If you pick 3 points at random in the unit disk, the probability that they lie on a perfectly straight line is zero.
However, the probability that the curvature of the circle they lie on is less than ε is, in the limit of small ε, equal to:
1024 ε / (525 π²)
≈ 0.2 ε
@gregeganSF I'm trying to think of a single, even hypothetical, practical use of this knowledge...
@gregeganSF About “they lie on a perfectly straight line is zero”，here is a slightly overengineered proof.
Write the three points as
\( P_i=(X_i,Y_i) \)
The sample space is
\[ D^3\subset\mathbb R^6 \]
where \(D\) is the unit disk.
The three points are collinear if and only if
\[\begin{vmatrix} X_1&Y_1&1\\ X_2&Y_2&1\\ X_3&Y_3&1 \end{vmatrix}=0 \]
In other words,
\[ f(X_1,Y_1,X_2,Y_2,X_3,Y_3)=0 \]
where \(f\) is a polynomial in six variables.
Thus the set of collinear triples is
\(A=f^{-1}(0) \)
an algebraic hypersurface in \(\mathbb R^6\). Its dimension is at most \(5\), and hence it has \(6\)-dimensional Lebesgue measure zero.
On the other hand, the joint distribution of the three independently and uniformly chosen points is absolutely continuous with respect to \(6\)-dimensional Lebesgue measure.
Therefore,
P(A)=0