You need a new staff assistant, and you have $n$ people to interview. You want to hire the best candidate for the position. When you interview a candidate, you can give them a score, with the highest score being the best and no ties being possible. You interview the candidates one by one. Because of your company's hiring practices, after you interview the $k$ th candidate, you either offer the candidate the job before the next interview or you forever lose the chance to hire that candidate. We suppose the candidates are interviewed in a random order, chosen uniformly at random from all $n$ ? possible orderings.
We consider the following strategy. First, interview $m$ candidates but reject them all; these candidates give you an idea of how strong the field is. After the $m$ th candidate, hire the first candidate you interview who is better than all of the previous candidates you have interviewed.
(a) Let $E$ be the event that we hire the best assistant, and let $E_{i}$ be the event that $i$ th candidate is the best and we hire him. Determine $\operatorname{Pr}\left(E_{i}\right)$, and show that
$$
\operatorname{Pr}(E)=\frac{m}{n} \sum_{j=m+1}^{n} \frac{1}{j-1}
$$
(b) Bound $\sum_{j=m+1}^{n} \frac{1}{j-1}$ to obtain
$$
\frac{m}{n}(\ln n-\ln m) \leq \operatorname{Pr}(E) \leq \frac{m}{n}(\ln (n-1)-\ln (m-1)) .
$$
(c) Show that $m(\ln n-\ln m) / n$ is maximized when $m=n / e$, and explain why this means $\operatorname{Pr}(E) \geq 1 / \mathrm{e}$ for this choice of $m$.