I came across the following claim and proof in this paper, and I really don't follow.
If $G$ is a vertex-transitive graph with independence number $\alpha$ and chromatic number $\chi$ then
$n/α(G) ≤ χ(G) ≤ n(1 + ln(n))/α(G)$
Proof (of the rightmost inequality): Let $A$ be an independent set of size $α = α(G)$; then the probability that $m = ⌈ln(n)⌉$ random translates of $A$ (by automorphisms) do not cover $V(G)$ is less than $n · (1 − α/n)^m < n · e^{−αm/n} ≤ 1$.
More specifically, I don't understand where the chromatic number comes into it, or why the final inequality leads to something of the form $n(1 + ln(n))/α(G)$. I understand the meaning of vertex transitivity, and both $\alpha$ and $\chi$, but the logic of the proof itself doesn't make sense to me.
Since the probability that $m$ random translates of $A$ by automorphisms do not cover $V(G)$ is less than $1$, there must be some cover of $V(G)$ by $m$ translates of $A$. Each translate of $A$ is an independent set, so each translate of $A$ has chromatic number $1$, and it follows that
$$\chi(G)\le m=\lceil\ln n\rceil<1+\ln n\le(1+\ln n)\frac{n}{\alpha(G)}\;,$$
since clearly $\alpha(G)\le n$ and therefore $\frac{n}{\alpha(G)}\le 1$.