Suppose we have an urn that contains $r$ red and $g$ green balls. At each time we draw a ball, we look at the color, replace it, and add $c$ more balls of that color.
Let $X_n$ be the ratio of green balls to the total number of balls after the $n^{th}$ draw.
It is not hard to see that $X_n$ is a martingale, and since $X_n \geq 0$, it will converge absolutely to some $X_\infty$.
It is also not hard to see that the probability of getting a green ball on the first $m$ draws and a red ball on the next $l = n-m$ draws is $$\frac{g}{g+r}\cdot \frac{g+c}{g+r+c} \cdots \frac{g+(m-1)c}{g+r+(m-1)c} \cdot \frac{r}{g+r+mc}\cdot \frac{r+c}{g+r+(m+1)c}\cdots\frac{r+(l-1)c}{g+r+(n-1)c}.$$
Also, any other outcome of $n$ total draws with drawing $m$ green and $l$ red balls will have the same probability.
Can anyone help me see why the above probability should be $$\frac{\beta(\frac{g}{g+r+c},\frac{r}{c+n-m})}{\beta({\frac{g}{c}, \frac{r}{c}})},$$ where $\beta(a,b) = \int_0^1 x^{a-1}(1-x)^{b-1} dx$?
I know that $\gamma(a+1)=a\gamma(a)$, but I can't figure out a nice way to compute the betas.
I think the required equality is wrong. For instance, when $g=r=c=m=l$ then the product of the fractions equals $\frac 16$, whereas $\frac {B\left(\tfrac 13,\tfrac 12\right)} {B\left(1,1\right)}= B\left(\tfrac 13,\tfrac 12\right)=4.2\dots$.
On the other hand, I obtained a different expression with $B$-function for the product.
Below we shall use a well-known equality that $B(x,y)=\tfrac{\Gamma(x) \Gamma(y)}{ \Gamma(x+y)}$ for any $x,y>0$. Thus
$$\frac{g}{g+r}\cdot \frac{g+c}{g+r+c} \cdots \frac{g+(m-1)c}{g+r+(m-1)c} \cdot \frac{r}{g+r+mc}\cdot \frac{r+c}{g+r+(m+1)c}\cdots\frac{r+(l-1)c}{g+r+(n-1)c}=$$ $$\frac{g/c}{g/c+r/c}\cdot \frac{g/c+1}{g/c+r/c+1} \cdots\frac{g/c+(m-1)}{g/c+r/c+(m-1)/c}\times$$ $$\frac{r/c}{g/c+r/c+m}\cdot \frac{r/c+1}{g/c+r/c+(m+1)}\cdots\frac{r/c+(l-1)}{g/c+r/c+(n-1)}=$$ $$\frac{\tfrac{\Gamma(g/c+m)}{\Gamma(g/c)}\cdot \tfrac{\Gamma(r/c+l)}{\Gamma(r/c)}} { \Gamma(g/c+r/c+n)/\Gamma(g/c+r/c)}=\frac{B(g/c+m,r/c+n-m)}{B(g/c,r/c)}.$$