A building has n floors numbered 1,2,...,n, plus a ground floor G. At the ground floor, m people get on the elevator together, and each gets off at a uniformly random one of the n floors (independently of everybody else). What is the expected number of floors the elevator stops at (not counting the ground floor)?

The way I approached this problem was the following:

Let $X$ be the number of floors the elevator stops at. $X = X_1 + X_2+...X_n$ where $X_i$ is $1$ if the elevator stops at the i-th floor and $0$ otherwise.

Therefore, $E(X_i) = Pr[X_i=1]$ and since each person gets off the elevator uniformly randomly, $Pr[X_i=1] = \frac{1}{n}$, so $E[X]=n\cdot \frac{1}{n}$

However, the answer was wrong, the book gives the following answer:

$Pr[A_i=1] = 1 - Pr[no\,one\,gets\,off\,at\,i]=1-(\frac{n-1}{n})^m$

I don't disagree with this notion but I'm confused as to why my method would be considered incorrect. By the way, I see that similar questions have been asked on this site before, but I particularly want to know why my original answer was wrong.

  • $\begingroup$ @amd Please check the last sentence of my question, I want to know why my original answer was wrong $\endgroup$
    – Carpetfizz
    May 28, 2017 at 23:45
  • $\begingroup$ Your solution fails the plausibility test, at the very least. The elevator can’t stop at no floors at all, so $X\ge1$, but we also know that $X$ can be greater than one, so $E[X]\gt1$. In other words, the only way for $E[X]=1$ to be true is for everyone to always get off on the same floor. $\endgroup$
    – amd
    May 28, 2017 at 23:46
  • $\begingroup$ @amd oh, so you're saying that $X$ isn't an indicator r.v. since it can never be $0$ ? $\endgroup$
    – Carpetfizz
    May 28, 2017 at 23:47

1 Answer 1


Your statement that $Pr[X_i=1]=\frac 1n$ is confused. $X_i$ is the chance that somebody gets off on floor $i,$ not the chance that a given person gets off on floor $i$. If you multiply out the consequence of this, you get $E[X]=1$, which says you expect the elevator to stop on only one floor. It would be closer to say $Pr[X_i=1]=\frac mn$ because there are $m$ people who might get off on floor $i$. This would lead to the expected number of floors the elevator stops at being $m$. In fact, it is lower than that because you could have more than one person get off on a given floor. The book answer takes this into account.


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