A question on probability

Question

The original question is:

Suppose a man got two matchboxes in his pocket, each matchbox contains n matchsticks. Each time this guy would randomly pick up one matchbox, then consume a matchstick. One by one, he will eventually pick up a box and find it empty. Ask: what's the probability of having m matchsticks in the other matchbox?

The answer to this question is easy to follow:

1. Since this guy picked up his matchboxes (2n+1-m) times, and every time he has two choices; overall there are $2^{2n+m-1}$ different ways

2. Consider two different circumstances. It is possible that he picks up matchbox A the last time, so in the previous 2n-m rounds, he has picked matchbox A n times, and this yields $\binom{2n-m}n$ ways to pick up matchboxes. Similarly, he might pick up matchbox B and realize it's empty, both circumstances are possible and the results are symmetric. Therefore, the overall probability would be $$2\binom{2n-m}n/2^{2n+1-m}=\binom{2n-m}n/2^{2n-m}$$

Strangely, the final equation seems that one can simply use the binomial distribution to solve the problem.

However, I think the coincidence happens because two matchboxes have equal probabilities to be picked up. If I alter the question to: if this guy has probability of 1/3 to pick up box A and 2/3 to pick up box B, when he picks up an empty box (A or B is not specified), what's the probability of having m matchsticks in the other box? How will the answer be different?

ADDED:

I had thought about a binomial solution before, but then I felt something wasn't right. If I calculate the probability of picking up A as empty box and had B still have m matchsticks, and the probability of picking up B as empty box, then add them up; then in the initial question, why shoudn't the final answer be:

$$2*\binom{2n-m}n/2^{2n-m}$$ insteand?

Because $\binom{2n-m}n/2^{2n-m}$ seems to me only indicate probability of hitting either box A or box B n times in (2n-m) trails. However, in the original answer, one seems had already incorporated both circumstances into the consideration.

Answer 1

If he finds out that $A$ is empty, then from beginning he would have picked up box $A$ $n+1$ times and box $B$ $n-m$ times, with picking up box $A$ at $2n-m+1$th pick. Thus if we consider all sequence of $A$s and $B$s which shows the order that he picked (except for the last one which is fixed as $A$), there would be total $\displaystyle{2n-m \choose n}$ such sequences containing $n$ $A$s and $n-m$ $B$s. From here we obtain that the probability is $\displaystyle{2n-m \choose n} (\frac{1}{3})^{n+1}(\frac{2}{3})^{n-m}$ (note that all sequences indicate distinct cases). Similarly the probability that he will find out $B$ being empty is $\displaystyle{2n-m \choose n}(\frac{1}{3})^{n-m}(\frac{2}{3})^{n+1}$. Thus the total probability is $\displaystyle{2n-m \choose n}\displaystyle\frac{2^{n-m}+2^{n+1}}{3^{2n-m+1}}$.