STEP I 2011, Q12. An interesting extension...

To paraphrase the question: There are n+m customers in a queue, each wanting to buy a £1 raffle ticket. m customers have a £1 coin, and n customers have a £2 coin, and the seller has no change to begin with. What is the probability he can serve all the customers without running out of change?

The question leads you through the cases n=1, where P(success) = $\dfrac{m}{m+1}$.
and then n=2, where P(success) = $\dfrac{m-1}{m+1}$.
and finally n=3, where P(success) = $\dfrac{m-2}{m+1}$.

So of course, I wondered if in general, P(success) = $\dfrac{m+1-n}{m+1}$?

I believe this to be true, and I have a fairly short proof based on a solution to the case m=n I found via google. Anyone else want to try?