expected number of balls withdrawn to get equal numbers of black and white balls There are $n$ black balls and $n$ white balls in a bin.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.
 A: Your conjecture is true. The following argument is not elegant, but it works!
The number of arrangements of $k$ white and $k$ black balls 
where the first equalization  occurs at $2k$ is $2 C_{k-1}$,
 where $C_{k-1}={1\over k}{2(k-1)\choose k-1}$ is the $k-1$th 
Catalan number.
The number of arrangements of $n$ white and $n$ black balls 
where the first equalization  occurs at $2k$ is therefore
$$ {2\over k}{2(k-1)\choose k-1} {2(n-k)\choose n-k}.\tag 1$$
All ${2n\choose n}$ arrangements are equally likely, so the chance that 
equalization first occurs at time $2k$ is
$$\mathbb{P}(T=2k)= {1\over{2n\choose n}}  {2\over k}{2(k-1)\choose k-1} {2(n-k)\choose n-k}.\tag 2$$
The expected time to equalization is therefore
$$\mathbb{E}(T)={1\over{2n\choose n}} 
\sum_{k=1}^n 2k\, {2\over k}\,{2(k-1)\choose k-1} {2(n-k)\choose n-k}.\tag 3$$
Cancelling the $k$'s in (3) and using the known identity
$$\sum_{k=1}^n {2(k-1)\choose k-1} {2(n-k)\choose n-k}=4^{n-1}\tag4$$
gives the result. 
A: Suppose that the first ball drawn is white and that you stop on draw number $2k$. Then the $2k$-th ball drawn was black, and the $2k-2$ balls drawn in positions $2$ through $2k-1$ form a Dyck word of length $2k-2$. Conversely, any sequence of $2k-2$ white and black balls in which the number of black balls never exceeds the number of white balls can occupy those $2k-2$ positions. Thus, there are $C_{k-1}$ sequences of draws beginning with a white ball that terminate with draw $2k$. If we imagine continuing until the bin is empty, there are $\binom{2n-2k}{n-k}$ ways to complete the draw, for a total of $C_{k-1}\binom{2n-2k}{n-k}$ full draws that start with a white ball and first balance (at equal numbers of white and black balls) on draw $2k$. There is an equal number starting with a black ball, so 
$$2C_{k-1}\binom{2n-2k}{n-k}=\frac2k\binom{2k-2}{k-1}\binom{2n-2k}{n-k}$$
of the $\binom{2n}n$ possible full draws first balance on draw $2k$. The expected number of draws to the first balanced sample is therefore
$$\binom{2n}n^{-1}\sum_{k=1}^n\frac2k\binom{2k-2}{k-1}\binom{2n-2k}{n-k}(2k)=4\binom{2n}n^{-1}\sum_{k=1}^n\binom{2k-2}{k-1}\binom{2n-2k}{n-k}\;,$$
and your conjecture is equivalent to 
$$4^{n-1}=\sum_{k=1}^n\binom{2k-2}{k-1}\binom{2n-2k}{n-k}=\sum_{k=0}^{n-1}\binom{2k}k\binom{2n-2k-2}{n-k-1}$$ or, after replacing $n-1$ by $n$, to
$$4^n=\sum_{k=0}^n\binom{2k}k\binom{2n-2k}{n-k}\;.$$
You can find a proof of this identity in this question together with an outline of a combinatorial proof; this answer gives a full combinatorial proof.
A: I would start by defining $f(n,m), n \ge m$ as the expected number of draws starting from $n$ white balls and $m$ black balls with $n-m$ black balls in hand.  Then we have $f(n,n)=1+f(n,n-1)$
$f(n,0)=n$
$f(n,n-1)=\frac n{2n-1}+\frac{n-1}{2n-1}(1+f(n,n-2))$
$f(n,m)=1+\frac m{n+m}f(n,m-1)+\frac n{n+m}f(n-1,m)$
where the cases are checked in this order and seek a solution.  No guarantees that this will work.
