Probability 3 points are collinear Given $n \ge 3$ points with real number coordinates, distributed randomly in a unit square, what is the probability $P$ that 3 points exist which are collinear?
My reading of this answer to a similar question indicates to me that $P=0$ for any finite $n$. Is this correct?
What happens when $n \to \infty$? Does $\lim \limits_{n \to \infty} P = 1$?
 A: Let $A$ be the event "$P_1,P_2,P_3$ are collinear". Then $A \iff 0, P_2-P_1, P_3-P_1$ are collinear (I just performed a translation). Then if we could prove that for $v,w$ varying uniformly in the plane, $P(B)=0$ for $B$ the event "$0,v,w$ are collinear", we would have $p(A)=0$. Now observe $B\iff det(v,w)=0$ (I first suppose we are talking about points in the plane, and look at $v,w$ as vectors or as points). But since $det(v,w)$ can take any value in $\mathbb{R}$ with the same probability $0$ as $v,w$ vary uniformly in the plane, this probability is just $\frac{\mu({0})}{\mu({\mathbb{R}})}=0$, where $\mu$ denotes the Lebesgue measure in $\mathbb{R}$, wich measures the "lenght" of subsets.
Now, for the general case, if you have $k$ points ($k\geq 3$) varying uniformly in Euclidean space $\mathbb{R}^n$ ($n\geq 2$), they being collinear implies their projections onto $\mathbb{R}^2\times 0_{\mathbb{R}^{n-2}}$ (or another suitable $2$-plane) are collinear in this plane, and we just proved that three of them being collinear has probability $0$, let alone more than three.
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$\textbf{Added}:$ With the reasoning above, you can treat a similar problem: compute the probability that $n+1$ points in $\mathbb{R}^n$ lie in a hyperplane. First solve the problem for $n=3$:
$P_1, P_2, P_3, P_4$ lie in a plane $\iff 0, P_2-P_1, P_3-P_1, P_4-P_1$ lie in a plane. Now construct the map $det(v_1, v_2, v_3)$ and treat this as before, for $v_1, v_2, v_3$ varying uniformly in $\mathbb{R}^3$. Then you can extend it to the general case again using projections onto $\mathbb{R}^3\times 0_{\mathbb{R}^{n-3}}$ (or another suitable $3$-plane).
