Lets say there is a group of people and we don't know how big is this group.
Lets say that we are told that 3 people had a birthday today.
For simplicity we can assume that birthdays are uniformly distributed and there is 365 days in a year.
Can I, based on this information only, estimate the number of people in this group?
Putting it other way I would like to answer the questions:
Given that 3 people had a birthday today what is the probability that the group has 100 people?
Given that 3 people had a birthday today what is the probability that the group has 251 people?
So lets X be a random variable representing a number of people in a group then
P(X = x | number_of_people_who_had_birthday_today = n)
is a probability of a group being of size x given number of the people having birthday today is n.
It is obvious that:
P(X = 0 | number_of_people_who_had_birthday_today = 3) = 0
P(X = 1 | number_of_people_who_had_birthday_today = 3) = 0
P(X = 2 | number_of_people_who_had_birthday_today = 3) = 0
since because 3 people had a birthday there must be at least 3 people in this group.
I would like to find formula for P. But, I struggle to put my head around it.
If I would somehow be able to model the initial number of people in a room as some distribution with a given mean \mu. Lets say I would be coming back on some number of consecutive days and asking how many people had birthday today. Lets say that after 5 days I would get a list of answers [2,3,2,3,4].
I believe that I could use my initial distribution and the above list to somehow update my initial believe using Bayes theorem. But for that I would need P(N | x)(if I am not mistaken) but I don't know what it is.
In the classical statistics formulation of this, you're asking the wrong question. You can't ask for the probability that there are $m$ people in the room unless you have an a priori distribution for the number of people there (in which case you can use Bayes' theorem). What the classical statistician may ask is, what is an unbiased estimator for the number of people in the room? That is:
Let $M$ be the number of people in the room, and $X$ the number of these whose birthday is today. For simplicity, I'll neglect February 29, and suppose that each person independently chooses a birthday among $365$ possibilities, with equal probabilities. Then $\widehat{M} = 365 X$ is the minimum variance unbiased estimator of $M$ that depends only on $X$.
Of course, in real life things are somewhat messier. Not all days are equally likely to be birthdays (e.g. there is seasonal dependence, and there are relatively few births on holidays and weekends, because elective Caesarians will not be scheduled then).