3

Several authors have proved that the optimal strategy in the secretary problem is a stopping rule whereby the interviewer rejects the first $r-1$ applicants (let $M$ be the best applicant among these $r-1$), then hires the first subsequent applicant who is better than $M$. The argument can be found here.

In the special case where there are only two applicants, this strategy clearly gives the interviewer a success probability of $\frac{1}{2}$. However, it can be shown (via a similar argument to this one) that there is actually a strategy that gives the interviewer a success probability of strictly greater than $\frac{1}{2}$.

Why is there a contradiction here? Why is the 'optimal' strategy not optimal when there are two applicants? Indeed, how do we know that we can't use some similar approach that improves (however slightly) upon the 'optimal' strategy when there are more than two applicants?

  • 3
    Without any knowledge about the topic - most likely the argument is made for case where $n$ is in some sense "large". Wiki article mention rejecting the first $n/e$ participants - it would therefore make sense to have $n>>e$. – Piotr Benedysiuk Feb 21 '19 at 14:57
  • 1
    Rejecting the first $n/e$ applicants is asymptotically the best approach when the stopping rule is implemented. The proofs I have seen for the optimality of this strategy are not considering $n$ large. See, for example, Gilbert & Mosteller's analysis (pp. 39-40) - in particular the table on p. 40. –  Feb 21 '19 at 15:10
  • I also have little knowledge about the topic, but if you don't have control over the value of each dowry/secretary, then in the $n=2$ case, the distance between the two values $a$ and $b$ can be arbitrarily small. The better solution has probability $\frac{1}{2}+\frac{1}{2}P(a<y<b)$, and since the second term can be arbitrarily small, then it isn't a better solution in general. – Kevin Long Feb 21 '19 at 16:43
  • I don't follow, sorry. The second term is always strictly greater than $0$ because it's the integral of a PDF on a finite interval. True, it'll inevitably be extremely close to $0$ - but it certainly won't be precisely $0$. It would only be $0$ if $a=b=\infty$, or if $a=b$, neither of which are true by assumption. –  Feb 21 '19 at 20:07

1 Answers1

1

In the most abstract version of the secretary problem, you cannot measure each candidate with a number. Instead, as each candidate comes in, you can insert him/her into the sorted list (total order) of all previous candidates. I don't see how you can use the envelop paradox trick in this scenario.

You can add your own model and assign numbers to the candidates you've seen, but those are your numbers, and the problem doesn't care. The initial random permutation is equivalent to this: suppose you have seen $t$ candidates and they are ranked. Now the $(t+1)$th candidate comes in. Then he/she is equally likely to be inserted into any of the $(t+1)$ places in the existing list. So for the $n=2$ case the 2nd candidate is equally likely to be better or worse. Imagine you flip a fair coin and to decide if the 2nd candidate is better or worse. That's the problem setting. Now suppose you model the first candidate as score $S$ and draw a random number from $N(0,1)$ and compare etc. This doesnt help at all - the probability is still 0.5. If you claim to do better than 0.5, then you are claiming you can predict a fair coin flip better than 0.5, which is of course impossible. The key is that the 2nd candidate did not have a number attached a-priori. Alternatively, you can imagine that given your choice of 1st candidate's score $S$, the universe then decides the 2nd candidate's score $S_2$ to be $S \pm 1$ based on a fair coin flip.

Yet another alternative view: suppose the candidates actually secretly have scores $1, 2, ... n$, but unfortunately for you, those are not revealed to you as they come along. Instead, the only info revealed to you is the ordering of everyone you've seen so far. Again, you assigning them some other arbitrary scores of your choice doesn't help. The $(t+1)$th candidate is still equally likely to be in any of the $(t+1)$ possible insertion locations into the existing list of $t$ previous candidates.

In other versions maybe each candidate is a number (dowry?) and the numbers are a-priori attached to the candidates and revealed to you as they come along, in which case you are right that the envelop paradox trick can apply. Indeed, I have not seen that version analyzed fully incorporating the envelop paradox trick. Hmm...

antkam
  • 15,363
  • Well you can model each applicant's competency by a (different) positive real number and it comes to the same thing. Anyway, this hasn't answered my question... –  Feb 22 '19 at 03:42
  • @GuyLough I have added more explanation. Does this help? – antkam Feb 22 '19 at 13:46
  • My question refers to your 'Hmm...' at the end of your answer. As far as I'm concerned, by assuming that each applicant has some measurable competency, and modelling these competencies by distinct real numbers, the problem doesn't change at all. The interviewer still doesn't know the relative rank of each applicant until he interviews them. I feel like the issue is something subtle about the notion of a 'random' real number (because there's no uniform distribution over the reals). –  Feb 22 '19 at 14:28
  • As far as I can see, this answer is correct. If all you know is the relative rankings of the candidates, the $n/e$ method is optimal. If you're given some numerical values for the candidates, the envelope method does better. If you're given only relative rankings but make up (consistent) numerical values as you go along, $n/e$ remains optimal. – Andreas Blass Feb 22 '19 at 14:53
  • @GuyLough - IMHO, the envelop paradox is about the subtle notion of a 'random' real number, but the secretary problem, as usually stated (i.e. only the relative ranking is revealed to you), is not about the same subtle notion. however, a modified secretary problem where the numbers are revealed to you, would involve the same subtle notion. as you have clearly pointed out, the modified problem is not equivalent to the standard problem. as far as i know, such a modified version (for $n>2$) has not been fully analyzed anywhere. – antkam Feb 23 '19 at 04:34
  • anyway, i hope i have answered the question you asked, even if it is not the question you meant. :) my answer: there is no contradiction because the standard secretary problem is not susceptible to the subtle issue exploited in the envelop trick. so all authors who analyzed the standard problem and claimed you can only get to 0.5 for $n=2$, are in fact correct. if any authors also claimed this solves the modified problem, then they are wrong because they forgot about the very subtle issue raised by the envelop trick. the two problems are not equivalent -- as you clearly pointed out. – antkam Feb 23 '19 at 04:42