Two urns of balls, expected amount of remaining balls. There are two urns with n red balls and n blue balls respectively.
You will first randomly choose a urn and then draw one ball from it. Repeat the process till one of the urns is empty for the first time. What is the expected amount of remaining balls, that is, how many balls are left in the non-empty urn on average?
 A: There is a close form for the expectation. As mentioned above, $Y=2n-X$ has probability mass function
$$
\mathbb{P}(Y=k)= {k-1\choose n-1} \frac{1}{2^k}, \; k\in\{n,n+1,\cdots,2n-1\}
$$
We can directly compute
$$
\mathbb{E}(Y) = \sum_{k=n}^{2n-1} k {k-1\choose n-1} \frac{1}{2^{k-1}} \\
              = 2n \sum_{k=n}^{2n-1}  {k\choose n} \frac{1}{2^{k}}
$$
Now we notice that $\sum_{k=n}^{2n}  {k\choose n} \frac{1}{2^{k}}=1$. (WHY? let $N$ be the number of tosses needed to get at least n+1 heads or tails. What is the probability $\mathbb{P}(N = k+1)$?) Therefore, we have
$$
\mathbb{E}(Y)=2n\left(1-{2n\choose n} \frac{1}{2^{2n}}\right) \quad \text{and} \quad
\mathbb{E}(X) = 2n {2n\choose n} \frac{1}{2^{2n}}
$$
Using Stirling approximation, we finally see that
$$
\mathbb{E}(X) = 2n {2n\choose n} \frac{1}{2^{2n}} \sim 2\sqrt{\frac{n}{\pi}}
$$
A: Let $X$=The number of trials until one of the urns is empty. We want to find the distribution of X. We know that $X$ must be at least $n$ and at most $2n-1$.
$$P(X=n)=2\left(\dfrac{1}{2}\right)^n$$
$$P(X=n+1)=2\left[ {n \choose n-1}\left(\frac{1}{2}\right)^{n-1}\left(\frac{1}{2}\right)\cdot \left(\frac{1}{2}\right)\right] = 2 \left[{n \choose n-1}\left(\frac{1}{2}\right)^{n+1} \right]$$
Continuing this we find that 
$$P(X=x) = 2\left[ {x-1 \choose n-1}\left(\frac{1}{2}\right)^x \right], \;x \in \{n,n+1,...,2n-1\}$$
We need to find the expected value of $X$ because the answer to the problem (what is the expected number of ball in the remaining urn) is $2n - E(X)$. I wasn't able to find a closed form for $E(X)$. If we let $Y=X-n$ then we get that 
$$ P(Y=y)=2{n+y-1 \choose n-1}\left(\frac{1}{2} \right)^{n+y} = 2{n+y-1 \choose y}\left(\frac{1}{2} \right)^{n+y}, \; y \in \{0,1,...,n-1\} $$
I thought it would be easier to find $E(Y)$ because $Y$ ranges from $0$ to $n-1$ versus $X$ which ranges from $n$ to $2n-1$ but I still wasn't able to come up with a closed form. If you could, then $E(X) = E(Y)+n$ and the answer to the original problem would be $2n- E(X) = n - E(Y)$. 
