Distribution like hypergeometric distribution, but with false replacements I'm trying to find probabilities for a scenario that works exactly like hypergeometric distribution, except that every time you make a draw, you replace it with a "failure"-object. So eventually you'd end up with no successes among the population. What I'd like to describe is the probability of a certain number of successes given a certain number of draws.
An example of this situation could be: You're pulling marbles out of an urn. There are green marbles (successes) and white marbles. Everytime you pull out a white marble (failure) you just put it back but every time you pull out a green marble (success) you replace it with a white marble. So the number of marbles remain constant, but the number of successes decreases each time you draw a success.
Do we have a probability distribution to describe this?
 A: Suppose an urn of $n$ balls begins with $s$ successes.  What are the chances it will end up with $t$ successes ($0 \le t \le s$) after $d$ draws?  
Ignoring the trivial case $s=0$, this is a Markov chain on the numbers of successes $s$.  The chance of a transition from $s$ to $s-1$ is $p(s,s-1)=s/n$; otherwise, the state stays the same: $p(s,s)=1-s/n$.
The $n+1\times n+1$ transition matrix $\mathbb{P} = p(s,t)$ (indexing from $0$ through $n$) readily decomposes as
$$\mathbb P = \mathbb{V}\operatorname{Diagonal}\left(\frac{n}{n}, \frac{n-1}{n}, \ldots, \frac{1}{n}, \frac{0}{n}\right) \mathbb{V}^{-1}$$
where $$\mathbb{V} = (v_{ij}, i=0,\ldots,n; j=0,\ldots, n) = \left(\binom{i}{j}\right)$$
is Pascal's Triangle and $$\mathbb{V}^{-1} = (v^{-1}_{ij}) = \left((-1)^{i+j}\binom{i}{j}\right)$$
contains the same values but with alternating signs.
The chance of a transition from $s$ to $t$ after $d$ steps, $p_n(s,t)$, is found from the $s,t$ entry in $\mathbb{P}^n$ (again indexing from $0$ through $n$).  But we easily compute
$$\mathbb{P}^n= \mathbb{V}\operatorname{Diagonal}\left(\frac{n^d}{n^d}, \frac{(n-1)^d}{n^d}, \ldots, \frac{1^d}{n^d}, \frac{0^d}{n^d}\right)\mathbb{V}^{-1}.$$
Consequently, omitting all terms obviously zero,
$$p_d(s,t) = n^{-d}\sum_{j=t}^s \binom{s}{j} (n-j)^d (-1)^{j+t}\binom{j}{t}.$$

These figures plot the chance of reaching an urn with $t$ successes starting with $s=100$ successes in $n=100$ total.  The numbers of draws are 10,50,150,and 250.  As they increase, the distribution moves to the left more and more slowly, first spreading and then contracting to zero.
In terms of $t$, these are generalized hypergeometric distributions.
