Problem with people and hats after lunch I am trying to solve interesting probability problem which I found in the book "50 Mathematical Ideas You Really Need to Know" by Tony Crilly:
A group of people go to lunch and afterwards pick up their hats at random. What is the probability that no one gets their own hat?"
So he claimed that It can be shown that this probability is $1/e$ (about $37\%$) so that the probability of at least one person getting their own hat is $1 – 1/e$ $(63\%$)
I was thinking about asymptotic cases where $n \to \infty$. The probabilities of "no one gets their own hat" and "at least one person getting their own hat" should be $0$ and $1$ respectively, right? So why it is $1/e$ and $1-1/e$?
Thank you in advance.

For exact answer you can see here: At a party $n$ people toss their hats into a pile in a closet.$\dots$ and here: Question on the 'Hat check' problem.
For $n\to\infty$ you can approximate the distribution with the Binomial distribution. Note that the probability that $k$th man will pick his own hat is $\frac{n-1}{n}\frac{n-2}{n-1}\cdots\frac{n-k+1}{n-k}\frac{1}{n-k+1}=\frac{1}{n} $ wich holds for each $k=1,...,n$, where $n$ is the number of people/hats. Hence, for $n\to \infty$ you can "neglect" the intrinsic dependency between the trials and state that whether the $k$th man picks the right hat is Bernoulli r.v. with $1/n$; $X_k \sim Ber(1/n)$. Hence, the total number of successes follows the Binomial distribution with $n$ and $1/n$, $\sum_{k=1}^nX_k\sim Bin(n, 1/n)$.  Consequently, 
$$
\lim_{n\to\infty}P\left(\sum_{k=1}^nX_k=0\right)= \lim_{n\to\infty}\binom{n}{0}\frac{1}{n^0}\left(1-\frac{1}{n}\right)^n =\lim_{n\to\infty} \left(1-\frac{1}{n}\right)^n = e^{-1}\,.
$$ 
