278 lines
8.1 KiB
TeX
278 lines
8.1 KiB
TeX
\section{The Secretary Problem}
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\definition{The secretary problem}
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Say we need to hire a secretary. We have exactly one position to fill,
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and we must fill it with one of $n$ applicants. These $n$ applicants,
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if put together, can be ranked unambiguously from \say{best} to \say{worst}.
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\vspace{2mm}
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We interview applicants in a random order, one at a time. \par
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At the end of each interview, we either reject the applicant (and move on to the next one), \par
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or select the applicant (which fills the position and ends the process).
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\vspace{2mm}
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Each applicant is interviewed at most once---we cannot return to an applicant we've rejected. \par
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In addition, we cannot reject the final applicant, as doing so will leave us without a secretary.
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\vspace{2mm}
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For a given $n$, we would like to maximize our probability of selecting the best applicant. \par
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This is the only metric we care about---we do not try to maximize the rank of our applicant. \par
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Hiring the second-best applicant is no better than hiring the worst.
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\problem{}
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If $n = 1$, what is the best hiring strategy, and what is the probability that we hire the best applicant?
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\begin{solution}
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This is trivial. Hire the first applicant, she's always the best.
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\end{solution}
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\vfill
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\problem{}
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If $n = 2$, what is the best hiring strategy, and what is the probability that we hire the best applicant? \par
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Is this different than the probability of hiring the best applicant at random?
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\begin{solution}
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There are two strategies:
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\begin{itemize}
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\item hire the first
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\item hire the second
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\end{itemize}
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Both are equivalent to the random strategy.
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\vspace{2mm}
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Intuitively, the fact that a strategy can't help us makes sense: \par
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When we're looking at the first applicant, we have no information; \par
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when we're looking at the second, we have no agency (i.e, we \textit{must} hire).
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\end{solution}
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\vfill
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\problem{}
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If $n = 3$, what is the probability of hiring the best applicant at random? \par
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Come up with a strategy that produces better odds.
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\begin{solution}
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Once we have three applicants, we can make progress.
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\vspace{2mm}
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The remark from the previous solution still holds: \par
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When we're looking at the first applicant, we have no information; \par
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when we're looking at the last, we have no choices.
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\vspace{2mm}
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So, let's make our decision at the second candidate. \par
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If we hire only when the second candidate is better than the first, \par
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we end up hiring the best candidate exactly half the time.
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\vspace{2mm}
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This can be verified by checking all six cases.
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\end{solution}
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\vfill
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\pagebreak
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%
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% MARK: Page
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%
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\problem{}<bestyet>
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Should we ever consider hiring a candidate that \textit{isn't} the best we've seen so far? \par
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Why or why not? \hint{Read the problem again.}
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\begin{solution}
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No! A candidate that isn't the best yet cannot be the best overall! \par
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Remember---this problem is only interested in hiring the \textit{absolute best} candidate. \par
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Our reward is zero in all other cases.
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\end{solution}
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\vfill
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\remark{}
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\ref{bestyet} implies that we should automatically reject any applicant that isn't
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the best we've seen. We can take advantage of this fact to restrict the types of
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strategies we consider.
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\remark{}
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Let $B_x$ be the event \say{the $x^\text{th}$ applicant is better than all previous applicants,} \par
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and recall that we only know the \textit{relative} ranks of our applicants: \par
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given two candidates, we know \textit{which} is better, but not \textit{by how much}.
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\vspace{2mm}
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Therefore, the results of past events cannot provide information about future $B_x$. \par
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All events $B_x$ are independent.
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\vspace{2mm}
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We can therefore ignore any strategy that depends on the outcomes of individual $B_x$.
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Given this realization, we are left with only one kind of strategy: \par
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We blindly reject the first $(k - 1)$ applicants, then select the next \say{best-yet} applicant. \par
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All we need to do now is pick the optimal $k$.
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\problem{}
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Consider the secretary problem with a given $n$. \par
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What are the probabilities of each $B_x$?
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\vfill
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\problem{}<seca>
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What is the probability that the $n^\text{th}$ applicant is the overall best applicant?
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\begin{solution}
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All positions are equally likely. $\nicefrac{1}{n}$.
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\end{solution}
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\vfill
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\pagebreak
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%
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% MARK: Page
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%
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\problem{}<secb>
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Given that the $x^\textit{th}$ applicant is the overall best, what is the probability of hiring this applicant \par
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if we use the \say{look-then-leap} strategy detailed above? \par
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\hint{
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Under what conditions would we \textit{not} hire this applicant? \par
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This probability depends on $k$ and $x$.
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}
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\begin{solution}
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Say that the $x^\text{th}$ applicant is the best overall. If we do not hire this applicant,
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we must have hired a candidate that came before them. \par
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\vspace{2mm}
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What is the probability of this? We saw $x-1$ applicants before the $x^\text{th}$. \par
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If we hired one of them, the best of those initial $x-1$ candidates did \textit{not} fall
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into the initial $k-1$ applicants we rejected.
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\note{(This is again verified by contradiction: if the best of the first $x-1$ applicants
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\textit{was} within the first $k-1$, we would hire the $x^\text{th}$)}
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\vspace{2mm}
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There are $x-1$ positions to place the best of the first $x-1$ candidates, \par
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and $k-1$ of these positions are initially rejected. \par
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Thus, the probability of the best of the first $x-1$ applicants being rejected is $\frac{k-1}{x-1}$.
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\vspace{2mm}
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Unraveling our previous logic, we find that the probability we are interested in is also $\frac{k-1}{x-1}$. \par
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\note{Assuming that $x \geq k$. Of course, this probability is 0 otherwise.}
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\end{solution}
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\vfill
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\problem{}<phisubn>
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Consider the secretary problem with $n$ applicants. \par
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If we reject the first $k$ applicants and hire the first \say{best-yet} applicant we encounter, \par
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what is the probability that we select the best candidate? \par
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Call this probability $\phi_n(k)$.
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\begin{solution}
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Using \ref{seca} and \ref{secb}, this is straightfoward:
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\[
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\phi_n(k)
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= \sum_{x = k}^{n}\left( \frac{1}{n} \times \frac{k-1}{x-1} \right)
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\]
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\end{solution}
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\vfill
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\problem{}
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Find the $k$ that maximizes $\phi_n(k)$ for $n$ in $\{1, 2, 3, 4, 5\}$.
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\begin{solution}
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Brute force. We already know that $\phi_1(1) = 1.0$ and $\phi_2(1) = \phi_3(2) = 0.5$. \par
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The maximal value of $\phi_4$ is $\phi_4(2) = 0.46$, and of $\phi_5$ is $\phi_5(3) = 0.43$.
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\end{solution}
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\vfill
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\pagebreak
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%
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% MARK: Page
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%
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\problem{}
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Let $r = \frac{k-1}{n}$, the fraction of applicants we reject. Show that
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\begin{equation*}
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\phi_n(k)
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= r \sum_{x = k}^{n}\left( \frac{1}{x-1} \right)
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\end{equation*}
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\begin{solution}
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This is easy.
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\end{solution}
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\vfill
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\problem{}
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With a bit of fairly unpleasant calculus, we can show that the following is true for large $n$:
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\begin{equation*}
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\sum_{x=k}^{n}\frac{1}{x-1}
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~\approx~ \text{ln}\Bigl(\frac{n}{k}\Bigr)
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\end{equation*}
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Use this fact to find an approximation of $\phi_n(k)$ at large $n$ in terms of $r$. \par
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\hint{If $n$ is big, $\frac{k-1}{n} \approx \frac{k}{n}$.}
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\begin{solution}
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\begin{equation*}
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\phi_n(k)
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~=~ r \sum_{x = k}^{n}\left( \frac{1}{x-1} \right)
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~\approx~ r \times \text{ln}\left(\frac{n}{k}\right)
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~=~ -r \times \text{ln}\left(\frac{k}{n}\right)
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~\approx~ -r \times \text{ln}(r)
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\end{equation*}
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\end{solution}
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\vfill
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\problem{}
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Find the $r$ that maximizes $\underset{n \rightarrow \infty}{\text{lim}} \phi_n$. \par
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Also, find the value of $\phi_n$ at this point. \par
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\note{If you aren't familiar with calculus, ask an instructor for help.}
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\begin{solution}
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Use the usual calculus tricks:
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\begin{equation*}
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\frac{d}{dr} \bigl( -r \times \text{ln}(r) \bigr)
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= -1 - \text{ln}(r)
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\end{equation*}
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Which is zero at $r = e^{-1}$. The value of $ -r \times \text{ln}(r)$ at this point is also $\frac{1}{e}$.
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\end{solution}
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\vfill
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Thus, the \say{look-then-leap} strategy with $r = e^{-1}$ should select the best candidate about $e^{-1} = 37\%$ of the time,
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\textit{regardless of $n$.} Our probability of success does not change as $n$ gets larger! \par
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\note{Recall that the random strategy succeeds with probability $\nicefrac{1}{n}$. \par
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That is, it quickly becomes small as $n$ gets large.}
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\pagebreak
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