**Modified Birthday Problem:** a bunch of people line up, and the winner is the first person who shares his birthday with someone lined up ahead of him. What position in the line is optimal?

**Three (similar) approaches:** Recall the well known Birthday Problem, and let $b(n)$ denote the likelihood that there is a shared birthday in a collection of $n$ people. Suppose you are the $(n+1)$st person in line, and you want to be the first person to share your birthday with someone ahead of you. 'None of the people ahead of you sharing a birthday' happens with probability $1 - b(n)$, and 'you sharing a birthday with one of them' happens with probability $n/365$. Therefore, we wish to maximize: $(1-b(n))(n/365)$. A computer search (e.g. in WolframAlpha) can find $n$ is about $19$, and so you want position $20$.

Alternatively, we can approximate $1-b(n)$ with $e^{-n(n-1)/730}$ pretty well. Now consider setting $\frac{d}{dx} e^{-x(x-1)/730} (x/365) = 0,$ which ends up requiring only that we solve a quadratic. In particular, we end up with the quadratic $x^2 - x/2 = 365.$ Then $x$ is a little over $19$, and again we guess the optimal position is at $20.$

Finally, we consider the ratio of probabilities of winning for consecutive people in line. Eliding over some details (e.g. showing initial ratio $> 1$, the ratios are decreasing) we can check to see when this ratio drops below $1.$ This will again culminate in solving a quadratic, namely, $x^2 - x = 365.$ Since $x$ is about $19$, we have (for the third time) found the optimal position to be at $20.$

**Question 0:** Why do the quadratics arrived at in our second and third approach differ? (I suspect this is just an artifact of the rough estimation used in the second approach, although I could not explain the reasoning behind this difference in any greater detail.)

**Question 1:** Since our third method indicates that solving for $x$ in $x(x-1) = 365$ ultimately leads to the solution, I am wondering: is there a way to have seen "at a glance" the importance of this quadratic? My intuition tells me that there is a much more straightforward way of thinking about this problem, but I am not sure what it would be.