At a movie theater, the whimsical manager announces that a free ticket will be given to the first person in line whose birthday is the same as someone in line who has already bought a ticket. You have the option of choosing any position in the line. Assuming that you don't know anyone else's birthday, and that birthdays are uniformly distributed throughout the year (365 days a year), what position in line gives you the best chance of getting a free ticket?

The probability that person number $N$ gets the free ticket is equal to the probability that: P(N) = (probability that none of the first N-1 people share birth dates) * (probability that you share birthday with one of the first N-1 people | Those N-1 are all distinct) So, we have that \begin{equation} P(N) = 1 * \frac{364}{365} * \frac{363}{365} * \frac{362}{365} \;...\; * \frac{365 -(N-2)}{365} * \frac{N-1}{365} \end{equation} Now, we need to look at the smallest $N$ for which holds that P(N) > P(N+1), or, we need to look at when \begin{equation} \frac{P(N)}{P(N+1)} > 1 \end{equation} To find the smallest N, we need to go through the following steps: \begin{equation} \frac{P(N)}{P(N+1)} = \frac{1 * \frac{364}{365} * \frac{363}{365} * \frac{362}{365} \;...\; * \frac{365 -(N-2)}{365} * \frac{N-1}{365}}{1 * \frac{364}{365} * \frac{363}{365} * \frac{362}{365} \;...\; * \frac{365 -(N-2)}{365} * \frac{365 -(N-1)}{365} * \frac{N}{365}} \end{equation} \begin{equation} \frac{P(N)}{P(N+1)} = \frac{\frac{N-1}{365}}{\frac{365 -(N-1)}{365} * \frac{N}{365}} \end{equation} \begin{equation} \frac{P(N)}{P(N+1)} = \frac{365(N-1)}{(365 -(N-1)) * N} \end{equation} \begin{equation} \frac{P(N)}{P(N+1)} = \frac{365N-365}{366N -N^2} \end{equation} So finally, we want to solve \begin{equation} \frac{P(N)}{P(N+1)} > 1 \end{equation} \begin{equation} \frac{365N-365}{366N -N^2} > 1 \end{equation} \begin{equation} 365N-365 > 366N -N^2 \end{equation} \begin{equation} N^2 - N-365 > 0 \end{equation} If you solve this quadratic equation, you get that $N \approx 19.6$. So, at $N=20$, we have for the first time that \begin{equation} \frac{P(20)}{P(21)} > 1 \end{equation} Therefore, you want to stand at the 20-th position!

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