Confusion about Banach Matchbox problem While trying to solve Banach matchbox problem, I am getting a wrong answer. I dont understand what mistake I made. Please help me understand.
The problem statement is presented below (Source:Here)

Suppose a mathematician carries two matchboxes at all times: one in his left pocket and one in his right. Each time he needs a match, he is equally likely to take it from either pocket. Suppose he reaches into his pocket and discovers that the box picked is empty. If it is assumed that each of the matchboxes originally contained $N$ matches, what is the probability that there are exactly $k$ matches in the other box?

My solution goes like this. Lets say pocket $1$ becomes empty. Now, we want to find the probability that pocket $2$ contains $k$ matches (or $n-k$ matches have been removed from it. I also note that wikipedia solution does not consider the $1^{st}$ equality -- maybe thats where i am wrong?).
Let
$p = P[k\ \text{matches left in pocket}\ 2\ |\ \text{pocket 1 found empty}]$
= $\frac{P[k\ \text{matches left in pocket}\ 2\ \text{and pocket 1 found empty}]}{\sum_{i=0}^{n}P[i\ \text{matches left in pocket}\ 2\ \text{and pocket 1 found empty}]}$
= $\frac{\binom{2n-k}{n} \cdot \frac{1}{2^{2n-k}}}
{\sum_{i=0}^{n}\binom{2n-i}{n} \cdot \frac{1}{2^{2n-i}}}$
In my $2^{nd}$ equality, I have written the probability of removing all matches from pocket $1$ and $n-k$ from pocket $2$ using Bernoulli trials with probability $\frac{1}{2}$. The denominator is a running sum over a similar quantity.
Now, my answer to the original problem is $2p$ (the role of pockets could be switched). I am unable to see whats wrong with my approach. Please explain. 
Thanks
 A: Apart from doubling $p$ at the end, your answer is correct: your denominator is actually equal to $1$. It can be rewritten as
$$\frac1{2^{2n}}\sum_{i=0}^n\binom{2n-i}n2^i=\frac1{2^{2n}}\sum_{m=n}^{2n}\binom{m}n2^{2n-m}=\frac1{2^{2n}}\sum_{i=0}^n\binom{n+i}n2^{n-i}\;,$$
and
$$\begin{align*}
\sum_{i=0}^n\binom{n+i}n2^{n-i}&=\sum_{i=0}^n\binom{n+i}n\sum_{k=0}^{n-i}\binom{n-i}k\\\\
&=\sum_{i=0}^n\sum_{k=0}^{n-i}\binom{n+i}i\binom{n-i}k\\\\
&=\sum_{k=0}^n\sum_{i=0}^{n-k}\binom{n+i}n\binom{n-i}k\\\\
&\overset{*}=\sum_{k=0}^n\binom{2n+1}{n+k+1}\\\\
&=\sum_{k=n+1}^{2n+1}\binom{2n+1}k\\\\
&=\frac12\sum_{k=0}^{2n+1}\binom{2n+1}k\\\\
&=2^{2n}\;,
\end{align*}$$
where the starred step invokes identity $(5.26)$ of Graham, Knuth, & Patashnik, Concrete Mathematics. Thus, your result can be simplified to
$$p=\binom{2n-k}n\left(\frac12\right)^{2n-k}\;.$$
And you don’t want to multiply this by $2$: no matter which pocket empties first, this is the probability that the other pocket still contains $k$ matches.
A: I just wanted to add something from myself. I think it is easier to solve it without calculating conditional probabilities. Let $X$ be a random variable that we are interested in. $\{X=k\}$ in one of two cases: 


*

*left pocket contains $k$ matches and right one is already empty 


or


*right pocket contains $k$ matches and left one is already empty 


The first event takes place if smoker has already taken $n-k$ matches from left pocket AND $n$ matches from right pocket AND is going to check his right pocket. Accordingly, it means that he checked his pockets $n-k+n=2n-k$ times in total, out of which $n-k$ are done at left pocket and $n$ are at right one. 
From binomial distribution, this probability is equal to
$$
C_{2n-k}^{n}\left(\frac{1}{2}\right)^{n-k}\times \left(\frac{1}{2}\right)^{n}\cdot \frac{1}{2}=C_{2n-k}^{n}\left(\frac{1}{2}\right)^{2n-k+1}
$$
The second case is tackled in the same way if we swap left and right pockets, therefore
$$
\mathbb{P}(X=k)=2\times C_{2n-k}^{n}\left(\frac{1}{2}\right)^{2n-k+1}=C_{2n-k}^{n}\left(\frac{1}{2}\right)^{2n-k}
$$
