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I believe that this question is an extension of this other one, and is vaguely reminscent of the German tank problem.

Let us assume I am dealing with an organisation, and on each encouter I deal with a random person. After (say) 100 encounters I note that I have met some people more than once so something like:

Times met same person       Number of occurences
1                           60
2                           12
3                            4
4                            1

(The figures above are made-up). What would be the best way of estimating the total number of people in the organisation?

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  • $\begingroup$ That looks like the results after 83*1 + 12*2 + 4*3 + 1*4 = 123 encounters. So it would help to say either "after 123 encounters" or "after 100 encounters with new people", depending on how you are determining the stopping time. $\endgroup$
    – user44143
    Commented Oct 24, 2016 at 21:34
  • $\begingroup$ Doh! Sorry @MattF. Well spotted. I can't even add up. Corrected the question. $\endgroup$
    – TimGJ
    Commented Oct 24, 2016 at 21:44
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    $\begingroup$ I think you should give an initial distribution of chances for each amount of people. Since this can't be evenly, there is no natural assumption. $\endgroup$
    – Lucas K.
    Commented Oct 24, 2016 at 22:03
  • $\begingroup$ This is a standard problem with many applications. If you search for "estimating population size" at Google Scholar, you will find that there is a very large literature on it. $\endgroup$
    – Brendan McKay
    Commented Oct 25, 2016 at 0:20
  • $\begingroup$ @BrendanMcKay It was actually Googling exactly that term which brought me to mathoverlow :). It did accur to me that Capture Recapture seems to be sort of what I am after, but not quite. If anyone can post the name of an algorithm or other suitable search term that would be great. $\endgroup$
    – TimGJ
    Commented Oct 25, 2016 at 6:29

1 Answer 1

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Here is an answer via maximum likelihood: choose the population size for which the observed distribution would be most likely.

Let $n$ be the population size. There are 77 people encountered, whom we identify as: person 1, the person encountered four times; person 2, the first person encountered three times; ... person 77, the sixtieth person encountered once. There are therefore ${n \choose 77}77!$ ways of choosing these people.

The probability of observing this distribution is therefore $$\frac{{n \choose 77}77!f(1,4,12,60)}{n^{100}}$$ where $f$ is the number of arrangements of those 77 people into 100 observations so that person 1 appears four times, person 2 appears three times and is the first to appear three times, ... and person 77 appears once and is the last person to appear only once. Since all we have to do is maximize the probability, we can ignore the $77!$ and the $f$.

Numerically, ${ n \choose 77}/n^{100}$ is maximized for $n=181$. So 181 is a reasonable estimate for the population size.

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