Drawing without replacement, a special case I have the following problem: you pick a set $x$ containing $|x|$ elements from a bag, containing $k$ marbles of $m$ possible types. Once a marble of a certain color is drawn, however, all other marbles of that color are removed from the bag.
Q: What is the probability of drawing a certain sequence $x$ from the bag?
At first sight I thought I'd be able to model this as a hypergeometric distribution but I can't make it fit my question. As such, my current approach is different: assume the length of the set to be fixed, then model this as a sequence of conditional distributions:
$P({x_1 ... x_{|x|}}) = \frac{1}{Z}\sum_{\sigma(x)} \prod_{j=1}^{|x|} P(x_j | x_1 \ldots x_{j-1})$
The sum over $\sigma$ represents the sum over all possible permutations of the set $x$. Z is a normalizing constant that normalizes over all possible selections of length $|x|$:
$Z = \binom{m}{|x|}$
How should I, however, represent the conditional probability? Currently I'm at:
$P(x_j | x_1 \ldots x_{j-1}) = \frac{P(x_j = 1)}{\sum_{k=1}^{m} P(x_k = 1) - \sum_{k=1}^{j-1} P(x_k = 1)}$
But I don't think that would lead to a proper distribution? Any suggestions?
 A: I may have misunderstood your notation: you seem to use $k$ in two different senses and I am not sure whether $x_j$ is the $j$th type drawn or something more general.   So I have slightly adapted your notation here. 
Suppose your ordered sequence of drawn types is $\left(x(1), x(2), x(3), \ldots, x(|x|)\right)$ and at the start there are $k(1), k(2), k(3), \ldots, k(|x|)$ of each drawn type respectively, effectively the weights associated with drawing the type. 
The conditional probability of the next being type $x(j)$ given the ordered types already drawn  is $$\Pr\left(x(j) | (x(1), \ldots x(j-1))\right) = \frac{k(j)}{k-\displaystyle\sum_{i \lt j}k(i)} = \frac{\Pr\left(\text{first drawn was }x(j)\right)}{1 - \displaystyle\sum_{i \lt j} \Pr\left(\text{first drawn was type }x(i)\right) }$$
where "was" really means "might have been".
Then the probability of this sequence is the product of these, giving $$\prod_{j=1}^{|x|} \left(\frac{k(j)}{k-\displaystyle\sum_{i \lt j}k(i)}\right)$$ where the product of the denominators depends on the order of the drawing though the product of the numerators does not. 
If you wanted a probability for an unordered sequence of drawn types, you would have to sum over all the possible orders of these types, and while not doubt there are approximations (you have something like a multivariate version of Wallenius' noncentral hypergeometric distribution when there is one ball of each type but each ball has a different weight), to get an exact result I suspect you may need to do the sum.   
I do not see a need for a normalising constant.
