I am trying to find a closed expression or at least a somewhat accessable solution to finding probabilities in the following die-urn setup:
We have five distinct urns (labeled $1$ through $5$ for simplicity), each containing balls in $n_{i}$ different colors (no individual color appears across multiple urns) and let $m_{i, j}$ be the number of balls of a given color ($j\in \{1, ..., n_{i}\}$). Additionally we are given five identical unfair dice with probabilities $p_{1}, ..., p_{5}$ to show the numbers $1, .., 5$ respectively.
One round consist of throwing the dice and randomly drawing as many balls from urn $i$ as there are dice showing $i$ (all values $m_{i, j}, p_{i}, n_{i}$ are known and some of them can be $0$).
Now the typical question would be to ask for the probability of a at least one of the drawn balls to have a specific color.
I can't really wrap my head around the dependencies here outside of actually counting cases in a hypothetically gigantic tree diagram. Above that this is only a simplified version of the setup where $k \leq 8$ players participate in the round one after the other and we try to calculate the same probability as before for the last player.
I would appreciate any insight into a dependent experiment of the given complexity as I have not found (I know of examples like coin-die etc. but those seem much more simplistic to me). Thanks in advance for everyone bothering.
Let us say that $j$, the color of interest, appears in urn $i$ only. The probability that exactly $k$ dice show up $i$ is $${5\choose k}p_i^k(1-p_i)^{5-k}$$ Now we need to sum the expected number of balls of color $j$ that will will drawn from an urn with $m_{i,j}$ balls of color $j$ among $n_i$ balls, if we draw $k$ balls without replacement. This is just $km_{ij}/n_i$ so altogether, we have $${m_{ij}\over n_i}\sum_{k=1}^5k{5\choose k}p_i^k(1-p_i)^{5-k}\tag{1}$$
The above assumes that each urn has at least $5$ balls in it. If this is not true, and we adopt the rule that if the number of dice showing $i$ is greater than $n_i$ then all balls in urn $i$ are drawn, we have just to amend $(1)$ to $${m_{ij}\over n_i}\sum_{k=1}^5\min(k,n_i){5\choose k}p_i^k(1-p_i)^{5-k}\tag{2}$$
Now, we don't really need the rule that each color appears in only one urn. If the colors are allowed to appear in more than one urn, the formula becomes $$\sum_{i=1}^5{m_{ij}\over n_i}\sum_{k=1}^5\min(k,n_i){5\choose k}p_i^k(1-p_i)^{5-k}$$