Four points are chosen independently and uniformly inside a unit square. What is the probability that they are in convex position, meaning none of them lies inside the triangle formed by the other three?

Four points are chosen independently and uniformly inside a unit square. What is the probability that they are in convex position, meaning none of them lies inside the triangle formed by the other three?

Approach: Four points fail to be convex exactly when one lies inside the triangle of the other three, and those four events are disjoint. Reduce the probability to the expected area of a random triangle.

25/36. The four events that a named point lies inside the triangle of the other three are disjoint events, so P(not in convex position) = 4 P(point 4 falls inside triangle 123), and by conditioning on the first three points that equals 4 times the expected triangle area for three uniform points in the unit square. That expected area is 11/144, so the failure probability is 44/144 = 11/36 and convex position has probability 1 - 11/36 = 25/36 = 0.6944. The same method in a disc of any radius gives 1 - 35/(12 pi^2) = 0.7045, so a rounder region makes convex position slightly more likely.

Follow-up: What is the probability that five uniform points in the square are in convex position?

Key concepts: convex position, expected triangle area, disjoint events, unit square.