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Aasma is a mathematician and devised an algorithm to find a husband. The strategy is -

  • Start interviewing a maximum of $1000$ prospective husbands. Assign a ranking $r$ to each person that is a positive integer. No two prospects will have same the rank $r$.
  • Reject the first $k$ men and $H$ be the highest rank of these $k$ men.
  • After rejecting the first $k$ men, select the next prospect with a rank greater than $H$ and then stop the search immediately. If no candidate is selected after $999$ interviews, the $1000th$ person is selected.

Aasma wants to find the value of $k$ for which she has the highest possibility of choosing the highest ranking prospect among all $1000$ candidates without having to interview all $1000$ prospects.

  • $(a)(6\; points)$ What is the probability that the highest ranking prospect among all 1000 prospects is the $(m+1)th$ prospect?
  • $(b)(6\; points)$ Assume the highest ranking prospect is the $(m+1)th$ person to be interviewed. What is the probability that the highest rank candidate among the first $m$ candidates is one of the first $k$ candidates who were rejected?
  • $(c)(6\; points)$ What is the probability that the prospect with the highest rank is the $(m+1)th$ person and Aasma will chose the $(m+1)th$ person using this algorithm?
  • $(d)(16\; points)$ The total probability that Aasma will chose the highest tanking prospect among the $1000$ prospects is the sum of the probability for each possible value of $m+1$ with $m+1$ ranging between $k+1$ and $1000$. Find the sum. To simplify your answer use the formula $$\ln N = \frac{1}{N-1} + \frac{1}{N-2} + \cdots + \frac{1}{2} + \frac{1}{1} $$
  • $(e)(6\; points)$ Find that value of $k$ that maximizes the probability of chossing the highest ranking prospect without interviewing all $1000$ candidates. You may need to know that the maximum of the function $x\ln \frac{A}{x-1}$ is approximately $\frac{A+1}{e}$ where $A$ is a constant and $e$ is Euler's number, $e = 2.718.....$

Actually I didn't even get much time to read the problem, let alone solving it. I didn't even get the meaning of this problem. Any partial solution (in comment) or full solution will be helpful.

Note: This is a problem from BdMO $2016$ National Secondary $7/8$ and Higher Secondary $7/9$.

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    parts $a$ and $b$ are easy right? – 2017-01-06
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    how is this the hardest problem? they help you a ton, P.S. this is normally called the sultan drowry problem. – 2017-01-06
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    Hard to me.. Maybe This is not that hard – 2017-01-06
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    I think it is easier than most of the problems you have posted before. – 2017-01-06
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    Not particularly hard or anything, but it gave me good laughs... :D – 2017-01-06
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    That maybe true. Actually I didn't even get what I am supposed to do in this problem – 2017-01-06
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    whats the answer for a? – 2017-01-06
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    @JorgeFernándezHidalgo I am absolute newbie in Probability. I guess the answer maybe something like $\frac{somethin'}{1000}$ ? – 2017-01-06
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    The answer for $a$ is $\frac{1}{1000}$, every position is equally likely. – 2017-01-06
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    Yes. I got that ONLY :( – 2017-01-06
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    ok, so what do you think is the answer for part $b$? – 2017-01-06
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    [Numberphile](https://youtu.be/ZWib5olGbQ0) did a video on this problem (although phrased in terms of 100 festival toilets). – 2017-01-06
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    I Think it might be $\frac{1}{2}$. :( – 2017-01-06
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    @JorgeFernándezHidalgo, don't you think, that the answer to part (a) should be 1/(1000-k) ? Because, looking at the whole comprehension, we see a sentence (it is situated in part c) where it is given, " value of (m+1) ranges from (k+1) to 1000" – 2017-01-18

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