Find probabilty of having r matchsticks in one box when other is empty A mathematician carries a matchbox one each in both left and right pockets of his shirt. Initially,each matchbox contains $N$ matchsticks. Each time he wants a matchbox, he selects at random from the right or left pocket to get it.
Consider the moment when the mathematician discovers that a box is empty.
Find the probability that there are exactly $r$ matchsticks in one box when the other box is found empty.
My Attempt
When he finds that one box is empty and other contains $r$ matchsticks then he would have taken out a total of $2N-r$ matchsticks. But now how to get that $N$ matchsticks have come from one pocket.
 A: The number of routes starting at $(N,N)$ and ending at $(0,r)$ equals $\binom{2N-r}{N}$ and the same is true for the number of routes starting at $(N,N)$ and ending at $(r,0)$.
For each route the probability that this route will be taken equals $2^{r-2N}$.
That leads to probability $$\binom{2N-r}{N}2^{r-2N}\frac12+\binom{2N-r}{N}2^{r-2N}\frac12=\binom{2N-r}{N}2^{r-2N}$$that one of the routes $(N,N)\to(0,r)\to(-1,r)$  and $(N,N)\to(r,0)\to(r,-1)$ will be chosen.
A: The only random variate in this problem is X, which indicates the box from which the mathematician picks each matchstick. Let be $K$, a new random variate that counts the number of picks from box A, ie the number of times X indicates "box A" in the experiment. Then $K$ has a binomial distribution (more about the binomial distribution here).
So we have by definition:
$P(K=k) =\binom nk p^n(1-p)^{n-k}$ 
with $n$ the total number of picks, $k$ the number of picks from box A, $p$ the probability of picking from box A.
Of course, the event "Box A is empty" (ie $K=N$) is equivalent to the event "$r$ matchsticks remain in box B" by definition, ie $P(r)=P(K=N)$.
Let's simplify $P(K=k)$ for $k=N$ and $p=1/2$:
$P(r)=P(K=N)=\binom nN \frac{1}{2^n}$
Substitute with $n=2N-r$ and we finally get:
$P(r)=\binom {2N-r}{N} \frac{1}{2^{2N-r}}$
Please tell me what you think.
A: Actually correct answer of this question is in picture
