Question: You are on the $10^\text{th}$ floor of a tower and want to exit on the first floor. You get into the elevator and hit $1.$ However, this elevator is malfunctioning in a specific way. When you hit $1,$ it correctly registers the request to descend, but it randomly selects some floor below your current floor (including the first floor). The car then stops at that floor. If it’s not the first floor, you again hit $1$ and the process repeats.
Assuming you are the only passenger on the elevator, how many floors on average will it stop at (including your final stop, the first floor) until you exit?
Solution
To avoid finicky bookkeeping, we can count the lobby as floor zero, instead of floor $1$. Also, it’s equivalent to think in terms of how many times the passenger will press the button, and that’s what we’ll do.
When the passenger presses the button from floor $(k+1),$ they can either go to floor $k,$ or go to floors $0$ through $(k-1),$ which is the same as starting from floor $k$ but without the extra button press.
This means that the expected number of button presses from floor $(k+1)$ is
\[\begin{align} \langle B_{k+1} \rangle &= \tfrac1k(1 + \langle B_k\rangle) + (1-\tfrac1k) \langle B_k\rangle \\ &= \tfrac1k + \langle B_k\rangle, \end{align}\]so
\[\langle B_k\rangle = 1 + \tfrac12 + \tfrac13 + \ldots + \tfrac1k,\]the $k^\text{th}$ harmonic number.
For the $9$ story building in question, this is $\frac{7129}{2520} \approx
2.829$ button presses.