@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
​