What is the probability that a marble from the urn has been picked up by exactly $n$ people? An urn starts with $m$ marbles and is then approached by $p$ people, each of which picks up $k$ marbles, discarding one and returning the rest to the urn.  The urn now has $m - p$ marbles remaining.  Let $f(n)$ be the probability that a marble randomly selected from the urn has been picked up by exactly $n$ people.
What is $f$? If it can't be solved exactly, can it be characterized?
What happens if we replace $k$ with a random variable?
Is there a more natural way to define this function, for example, fixing $n$ and defining $g(p)$?
 A: This is not a complete answer, but it might still be helpful.
From what I understand, $f(n)$ only considers the $m-p$ marbles that where never discarded. In other words, if we consider all $m$ marbles in the pot initially, we have
$$f(n)=P(\text{marble }i\text{ is selected }n\text{ times} \mid \text{marble }i \text{ is never discarded})$$
I put emphasis on this because I feel like the following is then a logical first step to compute $f(n)$
$$f(n)=\frac{P(\text{marble }i\text{ is selected }n\text{ times AND marble }i \text{ is never discarded})}{P(\text{marble }i \text{ is never discarded})}$$
The probability that a random marble is never discarded during the process is simply $\frac{m-p}{m}$, so the only difficult thing is the numerator.
First of all, let me shorten the notation:
$$P(\text{marble }i\text{ is selected }n\text{ times AND marble }i \text{ is never discarded})=P(\text{selected by }n \wedge ND)$$
where ND stands for Never Discarded.
The only viable strategy I see is now to consider all the different ways a marble can be selected $n$ times out of $p$. There are $\binom{p}{n}$ of those and let $C_n^p$ be the set of all of them, i.e.
$$C_n^p=\{(a_1,a_2,\ldots a_n)\subseteq \{1,2,3,\ldots,p\}\}$$
Hence we have
$$P(\text{selected by }n \wedge ND)=\sum_{(a_1,a_2,\ldots a_n)\in C_n^p}P(\text{selected by persons }(a_1,a_2,\ldots a_n) \wedge ND)$$
EXAMPLE
Before we go on, let's look at an example: Let there be $m=6$ marbles, $p=3$ people and each one selects $k=2$.
Now, let's first look at $n=2$, so we look at when 2 out of the 3 people select the marble. The $\binom{p}{n}=3$ possibilies are $(1,2),(1,3)$ and $(2,3)$.
Let's go step by step in the $(1,2)$ case
$$P(\text{selected by persons }(1,2) \wedge ND)=\\
P(\text{selected by 1})\cdot P(\text{not discarded by 1 after selected})\cdot P(\text{selected by 2})\cdot P(\text{not discarded by 2 after selected})\cdot P(\text{not selected by 3})=\\
\frac{2}{6}\cdot\frac{1}{2}\cdot\frac{2}{5}\cdot\frac{1}{2}\cdot\frac{2}{4}=\\
\color{blue}{\frac{k}{m}\cdot\frac{k-1}{k}\cdot\frac{k}{m-1}\cdot\frac{k-1}{k}\cdot\frac{m-2-k}{m-2}=}\\
\color{blue}{\frac{(m-p)!}{m!}(k-1)^n(m-2-k)}$$
The other ones can be calculated similarly and give
$$P(\text{selected by persons }(1,3) \wedge ND)=\frac{(m-p)!}{m!}(k-1)^n(m-1-k)\\
P(\text{selected by persons }(2,3) \wedge ND)=\frac{(m-p)!}{m!}(k-1)^n(m-k)$$
The total is now
$$P(\text{selected by }2 \wedge ND)=\frac{(m-p)!}{m!}(k-1)^n(m-k+m-k-1+m-k-2)$$
Now for $n=1$, we again split into 3 possibilities. The calculations are similar to before:
$$P(\text{selected by person }(1) \wedge ND)+P(\text{selected by person }(2) \wedge ND)+P(\text{selected by person }(3) \wedge ND)=\\
\frac{(m-p)!}{m!}(k-1)^n(m-1-k)(m-2-k)+\frac{(m-p)!}{m!}(k-1)^n(m-k)(m-2-k)+\frac{(m-p)!}{m!}(k-1)^n(m-k)(m-1-k)=\\
\frac{(m-p)!}{m!}(k-1)^n[(m-1-k)(m-2-k)+(m-k)(m-2-k)+(m-k)(m-1-k)]
$$
Back to theory
Given what we saw in the example, it seems to make more sense to enumerate the indices which are not selected, so we rewrite
$$P(\text{selected by }n \wedge ND)=\sum_{(a_1,a_2,\ldots a_n)\in C_n^p}P(\text{selected by persons }(a_1,a_2,\ldots a_n) \wedge ND)\\
=\sum_{(b_1,b_2,\ldots b_{p-n})\in C_{p-n}^{p}}P(\text{NOT selected by persons }(b_1,b_2,\ldots b_{p-n}) \wedge ND)$$
If in a given term, person $i$ selects the ball, they add a factor $\frac{k}{m-i+1}\frac{k-1}{k}$ and if they don't, they add the factor $\frac{m-i+1-k}{m-i+1}$.
All this combined leads me to the expression
$$P(\text{selected by }n \wedge ND)=\frac{(m-p)!}{m!}(k-1)^n \sum_{(b_1,b_2,\ldots b_{p-n})\in C_{p-n}^{p}}\prod_i (m-b_i+1-k)$$
I am not sure if this is correct and if it is, if it can be simplified...
