The Broken Stick

Question

A stick is broken at two points chosen independently and uniformly along its length. What is the probability the three pieces can form a triangle?

Accepted Answer

Normalize the stick to length 1 and let the cut points be X, Y sim U[0,1] independent. A triangle is possible iff no piece exceeds frac12 (that condition is equivalent to all three triangle inequalities). Take the case X < Y (probability frac12, symmetric). Pieces are X, Y - X, 1 - Y. The conditions: X < tfrac12, qquad Y - X < tfrac12, qquad Y > tfrac12. In the (X,Y) unit square, the region x < y is a triangle of area frac12; the three constraints carve out the inner triangle with vertices (0,frac12), (frac12,frac12), (frac12,1), of area frac18. By symmetry the X > Y case contributes another frac18: P = frac18 + frac18 = frac14. Common trap "Break the stick, then break the longer piece" is a different experiment, the answer becomes 2ln 2 - 1 approx 0.386... and "break a random piece" different again. Geometric probability answers depend entirely on the randomization procedure, which is precisely why interviewers ask the variants.

The Broken Stick

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