Ticket line with no change in the till

A ticket costs $5. Thirty-eight people queue for tickets: 19 of them carry only a single $10 note and the other 19 carry only a single $5 note. The seller starts with an empty cash box. If the queue is a uniformly random ordering of the 38 people, what is the probability that every person gets their ticket (and any change owed) in turn, without anyone having to swap places?

$1/20$. Treat a $5 payer as an up-step and a $10 payer as a down-step; the seller never runs out of change exactly when the path never dips below zero. The number of such orderings is the Catalan number $\tfrac{1}{n+1}\binom{2n}{n}$, so dividing by $\binom{2n}{n}$ gives $\tfrac{1}{n+1}=\tfrac1{20}$ for $n=19$.