Choosing $n$ times from $n$ objects I am given $n$ objects and for $n$ times, I pick one of them with uniform probability and put it back after picking it.
For $k\in\{1,\ldots,n\}$ let $f_k$ denote the number of times that I have picked object number $k$. So we have $f_k\in \{0,\ldots,n\}$ for all $k$.
We consider $M:= \max\big\{f_k: k\in\{1,\ldots,n\}\big\}$, so we have $M\in\{1,\ldots,n\}$. We are interested in the expected value $E_n:= E(M)$.
Does $\lim_{n\to\infty} E_n$ exist? If yes, what is its value?
 A: $M = \Theta(\log n / \log\log n)$ with high probability and in expectation.
An alternative way of thinking about this problem is that you're tossing $n$ balls uniformly at random into $n$ bins and looking at the size of the largest bin. In this formulation, this model is used in computer science to think about load-balancing; Gonnet 1981 (Expected length of the longest probe sequence in hash code searching, Journal of the ACM 28(2): 289-304, https://cs.uwaterloo.ca/research/tr/1978/CS-78-46.pdf) has a sketch of a proof. Alternatively, these lecture notes https://people.eecs.berkeley.edu/~sinclair/cs271/n14.pdf work through the argument more carefully, justifying why the Poisson approximation can be used there.
As an aside, if instead of picking your object uniformly at random you first eyeball two objects (uniformly at random), and then actually pick the less popular one of those, the maximum will drop down to $\log\log(n) / \log(2) + O(1)$. This is known as the "power of two choices" and makes for a nice term to Google.
A: This is the Balls and Bins setting.
A: Fix some $k$. The probability that the element $1$ is chosen exactly $k$ times is ${n\choose k}\frac1{n^k}\left(1-\frac1n\right)^{n-k}\to \frac1{ek!}$ as $n\to\infty$. So $1$ is chosen at least $k$ times with probability at least, say, $\frac1{2ek!}$
Under the condition that $1$ has been chosen less than $k$ times, the same probability for $2$ is at least ${n-k\choose k}\frac1{n^k}\left(1-\frac1{n-1}\right)^{n-k}\to \frac1{ek!}$, so both $1$ and $2$ are not chosen with the probability at most $\left(1-\frac1{2ek!}\right)^2$ if $n$ is large.
Iterating this, we get that for large enough $n$ the probabitily of $M<k$ is at most $\left(1-\frac1{2ek!}\right)^d$, where $d$ is an arbitrary fixed number. Thus this probability tends to $0$ as $n\to\infty$, and $E_n\to\infty$.
