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For the binomial random graph $G(n,1/2)$, the chromatic number is $(1+o(1))n/(2{\log }_{2}n)$ with high probability. A first moment method bounds the independence number above. Edge-disjoint clique packing in the complement graph, followed by an edge-exposure martingale and a union bound over moderately large vertex sets, supplies independent sets for greedy colouring by removing independent sets.

We need independent sets in every sufficiently large remaining induced subgraph, rather than just one in the initial random graph. The useful concentration variable is the maximum edge-disjoint clique packing in the complement graph; changing one edge changes that variable by at most one. This avoids the much larger sensitivity of the ordinary clique count in a binomial random graph.

First work in a binomial random graph $H\sim G(m,1/2)$, take $r=r_0(m)$, and let $X$ count its $r$-cliques. Put $\mu =\mathbb{E}X\ge m^{9/5}$ and let $Z$ be its maximum edge-disjoint clique packing size. Form the clique conflict graph whose vertices are these cliques, adjacent when they share an edge. If $Q$ counts unordered conflicting pairs, the Caro-Wei bound gives, for each realization,To verify the bound, randomly order the vertices of a finite graph and retain each vertex preceding all its graph neighbours. These retained vertices are an independent set, with expected size ${\sum }_v(1+\mathrm{deg}v{)}^{-1}\ge X^2/(X+2Q)$ by the Cauchy-Schwarz inequality. Interpret the ratio as zero when $X=0$. A second application of the Cauchy-Schwarz inequality, now to expectations, givesCounting ordered pairs of $r$-sets with $j$ common vertices, including identical sets when $j=r$, yieldsThe shared edges account for the factor $2^{\left(\genfrac{}{}{0ex}{}{j}{2}\right)}$. We claimHere are the estimates, including the large overlaps. For $2\le j\le \lfloor r/2\rfloor$, the probability that a random $r$-set contains any specified $j$ elements is at most $(r/(m-r){)}^j$. Summing over their choices inside the first $r$-set givesThe logarithm of $T_j$ is a convex function, a quadratic in $j$, so the maximum on this interval is at an endpoint. We have $T_2=O(r^4/m^2)$. As $r=(2+o(1)){\log }_{2}m$, the other endpoint satisfiesand is smaller than $T_2$ for sufficiently large $m$. Summing at most $r$ terms gives $O(r^5/m^2)$.

For the other half of the overlaps, write $j=r-s$, where $0\le s\le \lceil r/2\rceil$. The exact identity and an upper bound areThe quantity in parentheses is $m^{-1/2+o(1)}$, which tends to zero. The geometric series is thus at most $2/\mu$ for sufficiently large $m$. This proves the claim. In fact $r^5/m^2=o(m^{-9/5})$ and $1/\mu \le m^{-9/5}$, so eventuallyNow expose the $N=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ independent edge indicators one at a time. The edge-exposure martingale $M_i=\mathbb{E}[Z\mid \text{first }i\text{ indicators}]$ starts at $\mathbb{E}Z$ and ends at $Z$. Deleting one edge destroys at most one member of any edge-disjoint clique packing. Hence changing one indicator changes $Z$ by at most one; coupling the remaining indicators shows $|M_i-M_{i-1}|\le 1$.

Precisely, the Azuma-Hoeffding inequality says that if $(M_i{)}_{i=0}^N$ is a martingale and $|M_i-M_{i-1}|\le c_i$ almost surely for deterministic $c_i$, then, for $t>0$,If all $c_i=0$, the martingale is constant. Applying the lower-tail inequality with $c_i=1$ and $t=\mathbb{E}Z$ givesfor sufficiently large $m$. This is the only martingale concentration inequality needed.

Return to the original binomial random graph $G\sim G(n,1/2)$ and set $m=\lceil n/({\log }_{2}n{)}^2\rceil$. For each fixed $m$-element vertex set $U$, the complement graph of $G[U]$ has distribution $G(m,1/2)$. The union bound therefore givessince $m^{8/5}/n\to \infty$. Thus with high probability every vertex set of size at least $m$ contains an independent set of sizeApply greedy colouring by removing independent sets: remove such an independent set, give it a fresh colour, and repeat while at least $m$ vertices remain. Give each of the fewer than $m$ final vertices its own colour. This uses at mostcolours, since $m=o(n/{\log }_{2}n)$. Together with the lower bound, this gives the chromatic number of the half-density binomial random graph: