Fermi Estimates and XKCD
I was trying to do a [Fermi estimate](https://en.wikipedia.org/wiki/Fermi_problem) about the number of comics on [XKCD](https://xkcd.com/). I saw a comic again today, which I had sent to a friend on 9th September last year, and hadn't seen again before today. That's about $100$ days ago. I see about $30$ comics a day, clicking on the "Random" option. So after $3000$ views, I found the same comic again. Let there be $N$ comics in total. Let $p$ be the number of (random) views needed to see a specific comic after having selected one. $$\mathbb{E}(p)=\frac{1}{N}\cdot1+\frac{1}{N}\left(1-\frac{1}{N}\right)\cdot2+\frac{1}{N}\left(1-\frac{1}{N}\right)^2\cdot3+...$$ This boils down to $\mathbb{E}(p)=N$. So I estimate the number of comics on XKCD to be $3000$, roughly the expected value of $p$. And the latest comic is numbered $3032$. I knew Fermi estimates work, but I still can't believe that they can work _so_ well.