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.

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 $$10-\left(\frac{9}{10}\right)^{12} - \left(\frac{8}{10}\right)^{12} \approx 9.6$$
