TITLE: Banach Matchbox Problem
QUESTION [15 upvotes]: Banach Matchbox Problem:
  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 for the first time 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?

I'm wondering whether the following reasoning is right, because it doesn't match up with the correct probability. But here goes my reasoning:
Assuming there are $k$ matches left in the other box, we have had to take $2n-k+1$ matches to notice there are none left. The total number of ways in which we could have picked those is $2 {2n - k \choose n}$, for either the matchbox in the left or right pocket has $k$ matches left inside, and in the $(2n-k+1)$th pick we would have found an empty box. 
The total number of possibilities, taken over all possible sizes $k$, would then be $\sum_{m=0}^{n} 2{2n-m \choose n}$. So I'd assume the overall probability would be $\frac{2 {2n - k \choose n}}{\sum_{m=0}^{n} 2 {2n-m \choose n}} =\frac{{2n - k \choose n}}{\sum_{m=0}^{n} {2n-m \choose n}}$.
However, the mentioned solution is ${2n - k \choose n} (\tfrac{1}{2})^{2n-k}$. Where is my reasoning wrong? Thanks in advance.

REPLY [5 votes]: Neither answer directly addresses the question posed: What is wrong with the reasoning proposed?
Here it is: In order to apply the "ratio" formula (nr. of favorable possibilities divided by total nr. of possibilities), those possibilities need to be equally likely, but this is not the case. Each of the  2C(2n-k, n) possibilities has a probability of (1/2)^(2n-k+1) to occur (n draws from one pocket, n-k from the other, one last from the first), which depends on k.
user76844 alludes to this in a comment to his answer: "Also, you are assuming that each possibility has the same probability (implicit in your use of strictly counting methods), whereas it is more likely that k will be near n than near 0..you didn't correct for the probability of a given k...its not a simple counting exercise."