Pólya urn model and counting 'successes'

Question

Let's consider the Pólya urn model. Inside an urn there are two kinds of balls in two different colours - black and white (b - the amount of black ones, w - the amount of white ones.)
We draw one ball, then put it back with and add $d$ more balls in the same colour.
For example:
the possibility of drawing a white ball is equal to:
1) $P_1^w =\frac{w}{b+w}$ in the first draw,
2) $P_2^w = \frac{b+d}{b+w+d}$ in the second draw.
And so on.
I am to find out the possibility of extraction exactly $k$ black balls in $n$ trials.
I thought of using Bernoullie's scheme: $$P(S_n = k) = {n\choose k}p^k(1-p)^{n-k}$$ where the success is pulling out the black ball of course.
There is a problem however. It is easy to say how the denominator changes after each extraction, the k-th denomitor will look like this: $$p_k = \frac{\alpha}{b+w+(k-1)d}$$ The counter is the problem. I think that: $$p_k^b = \frac{b + md}{b+w+(k-1)d}$$ where $m$ can be any number from the set $\{ 0, 1, 2, 3, 4,...k-1\}$
I would appreciate any tips.

Answer 1

Note that Polya’s urn give rise to exchangeable sequences--- conceptually the probability measure of the outcome is invariant under permutation. For example, let $W_t$ and $B_t$ be the event that the $t$-th draw is white and black respectively, then observing $2$ white balls in $3$ draws is same regardless of the sequence it is observed: \begin{align} \Pr[W_1W_2B_3] = \frac{w}{w+b}\frac{w+d}{w+b+d}\frac{b}{w+b+2d}\\ =\Pr[W_1B_2W_3] = \frac{w}{w+b}\frac{b}{w+b+d}\frac{w+d}{w+b+2d}\\ =\Pr[B_1W_2W_3] = \frac{b}{w+b}\frac{w}{w+b+d}\frac{w+d}{w+b+2d}. \end{align}

Therefore, observing $k$ black balls in $n$ draws is equal to $$\sum_{\text{possible sequence}}\frac{\prod_{i=0}^{k-1} (b+id)\cdot \prod_{j=0}^{n-k-1} (w+jd)}{\prod_{i=0}^{n-1} (w+b+id)} = {n\choose k}\frac{\prod_{i=0}^{k-1} (b+id)\cdot \prod_{j=0}^{n-k-1} (w+jd)}{\prod_{i=0}^{n-1} (w+b+id)}$$