Suppose we have $N$ people, $N\in\mathbb{Z}_+$. These people arrive at a place at a random time $t_1, t_2, \dots, t_N$, where $t_k\in(0,1)$, following certain distribution (it is unknown in my case, but maybe you can assume it is a uniform distribution). The distributions for different people are i.i.d.
Now, let $\tau=max(t_k)-c$, and $c\in(0,0.5)$ is a known constant. Let $S=\{t_k ~|~ t_k\le \tau\}$ be the set that people arrive earlier than $\tau$. Then, what is the probability $p(N)$ such that the cardinality $|S| \ge M$? Here $M>1$ is a given constant.
Note that, $N$ is the argument for $p(N)$. Clearly, we can see that $p(N)\to 1$, as $N\to \infty$.
I'm not very good at probability theory, I think the key difficulty comes from the max function. Or can we at least get a lower bound for $p(N)$?
This problem can be solved using order statistics and the (very interesting) Bapat-Beg Theorem.
https://en.wikipedia.org/wiki/Bapat%E2%80%93Beg_theorem
If you let $t_{(1)},...,t_{(N)}$ denote the ordered times, so that $t_{(N)} = \max\{ t_1,...,t_N\}$. Then $P(|S|>M) = P(t_{(N)}-t_{(N-M)} < c)$. Using the Bapat-Beg theorem, you can calculate the joint distribution of $(t_{(N)},t_{(N-M)})$ as a function of the CDFs of the respective times, and hence calculate $P(t_{(N)}-t_{(N-M)} < c)$. This distribution is much easier to calculate when the times are uniform.