ of their expected value, and so the ratio $\frac{X}{Y}$ satisfies
$$
\frac{3\binom n3 p^3 - n^{2.5}}{3\binom n3 p^2 + n^{2.5}} \le \frac{X}{Y} \le \frac{3\binom n3 p^3 + n^{2.5}}{3\binom n3 p^2 - n^{2.5}}
$$
and therefore $\frac XY = p + O(n^{-0.5})$ in these cases. The remaining cases occur with probability $4e^{-4n/9}$, and the ratio is between $0$ and $1$ even then, so they can affect the expected value by at most $4e^{-4n/9}$ as well. Therefore $\mathbb E[\frac XY] = p + O(n^{-0.5})$, which approaches $p$ as $n \to \infty$.
Best Answer
According to the paper titled "The Maximum Degree of a Random Graph" by Oliver Riordan and Alex Selby, given $b \geq 0$, the probability that the maximum degree of an Erdos--Renyi random graph $ER(N,p)$ is at most $Np+b\sqrt{Npq}$ is $\left(c+o(1)\right)^N$. Here, $q=1-p$ and $c=c(b)$ is the root of a certain equation. In particular, for $b=0$, $c=c(0)=0.6102\ldots$.