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.

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:

$E(X)=$

$\sum\limits_{n=1}^{10}n\cdot P(X=n)=$

$\sum\limits_{n=1}^{10}n\cdot\left(\left(\frac{n}{10}\right)^{12}-\left(\frac{n-1}{10}\right)^{12}\right)=$

$\sum\limits_{n=1}^{10}n\cdot\left(\frac{n^{12}-(n-1)^{12}}{10^{12}}\right)=$

$9.632571463867$