How does one prove this?
Let $\{X_r: r\ge1\}$ be independent and identically distributed with distribution function $F$ satisfying $F(y)<1$ for all $y$, and let $Y(y)=\min\{k:X_k>y\}$. Show that
$$ \lim_{y\to \infty} \mathbb P(Y(y)\le \mathbb E(Y(y)))=1-\exp(-1) $$
A start:
\begin{align} \Pr(Y(y) = k) &= \Pr(X_1 \leq y, ..., X_{k-1} \leq y, X_k >y) = p(y)^{k-1}(1-p(y))\\ \mathrm{E}(Y(y)) &= \sum_{k=1}^\infty k p^{k-1}(1-p(y)) = (1-p(y)) \frac{\mathrm{d}}{\mathrm{d}p(y)} \frac{1}{1-p(y)} = (1-p(y)) \frac{1}{(1-p(y))^2}\\ &= \frac{1}{1-p(y)}\\ \Pr(Y(y) \leq \mathrm{E}(Y(y))&= \Pr\left(Y(y) \leq \frac{1}{1-p(y)}\right)\\ &= \sum_{k=1}^{\lfloor 1/(1-p(y)) \rfloor} p(y)^{k-1}(1-p(y))\\ &= (1-p(y)) \frac{1-p(y)^{\lfloor 1/(1-p(y)) \rfloor+1}}{1-p(y)}\\ &= 1-p(y)^{\lfloor 1/(1-p(y)) \rfloor+1} \end{align} I have used the notation $$ p(y) = \Pr(X_1 \leq y). $$