Alignment of random points Whenever I draw randomly about ten points, I see that there will be always 3 points that are "almost" collinear. This observation leads me to considering the following questions:
Question 1: Suppose $n$ points are generated uniformly randomly in a square, let $\phi_n$ be the largest angle that is formed by three of them. How quickly does $E(\phi_n)$ tend to $\pi$?
Question 2: For $n>2$ and $\epsilon > 0$, estimate $P(\phi_n > \pi-\epsilon)$?
 A: The intuition in the comment by  Ori Gurel-Gurevich appears to be correct. 
Indeed, let us show that $\pi-\phi_n$ is on the order of $1/n^3$ in probability. 
Let $T$ denote the set of all triangles with vertices at some of the $n$ points, with $N:=|T|=\binom n3$, the cardinality of $T$. For each $t\in T$, let $X_t$ denote the largest angle in the triangle $t$, so that $\phi_n=\max_t X_t$. It should be not hard to show that (i) the random variable (r.v.) $X_t$ has a density $f$ left-continuous at point $\pi$, with $f(\pi-)=:c\in(0,\infty)$ and (ii) for any distinct $t$ and $s$ in $T$, the random pair $(X_t,X_s)$ has a joint density bounded by some $C<\infty$. 
Now we can use the key result by Galambos \url{https://www.jstor.org/stable/2239989?seq=1#page_scan_tab_contents}, based on a combinatorial graph sieve theorem by Renyi, with $N$ in place of the symbol $n$ in that paper, and with 
$E$ defined as the set of pairs $(t,s)\in T^2$ such that the triangles $t$ and $s$ have at least one common vertex. Then clearly $N_E:=|E|=O(n^3\cdot n^2)=o(N^2)$. 
For any fixed positive real $a$, let $c_N=c_N(a)$ be the root of the equation 
$$NP(X_t>c_N)=a.$$ 
Since $P(X_t>x_N)\sim c(\pi-x_N)$ if $x_N\uparrow\pi$, we have
\begin{equation}
c_N=\pi-\frac{a}{cN}\,(1+o(1));   
\end{equation}
all the limit relations here are for $n\to\infty$. 
Then it is easy to see all the conditions in Galambos's theorem hold, with $r(\dots)=0$ in his condition (iii) and $d_k=1$ in his condition (iv). 
It follows that $P(\phi_n<c_N)\to e^{-a}$, which is equivalent to 
\begin{equation}
 P(cN(\pi-\phi_n)>a)\to e^{-a}, 
\end{equation}
for each real $a>0$. That is, the r.v. $\frac c{3!}\,n^3(\pi-\phi_n)$ converges to an exponential r.v. in distribution. 
Additional efforts are needed to show that the corresponding convergence of the means holds, so that $\frac c{3!}\,n^3\mathbb{E}(\pi-\phi_n)\to1$, whence $\mathbb{E}(\pi-\phi_n)\sim\frac{3!}c\,n^{-3}$.
A: Concerning Question 2, $P(\phi_n > \pi-\epsilon)$ is at least exponentially close to $1$. Indeed, let $(p_1,\dots,p_n)$ be the random sequence of $n$ points drawn independently and uniformly from the square. Let $k:=\lfloor n/3\rfloor$. For $j=1,\dots,k$, let $A_j$ denote the largest angle in the triangle with vertices $p_{3(j-1)+1},p_{3(j-1)+2},p_{3(j-1)+3}$. Then $\phi_n\ge A_1\vee\cdots\vee A_k$, and so, 
\begin{equation}
 P(\phi_n > \pi-\epsilon)\ge1-P(A_1\vee\cdots\vee A_k \le \pi-\epsilon)
 =1-q^k=1-q^{\lfloor n/3\rfloor}, 
\end{equation}
where $q:=P(A_1\le\pi-\epsilon)<1$. 
A: On question 2, if $n > 2\pi/\epsilon$, then $P(\phi_n > \pi-\epsilon) = 1$.
Arrange the points in a polygon. The total of the interior angles is $(n-2)\pi$, so at least one must be at least $\pi - 2\pi/n$. 
A: This is not an answer to your original specific question, but 
this recent related result (and its references) may help:

"Finding Points in General Position."
  Vincent Froese, Iyad Kanj, André Nichterlein, Rolf Niedermeier.
  May 2016.
  arXiv abstract.

"Given a set of points
in the plane, find a maximum-cardinality subset of points in general position. We prove
that [this problem] is NP-hard, APX-hard, and present several
fixed-parameter tractability results for the problem."
