The number of people, $n$, who enter an elevator with $k$ floors follows a Poisson distribution with a mean of $10$. Each person has an equal probability of getting off on any given floor, independent of the others. What is the expected value of the last floor that the elevator stops at?
I first tried having $X=\text{number of floors}$ and $X \sim NB(n, \frac{1}{k})$, except that isn't correct, since $E(X)$ would end up being $nk$, and $k$ is the maximum last floor.
Assume $N$ passengers enter the elevator.
If $M\in \{1,...,k\}$ is the last floor the elevator stopped at, then at least one passenger got off at floor $M$ and those passengers who didn't exit floor $M$ got off at an earlier floor. So, for $m\in \{1,...,n\}$ $$\mathbb{P}(M=m)=\sum_{l=1}^N{N \choose l}\Big({1 \over k}\Big)^l\Big(\frac{m-1}{k}\Big)^{N-l}$$ The index $l$ signifies the number of passengers who exited the $m^{\text{th}}$ floor. Using the binomial theorem we can write this as $$\mathbb{P}(M=m)=\Big(\frac{m}{k}\Big)^N-\Big(\frac{m-1}{k}\Big)^{N}$$ The above pdf is the conditional pdf of $M$ given $N\sim \text{Poisson}(10)$ whose expected value is $$\mathbb{E}(M|N)=\sum_{m=1}^k m\Bigg[\Big(\frac{m}{k}\Big)^N-\Big(\frac{m-1}{k}\Big)^{N}\Bigg]=k-\sum_{m=1}^k\Big(\frac{m-1}{k}\Big)^N$$ Using double expectation and a whole bunch of calculus, $$\begin{eqnarray*}\mathbb{E}(M)&=&\mathbb{E}\Big(\mathbb{E}(M|N)\Big)\\&=& \mathbb{E}\Bigg(k-\sum_{m=1}^k\Big(\frac{m-1}{k}\Big)^N\Bigg) \\ &=& k-\sum_{m=1}^k\mathbb{E}\bigg(\Big(\frac{m-1}{k}\Big)^N\bigg) \\ &=& k - \sum_{m=1}^k \sum_{n=0}^{\infty}\Big(\frac{m-1}{k}\Big)^n\cdot e^{-10}\frac{10^n}{n!} \\ &=& k-\sum_{m=1}^k \exp\Big(\frac{10(m-k-1)}{k}\Big)\\ &=& k+\frac{e^{-10}-1}{e^{10/k}-1}\end{eqnarray*}$$