There are $n$ black balls and $n$ white balls in a bin. I withdraw the balls one at a time without replacement until I have an equal number of white and black balls. What is the expected number of balls that I have to withdraw?
It appears that the answer should be $4^n\left/{2n \choose n}\right.$. So for $n = 3$ it would be:
$$4^3\left/{6 \choose 3}\right. = \frac{64}{20} = \frac{16}{5}$$
I have verified the answers for $n = 2$ and $n = 3$, but I am not able to prove the general result.