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The secretary problem is a well-studied optimal stopping problem with a simple solution. Suppose a set of $N$ candidates are interviewed for a secretarial problem, one at a time, in random order. Each interviewee must be either accepted on the spot or rejected for good. The goal is to select the single best candidate (i.e., the payoff is 1 if the best candidate is hired, and 0 otherwise). Under these conditions, what is the best hiring policy?

As it turns out, the optimal strategy is to interview the first $N/e$ candidates and reject them, and then hire the next candidate that is better than all of those. Surprisingly, this strategy leads to hiring the single best candidate one out of every $e$ times (or about $37\%$ of the time), regardless of $N$.

Now suppose that instead of being interviewed one at a time, the secretaries can be interviewed in groups of up to $k$ (for some fixed $k>1$), after which one of the group may be accepted on the spot or the entire group may be rejected for good. It is not hard to see that this strictly increases the probability of hiring the best candidate. The question is, by how much? Does the probability of making the best hire still tend to $1/e$, or is there a non-zero advantage over the $k=1$ case in the limit of large $N$?

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2 Answers 2

up vote 4 down vote accepted

This paper explains it better than I could.

http://www3.stat.sinica.edu.tw/statistica/oldpdf/A10n216.pdf

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I would have thought that for fixed $k$, the probability of getting the best candidate from a total of $kN$ candidates interviewed $k$ at a time would be the same as that of getting the best of $N$ candidates interviewed $1$ at a time, i.e. converging on $1/e$ as $N$ increases.

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That's a nice reduction: From each group of $k$ candidates, consider the best, and ignore the others. –  Douglas Zare Mar 21 '11 at 7:47

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