Consider the queue of size $2n$ that goes to the cash register. Each person in the queue have, with equal probability, either a coin worth 5 bucks or a coin worth 10 bucks, and wants to buy a ticket. The ticket costs 5 bucks. In total, there are $2m$ coins in the cash register, 5 bucks each.
Find the probability of the queue not stopping due to the lack of change.
What if $ m = 0$?
Honestly, I don't even know where to start. I assumed it's a variation of Bertrand's ballot theorem. We might assume that we start from $(0, 2m)$ grid point with probable step $(+1, -1)$. I guess we should calculate the possible ways to reach $(0, 0)$ from $(0, 2m)$, which is $$\frac{1}{2m-1}\frac{1}{2^{2m}}{{2m}\choose{m}}$$ But here we don't take anything given in the task into consideration. How to proceed?
Partial answer. I think I know how to do in case $m = 0$
Probability queue will stop at first person = $\frac{1}{2}$
Probability queue will stop at $k = 2$ = $0$. Because we know that queue didn't stop at first person, so there must be 5$ in cash.
Probability queue will stop at $k = 3$ = $\frac{1}{2}^2$. Because before 3rd person comes there is equal probability of having $2$ 5 dollars or $0$. Multiplying it by probability that 3rd person has no 5dollar we get $\frac{1}{4}$
Probability queue will stop at $k = 4$ = $0$
And so on we get $\frac{1}{2} + \frac{1}{2}^2 + ... + \frac{1}{2}^n$