Let $H$ be a $k$-uniform hypergraph with $m$ edges. For each vertex $v \in V(H)$, we independently sample $x_v \sim U([0,1])$, a uniformly random number in $[0,1]$. We then order the vertices in increasing order of $x_v$, and run a 2-color greedy coloring algorithm in that order. That is, we color a vertex blue unless it is the last vertex in an all-blue edge, in which case it is colored red.
For two edges $e,f \in E(H)$, let $\delta \in (0,1)$, and $\mathcal{E}_{e,f}$ be the event
$\mathcal{E}_{e,f} = \{|e \cap f|=1, \text{ the last vertex $v$ of $e$ is the first vertex of $f$, and } \qquad \qquad x_v \in [\frac{1}{2}(1 - \delta),\frac{1}{2}(1 + \delta)] \}$
Show that $\mathbb{P}(\mathcal{E}_{e,f}) \leq \delta2^{2−2k}$.
This is an exercise that I'm doing and I'd like to ask for some hints. I get quite confused and don't see how to go about analyzing the situation.
Another proof that I've read contains a somewhat simpler statement that, for the event $\mathcal{E}'_{e,f} = \{\text{the last vertex $v$ of $e$ is the first vertex of $f$} \}$, $\mathbb{P}(\mathcal{E}'_{e,f}) = \frac{(k-1)!(k-1)!}{(2k-1)!}$. I'd like to see the clarification for this as well. Thanks.
We may suppose $|e\cap f| = 1$, for otherwise $\mathbb{P}(\mathcal{E}_{e,f}) = 0$. Let $v \in e \cap f$. The probability that $x_v \in [\frac{1}{2}(1-\delta),\frac{1}{2}(1+\delta)]$ is $\delta$. With $x_v$ fixed, the probability that it is the last vertex of $e$ and the first vertex of $f$ is simply $x_v^{k-1}(1-x_v)^{k-1}$, which is at most $\frac{1}{4}^{k-1}$, since $y(1-y) \le \frac{1}{4}$ for all $y \in (0,1)$. So, you end up with $\delta 2^{2-2k}$.
For $\mathcal{E}_{e,f}'$, there are $(2k-1)!$ orderings of the vertices in $e\cup f$ and $(k-1)!(k-1)!$ have the vertices of $e$ coming first with $v$ last, followed by the vertices of $f$.