Published 13 September 2026

Broken stick

Break a unit stick into three pieces. What is the probability they can form a triangle? With two independent uniform cut positions on the original stick, the answer is one quarter. Breaking once and then choosing a piece to break again is a different experiment.

Three positive lengths summing to one form a nondegenerate triangle exactly when every length is less than one half. This is the triangle inequality: the longest side must be shorter than the sum of the other two. A piece of length exactly one half gives a straight, degenerate configuration and is excluded.

Inspect the cuts and pieces

The horizontal coordinate is the smaller cut and the vertical coordinate is the larger cut, so all points lie above the diagonal. A point in the shaded region has three pieces short enough to form a triangle. The large ring selects one stick; its three lengths are listed below the chart.

All protocols start with a uniform first cut. They then differ in how the second location is obtained. The sample counts use every generated stick, including failures and boundary cases. Increasing sample size preserves the previous sticks for the same seed and protocol.

Two independent uniform cuts

Write x for the smaller cut and y for the larger. The pieces have lengths x, y − x, and 1 − y. Triangle formation requires x < 1/2, y > 1/2, and y − x < 1/2. Those inequalities describe the shaded triangle with boundary vertices (0,1/2), (1/2,1/2), and (1/2,1).

The possible ordered-cut region has area 1/2. Its density is constant because each distinct ordered pair can come from the two possible orders of the independent cuts. The favorable triangle has area 1/8, so the probability is (1/8)/(1/2) = 1/4. Sorting the cuts changes the drawing’s domain, not the underlying probability.

Choose one piece with equal probability

After the first cut, select one of the two resulting pieces by a fair coin, regardless of their lengths. Cut the chosen piece uniformly along its length. For success, the uncut piece must already be shorter than one half, meaning the selected piece must be the longer one.

Let l be the selected piece’s length. Under this protocol l is uniform on [0,1]: either the first cut position or its complement is selected with equal probability. When l ≤ 1/2, success is impossible. When l > 1/2, the second cut must leave both new pieces shorter than one half. Its allowed interval has length 1 − l within a piece of length l, so conditional success probability is (1 − l)/l.

Integrating from l = 1/2 to one gives ∫(1/l − 1) dl = ln 2 − 1/2, approximately 0.1931472. Equal chances per piece are not equal chances per unit length of stick.

Always choose the longer piece

Now choose the longer piece after the first cut. Its length lies between one half and one with density two. The same conditional success calculation applies, but averaging uses that new density. The result doubles to 2 ln 2 − 1, approximately 0.3862944.

This is larger than one quarter. It does not contradict the independent-cut calculation: using the first result to decide where to cut next changes the distribution of final pieces.

Geometry and sampling both matter

The shaded success region stays the same for every protocol. Only the density of sampled points changes. Area ratios alone are valid when the sample is uniform over the relevant region; changing the cutting mechanism can destroy that uniformity. The exact table and reproducible sample are shown separately so finite fluctuations do not masquerade as a new theoretical answer.

Reference

The Broken Stick Project examines the classical two-point construction. The sequential-piece calculations above follow their stated selection rules. Continue with Bertrand’s random chords for another geometric question whose answer changes with its sampler.