On the set $[n]:= \{1,\ldots,n\}$ we consider the set $${\cal P}\_2([n]) = \big\{\{a,b\}: a,b \in [n], a\neq b\big\}.$$ 
Since $$|{\cal P}\_2([n])| =2^{n \choose 2}$$ there are exactly $2^{n\choose 2}$ graphs on $n$ points. 


We set the clique number $\omega(G)$ to be the largest $n$ such that the complete graph $K\_n$ is a subgraph of $G$. For $n \geq 1$, let $$R\_n = \{ G = ([n], E): E\in {\cal P}\_2([n]) \text{ and } \omega(G) = \chi(G)\},$$ and set $q\_n = \frac{R\_n}{2^{n\choose 2}}$.


Does $\lim\_{n\to\infty} q\_n$ exist, and is its value equal to $1$?



Do graphs with $\omega(G) = \chi(G)$ grow "common" as $|V|$ grows large?
I would say that this limit is zero. Most of the graphs on $n$ vertices have $\sim n^2/4$ edges. For such graph, the expected number of complete subgraphs of size $k\sim2\log\_2n$ is
$$
 {n\choose k}\cdot \frac1{2^{k\choose 2}}\approx C\left(\frac{ne}{k2^{(k-1)/2}}\right)^k\to 0.
$$
So most of our graphs have $\omega(G)\lesssim 2\log\_2n$. Their complements $G^c$ also have $\sim n^2/4$ edges, and the same estimate for $\omega(G^c)$ holds. This means that most of such graphs have $\chi(G)\gtrsim n/(2\log\_2n)$.


