Birthday Line

At a cinema the manager gives a free ticket to the first person in line whose birthday matches that of someone already ahead of them. You may take any spot in line. Birthdays are uniform over 365 days and you know no one else's. Which position gives you the best chance?

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#When I win

I win at position $k$ on two conditions. The $k-1$ people ahead of me must all have different birthdays, or the prize is already gone. And my birthday must match one of those $k-1$ . The two are independent, so

\PP(\text{win at } k) = \underbrace{\prod_{i=1}^{k-1}\frac{365-(i-1)}{365}}_{\text{first } k-1 \text{ distinct}} \cdot \underbrace{\frac{k-1}{365}}_{\text{I match one}}. \tag{1}

#Where it peaks

Early in line the match term $(k-1)/365$ is tiny. Late in line a clash ahead of me has probably already claimed the prize. The product peaks in between. Comparing position $k+1$ with position $k$ , the ratio exceeds one exactly while $k^2 - k - 365 < 0$ , that is $k < \tfrac{1 + \sqrt{1461}}{2} \approx 19.6$ . So the chance climbs through position 20 and falls after.

Early positions are likely still all distinct but give you few birthdays to match, while late positions are likely already spoiled by a clash ahead of you. The chance peaks at position 20.

Stand twentieth.