Best queue position for a shared-birthday prize

A cinema gives a free ticket to the first person in the queue whose birthday is the same as someone already ahead of them in the queue. Birthdays are independent and uniform over $365$ days. You may pick your place in line. Which position maximises your chance of winning?

Position $20$. Position $n$ wins if the first $n-1$ birthdays are all different and yours matches one of them: $p_n=\tfrac{n-1}{365}\prod_{k=0}^{n-2}\tfrac{365-k}{365}$. The ratio $\tfrac{p_{n+1}}{p_n}=\tfrac{n(366-n)}{365(n-1)}$ exceeds $1$ only while $n\le19$, so the maximum ($\approx3.2\%$) is at $n=20$.