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We have a hall with $N$ secretaries. We choose randomly one secretary at a time and interview them. After the interview either we hire the secretary (and all others go home) or move to the next one (we can't go back and hire them later).

The interviewer wants to hire either the best or second best secretary with the highest probability

My approach:

I know we should use this strategy: after we interview the first $aN$ candidates (which we don't hire), then we choose the first one that's better than the previous $aN$ candidates. I am just stuck knowing what is the value of $a$. What is the probability of this value? and how we get to it?

I have seen the "Secretary Problem" where we want to choose the best one, the difference here is that we are satisfied with the best or second best one equally

Xander Henderson
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Adddison
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  • I have seen the 'secretary problem' ,this is where I got my approach from. I just want to know how things go when we want to choose the best or second best one. @hardmath – Adddison Nov 25 '18 at 17:36
  • I take your point, that you are asking about a variant of the Secretary Problem in which we replace the goal of choosing the best candidate with choosing the best or the second best. Variants of the Secretary Problem have been considered here as well, and there is a Comment by Michael Chernick that includes your version. I will vote to reopen, but you might explain the reasoning for having the same strategy as in the classic version. – hardmath Nov 25 '18 at 18:38
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    The comment does not have an answer,he is just asking.I could not find any question similar to mine.I don't why it got marked as duplicate and thank you for reopening.@hardmath – Adddison Nov 25 '18 at 21:00
  • Now that the Question has been reopened, I'd like to post an Answer. However your Question does contain a misconception that the optimal probability is achieved by determining coefficient $a$ and choosing "the first one that's better than the previous $aN$ candidates." In order to maximize the chances of getting the best or the second best of those available, one need to determine not only an initial number $aN$ of interviews to conduct, but a second number of interviews after which one hires someone who is better than all but at most one of the previous secretaries. – hardmath Nov 27 '18 at 03:40
  • Please advise me if the problem includes the restriction to making the job offer strictly on the basis of "one threshold" $aN$ as described, or if you are interested instead in the optimal way to make the job offer (subject to the usual conditions of the Secretary Problem apart from the goal of best or second best hire). – hardmath Nov 27 '18 at 19:22

1 Answers1

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I'd started preparing a response, but the OP seems to have lost interest. In any case for the benefit of future Readers here are my initial thoughts. If there are Comments I'll be glad to fill in more details.

The variant of the Secretary Problem considered here appears in the literature as early as 1966 with publications by:

Although both of these links lead to papers behind paywalls, as far as I can tell, a good summary of the results by Gusein-Zade can be gleaned from this open access technical report by Frank and Samuels (1979) . In particular as the number of secretaries $N \to \infty$, the asymptotic chance of success for the classic problem $1/e \approx 0.3679$ rises to roughly $0.5736$ when choosing either the best or second-best candidate counts as a success.

The general rule that maximizes the chance of getting one of the $b$-best secretaries is shown by Gusein-Zade to have the form of $b+1$ consecutive intervals subdividing the interviews $1,\ldots,N$.

In the present case $b=2$ and there are three phases of the rule. In the first phase the interviews $1 \le i \le N\cdot m_1(N)$ are conducted and the relative rankings observed without stopping, In the second phase the interviews $N\cdot m_1(N) \lt i \le N\cdot m_2(N)$ are conducted and only stop if one interview results in a relative "top" ranking (better than all previous interviews) happens, in which event the interviews are ended with a job offer being made. If the second phase finishes without a job offer, the third and final phase of interviews $N\cdot m_2(N) \lt i \le N$ applies the rule that a job offer is made if and only if an interview results in a relative ranking of "top" or "next to top" (better than all but at most one of the previous interviews).

It is useful to express the boundaries between phases as coefficients $m_1(N),m_2(N)$ because these multipliers have fairly simple limits as $N\to \infty$, namely as Gusein-Zade showed:

$$ \lim_{N\to \infty} m_2(N) = \frac{2}{3} $$

and:

$$ \lim_{N\to \infty} m_1(N) = \varphi $$

where $\varphi$ is the unique solution in $(0,1)$ of:

$$ \varphi - \ln \varphi = 1 - \ln(2/3) $$

Now $\varphi \approx 0.3470$, and the exact limiting probability of success for large $N$ is $\varphi (2-\varphi)$, whose approximate value $0.5736$ was mentioned before.

hardmath
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