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For any set $X$, let $[X]^2 = \{\{x,y\}:x\neq y\in X\}$. For $n\in\mathbb{N}$, let $[n]^2 = [\{1,\ldots,n\}]^2$. If $n\in\mathbb{N}$ is a positive integer and $S\subseteq [n]^2$, we set $$\chi(S) :=\min\{\kappa\in\omega\setminus\{0\}:(\exists f:\mathbb{N}\to\kappa)(\forall t\in S)\big|\text{im}(f|_t)\big| =2\}.$$ The expected chromatic number of a graph on $n$ vertices is defined by $$E_n:= \frac{1}{|{\cal P}([n]^2)|}\sum\{\chi(S):S\subseteq [n]^2\},$$ where of course $|{\cal P}([n]^2)| = 2^{n(n-1)/2}$.

What is the value of $\lim\inf_{n\to\infty}E_n/n$?

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  • $\begingroup$ You should really write ${X \choose 2}$ instead of $[X]^2$. Anyways, the chromatic number of $G(n,\frac{1}{2})$ is very concentrated around $\frac{\log 2}{2}\frac{n}{\log n}$, so much so that $\bigl(1+o(1)\bigr)\frac{\log 2}{2}\frac{n}{\log n}$ is its mean. So the answer to your question is $0$. (Look up "Erdos-Renyi" graph). $\endgroup$ Commented 9 hours ago

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