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 ${n \choose 77}$ ways of choosing these people.

The probability of observing this distribution is therefore $$\frac{{n \choose 77}f(1,4,12,60)}{n^{100}}$$ where $f$ is the number of arrangements of those 77 people in which person 1 appears four times, person 2 appears three times and appears three times before anyone else, ... 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 $f$.

Numerically, the probability is maximized for $n=181$. So 181 is a reasonable estimate for the population size.

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.

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: 1) the person encountered four times, 2) the first person encountered three times, ... 77) the sixtieth person encountered once. There are therefore ${n \choose 77}$ ways of choosing these people.

The probability of observing this distribution is therefore $$\frac{{n \choose 77}f(1,4,12,60)}{n^{100}}$$ where $f$ Is the number of ways of getting the first person to appear four times, the second to appear three times, ... the seventy-seventh to appear once. Since all we have to do is maximize the probability, we can ignore $f$.

Numerically, the probability is maximized for $n=181$. So 181 is a reasonable estimate for the population size.

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 ${n \choose 77}$ ways of choosing these people.

The probability of observing this distribution is therefore $$\frac{{n \choose 77}f(1,4,12,60)}{n^{100}}$$ where $f$ is the number of arrangements of those 77 people in which person 1 appears four times, person 2 appears three times and appears three times before anyone else, ... 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 $f$.

Numerically, the probability is maximized for $n=181$. So 181 is a reasonable estimate for the population size.