Repeated samples of multiple draws without replacement between draws or samples We have a urn containing $N$ marbles of $m$ colours, i.e. $N = \sum_{i=1}^{m} K_i$.
If we draw $n$ marbles from the urn without replacement, the probability that we have a specific distribution $(k_1,k_2, ..., k_m)$ (where $n = \sum_{i=1}^{m}k_i$) is given by the multivariate hypergeometric distribution:
$$
P(k_1,k_2, ..., k_m) = \frac{\prod_{i=1}^{m}\binom{K_i}{k_i}}{\binom{N}{n}} 
$$
What happens if i then take additional draws without replacing the marbles removed by the first draw?
I know conditional probability can be used if we assume that the first draw was the same composition as the second draw as we know the composition of what we removed in the first draw, i.e.
$$
P((k_1,k_2, ..., k_m),(k_1,k_2, ..., k_m)) = P(k_1,k_2, ..., k_m) \frac{\prod_{i=1}^{m}\binom{K_i-k_i}{k_i}}{\binom{N-n}{n}} = \frac{\prod_{i=1}^{m}\binom{K_i}{k_i}}{\binom{N}{n}}\frac{\prod_{i=1}^{m}\binom{K_i-k_i}{k_i}}{\binom{N-n}{n}}
$$
What I really want is the probability that sample $j$ consisting of $n$ draws without replacement from the urn will be of composition $(k_1,k_2, ..., k_m)$ but so that I can use it to compute a bulk property. If I was just looking for the single sample we could re-arrange so that sample $j$ is sample 1 and just use a hypergeometric.
I have a function $f(k_1,k_2, ..., k_m)$ that computes a scalar property from the sample and so I want to compute the expected value of $f$ when we exhaustively sample the urn:
$$
\hat{f} = \sum f(k_1,k_2, ..., k_m)\hat{Z}(k_1,k_2, ..., k_m)
$$
Where $\hat{Z}$ is the expected number of draws of composition $(k_1,k_2, ..., k_m)$ which I am assuming is going to be $\frac{N}{n}P'(k_1,k_2, ..., k_m)$ where $P'$ is the probability I am after.
(and if that turns out to be the multivariate hypergeometric distribution too that's fine... but I would need to know why it is valid as intuitively to me that doesn't seem correct)
Update
Here are some examples:
If we have an urn with 3 red, 5 blue and 4 green marbles
The sum for $\hat f$ when the sample size is 4 will need to consider $(3,1,0)$, $(3,0,1)$, $(2,2,0)$, $(2,1,1)$, $(2,0,2)$, $(1,3,0)$, $(1,2,1)$, $(1,1,2)$, $(1,0,3)$, $(0,4,0)$, $(0,3,1)$, $(0,2,2)$, $(0,1,3)$, $(0,0,4)$
In the case of $(2,1,1)$ only one sample of 4 per emptying of the urn can match because once 2 red marbles have been removed there is only 1 red marble remaining in the urn
In the case of $(1,2,1)$ at most two samples of 4 in any given emptying of the urn can match. An emptying of the urn that has two of the three emptyings matching $(1,2,1)$ will contribute twice to $\hat f$ in the brute force evaluation
 A: The distribution is the same for every draw.  (I'm assuming that the number $n$ of items in a draw divides the total number $N$ of items.)  Imagine that the $N$ items are drawn one at time, without replacement, and placed in a line.  What is the probability that items $kn+1$ through $k(n+1)$ have composition $(k_1,k_2, ..., k_m)$?  Obviously this is just a random selection of $n$ of the $N$ items.
Therefore the average value of $f$ on any draw of $n$ marbles is $$A := \frac{n!(N-n)!}{N!}\sum f(k_1,k_2, ..., k_m)\binom{K_1}{k_1}\binom{K_2}{k_2}\cdots\binom{K_m}{k_m}$$ where the sum is over all compositions $k_1+k_2+\cdots+ k_m=n$.
Since we are drawing without replacement, top compute the probability of the composition $(k_1,\dots k_m)$ we want the number of ways to to pick $k_1$ items from  $K_1$, $k_2$ from $K_2$, and so on, divided by the number of ways to pick $n$ items from $N$.
We make $\left\lfloor\frac Nn\right\rfloor$ draws, so the expected value of $f$ on all the draws is $$\left\lfloor\frac Nn\right\rfloor A$$
Although, as indicated in the update, the outcome of a draw of $n$ items is dependent on the prior draws, we are only concerned with the expected value, and expectation is linear, even if the random variables are dependent, so we can simply add up the expectations of all the draws.
