# Multivariate distribution from drawing balls with elimination of previously picked colors from urn at each step

Consider an urn with n balls of k colors ( labelled 1 through k ) where n >= k. You get to draw j times (j <= k) with the following action between draws. Each time you draw a ball belonging to a color, you remove all balls in the urn having that color. How can we characterize the multivariate distribution (an ordered set of j random numbers) resulting from this experiment?

To clarify a consequence of the rules of the experiment, the resulting multivariate whose properties I am seeking to understand will be of the form $X = \{ X_1, X_2 , ... X_j\}$ and will have the following obvious characters:

1. $X_i$ is an integer for all $1 \leq i \leq j$
2. $X_i \in [1,k]$
3. $X_i \neq X_j$ when $i \neq j$.

I am trying to understand if this falls under some known multivariate distribution families.

Probability of drawing first ball of $i$-th color is $p_i = n_i / n$ where $n_i$ is the number of balls of $i$-th color. In the next draw you have $n - n_i$ balls in the urn, so probability of drawing $j$-th ball is $\tfrac{p_j}{\sum_{m \notin \{i\}} p_m} = \tfrac{p_j}{1 - p_i} = \tfrac{n_j}{n-n_i}$, etc. The procedure is continued until you draw expected number of balls. The probability of drawing sequence $i,j,k$ is
\begin{align} \Pr(X_1 = i, X_2 = j, X_3 = k) & = p_i \cdot \frac{p_j}{1 - p_i} \cdot \frac{p_k}{1 - p_i - p_j} \\ & = \frac{n_i}{n} \cdot \frac{n_j}{n-n_i} \cdot \frac{n_k}{n-n_i-n_j} \end{align}
If you need formula for obtaining sample $i,j,k$ in any order, then you simply need to sum over probabilities of all the possible permutations of those items (this may not be managable for larger number of items).
If you are looking for a general, named, distribution, then the closest thing I recall is multivariate Wallenius' noncentral hypergeometric distribution with numbers of balls equal to $m_1 = m_2 = \dots = m_k = 1$ and weights equal to a priori probabilities of drawing each ball, i.e. $n_i/n$ for $i$-th ball.