Suppose we have an urn with $x$ white and $y$ black balls, with frequencies $p$ and $q$, respectively. There are 2 players who draw 2 balls each and write down the number $k$ of black balls they got, i.e. $k \in \lbrace 0,1,2 \rbrace$, such that we record $k_1$ for player 1 and $k_2$ for player 2.
I want to know the probability to draw each of the possible pair combinations of $(k_1,k_2)$. However, I also want to distinguish 3 cases:
(a) Each draw happens independently.
(b) Only one draw is independent, but in the other draw both players accidentally draw the same ball while they grab it (it's a narrow urn). It's not distinguished if this happened on the first or second draw, and both players still write down $k$ as if there were 4 balls drawn.
(c) Both players accidentally draw the same ball both times.
If I am not mistaken, I can model (a) as the joint probability of the binomial $$f(k) = {2\choose{k}} p^{2-k} q^k $$ in the multinomial $$ f_pair(a_0, a_1, a_2) = {2\choose{a_0, a_1, a_2}} \prod^{2}_{k=0} f(k)^{a_k} $$ where $a_k$ is the number of $k$'s in a $(k_1,k_2)$ pair.
But I do not have the right idea for (b) or (c). Perhaps somehow according to the Polya urn model?
Case A:
The number of black balls drawn by each player follows a binomial distribution, i.e.
$ f(k_1) = {2\choose k_1} p^{2-k_1} q^{k_1} $
$ f(k_2) = {2\choose k_2} p^{2-k_2} q^{k_2} $
Since the players' draws are independent, then $ f(k_1,k_2) = f(k_1) \times f(k_2) $.
Case B:
We can think about this situation as three independent events: 1) player 1 and player 2 select the same ball, 2) player 1 draws a ball during the independent draw, and 3) player 2 draws a ball during the independent draw.
Let $ t $ be the number of black balls drawn during the dependent draw. If both players draw the same ball, then there are only two possible outcomes: $ (k_1,k_2) = (0,0) $ and $ (k_1,k_2) = (1,1) $. Then we get:
$ f(t) = f(k_1=t,k_2=t) = p^{1-t} q^{t} $, where $ t \in \lbrace 0,1 \rbrace$
Let $ t_1 $ and $ t_2 $ be the number of black balls drawn by player 1 and player 2, respectively, during the independent draw. We have:
$ f(t_1) = p^{1-t_1} q^{t_1} $, where $ t_1 \in \lbrace 0,1 \rbrace$
$ f(t_2) = p^{1-t_2} q^{t_2} $, where $ t_2 \in \lbrace 0,1 \rbrace$
Since these three events are independent, we can multiply them to get the joint probability of these three events, i.e.
$ f(k_1 = t+t_1, k_2 = t+t_2) = p^{1-t} q^{t} \times p^{1-t_1} q^{t_1} \times f(t_2) = p^{1-t_2} q^{t_2} = p^{1-(t+t_1+t_2)} q^{t+t_1+t_2} $
If you try all values of $ t $, $ t_1 $, and $ t_2$, you get the following probability distribution:
Case C:
If both players draw the same ball both times, then there are only three possibles outcomes: $ (k_1,k_2) = (0,0) $, $ (k_1,k_2) = (1,1) $, and $ (k_1,k_2) = (2,2) $. It is easiest to think about this as a binomial experiment for one draw (as we did for case A).
$ f(k) = f(k_1,k_2) = {2\choose k} p^{2-k} q^{k} $