In a queue for £1 tickets, there are $m$ people with a £1 coin and $n$ people with a £2 coin. What is the probability that everyone receives change? I am selling raffle tickets for £1 per ticket. In the queue for tickets, there are $m$ people each
with a single £1 coin and $n$ people each with a single £2 coin. Each person in the queue
wants to buy a single raffle ticket and each arrangement of people in the queue is equally
likely to occur. Initially, I have no coins and a large supply of tickets. I stop selling tickets
if I cannot give the required change.
Show that the probability that I am able to serve everyone in the queue is $\frac{m+1-n}{m+1}$

This problem comes from a STEP question (see Q3 here) where the solution is shown in the cases $n=1,2$ or $3$. However they involve conditioning on permutations of the first couple of people in a way that I don't see how to generalise.
 A: This problem is equivalent to Bertrand's ballot theorem and this answer follows the argument found on Wikipedia. Thanks to aman_cc for pointing out this solution.

This problem can be visualised as counting the number of paths from $(0,0)$ to $(m,n)$ where:

*

*a step right represents a person with a £1 coin

*a step up represents a person with a £2 coin

All the tickets are sold if and only if the number of £1 coin customers is greater than the number of £2 coin customers - i.e the path does not intersect the line $y=x+1$.
To count the number of paths that intersect the line $y=x+1$, the following correspondence is helpful. If a path intersects the line $y=x$, then reflect the path up to the first point of intersection. This creates a new path from $(-1,1)$ to $(m,n)$. In the picture below the blue path is 'reflected' into the red path.

Thus the probability of failure is:
$$ \frac{ \text{  $\#$ {paths from $(-1,1)$ to $(m,n)$ } } }{ \text{  $\#$ {paths from $(0,0)$ to $(m,n)$ } } }
= \frac {\binom{m+n}{n-1} }{ \binom{m+n}{n} }
=\frac{n}{m+1}
 $$
Hence the probability of success is
$$ \frac{m+1-n}{m+1} $$
