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? (see, https://en.wikipedia.org/wiki/Banach%27s_matchbox_problem)
The solution on wikipedia includes the following equation:
Why does $2\cdot P(M=N-k)={2N-k\choose N-k}\left(\frac{1}{2}\right)^{2N-k}$ hold?
My idea:
If we forget for a moment the matchbox problem and simply consider a negative binomial distribution $P_{nb}$ where we have $N+1$ successes and ask for the probability of $m$-many failures, then we can define a conditional probability $P_{nb}(M=m\mid m<N+1)$. If we replace $m$ by $N-k$ then this conditional probability describes the matchbox problem. In other words \begin{align*} &\underset{\text{of matchbox problem}}{\underset{\text{probability distribution }}{\underbrace{P(K=k)}}}=\dots=P_{nb}(M=m\mid m<N+1)\\ &=\frac{P_{nb}(M=m~\cap ~m<N+1)}{P_{nb}(m<N+1)}=\frac{{m+N\choose m}\frac{1}{2^{m+N+1}}}{P_{nb}(m<N+1)}. \end{align*} Now a problem remains: How do we get $P_{nb}(m<N+1)$ without further information?