First colour to be drawn $n$ times from a hypergeometric distribution Consider a set up of drawing balls without replacement from a population of $R$ red balls and $B$ black balls. Is anyone aware of an efficient way to calculate the probability, $p$, that I draw $n$ red balls before I draw $n$ black balls?
The simple answer is to just sum over the possible ways of getting to $n-1$ red balls before $n$ black balls and then drawing another red ball,
$$
p = \sum\limits_{b=0}^{n-1} f_{hyp}(n-1; n+b-1, R+B) \frac{R-n+1}{R+B-n-b+1} 
$$
where
$$
f_{hyp} (r; r+b, R+B) = \frac{{R \choose r} {B \choose b}}{R+B \choose r+b}
$$
is the probability of drawing $r$ red balls from a total of $r+b$ draws from a population of size $R+B$.
However, for increasing $n$, this approach slows rather quickly. Can anyone provide some insight into finding either:


*

*a good approximation to the problem

*a more computationally efficient representation of the exact solution

 A: Following nsrt's suggestion of drawing $2n-1$ balls and checking whether there are at least $n$ red balls (or equivalently fewer than half the balls are black), you can produce a simpler expression $$\dfrac{\displaystyle\sum_{r=n}^{2n-1}{{R \choose r} {{B} \choose {2n-1-r}}}}{{R+B \choose 2n-1}} \text{ or equivalently } \dfrac{\displaystyle\sum_{b=0}^{n-1}{{R \choose {2n-1-b}} {{B} \choose {b}}}}{{R+B \choose 2n-1}} $$ 
This partial sum involves a generalised hypergeometric function which probably adds little to the ease of calculation.
For an approximation to the hypergeometric cumulative distribution, the mean of the number of black balls from a sample of $2n-1$ is $\mu_\text{black}= (2n-1)\tfrac{B}{B+R}$ and the standard deviation is $\sigma=\sqrt{(2n-1) \frac{ B  R (B+R-2n+1) }{(B+R)^2  (B+R-1)  }}$, and you might assume that a normal distribution with a continuity correction gave a reasonable approximation so look at $$\Phi^{-1}\left(\dfrac{n-\frac12 - \mu_\text{black}}{\sigma}\right)$$ where $\Phi^{1}()$ is the inverse of the cumulative distribution function of a standard normal.
For example, if $R=10$, $B=25$ and $n=5$ then the R language function phyper gives the result 
phyper(n-1, B, R, 2*n-1)
[1] 0.05251478

while my suggested approximation gives 
pnorm(n-1/2, mean=(2*n-1) * B / (B+R), 
      sd=sqrt((2*n-1) * B * R * (B+R-2*n+1) / ((B+R)^2 * (B+R-1))) )
[1] 0.05183762

which looks reasonably close to me. Some experimentation suggests the absolute error in the approximation may often be worst for given $R$ and $B$ when $n=1$, but this can be avoided by taking the obvious exact result $\frac{R}{R+B}$ instead. 
