Calculate the expected value of the highest floor the elevator may reach. I've been to solve this exercise for a few hours now, and all the methods I use seems wrong, I'll be glad if someone could solve this for me, since I don't know how to approach this correctly.
Given a building with 11 floors while the bottom floor is the ground floor (floor 0), and the rest of the floors are numbered from $1-10$, $12$ people gets into an elevator in the ground floor, and choose randomly and in independent way the floor they wish to go (which one of them has the probablility of $\frac{1}{10}$ to choose any floor in independent matter of the others).
Calculate the expected value of the highest floor the elevator may reach?
Thank you.
 A: Let $X$ denote the highest floor that the elevator reaches, then:


*

*$P(X=1)=\left(\frac{1}{10}\right)^{12}$

*$P(X=n)=\left(\frac{n}{10}\right)^{12}-P(X=n-1)$


Hence:
$E(X)=$
$\sum\limits_{n=1}^{10}n\cdot P(X=n)=$
$\sum\limits_{n=1}^{10}n\cdot\left(\left(\frac{n}{10}\right)^{12}-\left(\frac{n-1}{10}\right)^{12}\right)=$
$\sum\limits_{n=1}^{10}n\cdot\left(\frac{n^{12}-(n-1)^{12}}{10^{12}}\right)=$
$9.632571463867$
A: Hint: If $X_i$ is the floor that the $i$-th person chooses, you need to calculate the expectation of $X =\max \limits_{i=1}^{12}X_i$. Start by calculating $P(X \le k)$.
A: The highest floor that the elevator reaches is the maximum $M$ of the floors chosen by the 12 people.
For $m = 1,\dots,10$, the probability that $M \leq m$ is just the probability that all 12 people chose floors less than or  equal to $m$, $$\left(\frac {m}{10}\right)^{12}$$
So the probability that $M=m$ is $$\left(\frac {m}{10}\right)^{12} - \left(\frac {m-1}{10}\right)^{12}  = \frac{m^{12}-(m-1)^{12}} {10^{12}}$$
Hence the expected value of $M$ is $$\sum_{m=1}^{10} m \ \frac{m^{12}-(m-1)^{12}} {10^{12}} = \frac{1}{10^{12}}\left(\sum m^{13} - \sum (m-1)^{13} -\sum 
(m-1)^{12} \right) 
 $$
By cancellation, this equals
$$\frac{1}{10^{12}}\left(10^{13} - (1^{12} + \dots + 9^{12}) \right) 
 $$
Putting this into a computer gives the answer. It can be approximated by noting that $\frac{7^{12}}{10^{12}} \approx 0.014$ is small, so throwing away terms smaller than this gives $$10-\left(\frac{9}{10}\right)^{12} - \left(\frac{8}{10}\right)^{12} \approx 9.6$$
