The Game of Googol

In the Game of Googol, someone writes down $n$ different numbers (any numbers, however large or small) on slips of paper and turns them face down in random order. You flip them over one at a time. At any point you can stop and declare "this is the largest one" — but once you flip past a slip, you can't go back to it. You win if you correctly stop on the single largest number. What strategy maximizes your odds?

This is the classic secretary problem dressed up with numbers instead of job candidates. The surprising result: you should never guess on any of the first roughly $n/e$ slips (about 37% of them, where $e \approx 2.718$) — just look and pass, using them to learn the scale of numbers involved. After that point, stop on the very first slip that beats every number you've seen so far. This strategy wins about $1/e \approx 37\%$ of the time regardless of how large $n$ is, which is remarkably high given you're picking blind out of $n$ candidates in a single pass.

Worked intuition. With $n = 10$ slips, the rule says observe-only through the first $\lfloor 10/e \rfloor = 3$ slips, then pick the first slip after that which is bigger than all 3 you've already seen. Even though you're committing without ever seeing what's left, this beats naive strategies (like always guessing the first card, which wins only $1/10$ of the time) by a wide margin.