Broken Stick Problem

A stick of length 1 is broken at two points, chosen uniformly at random and independently. This creates three segment lengths.

Question: What is the probability that the three pieces can form a triangle?

1) Interactive stick view

Move the two break points. The stick is reordered automatically so the left break is first.

0 1 0.31 0.29 0.40
Triangle possible

2) Key triangle condition

Three lengths form a triangle iff the sum of any two is greater than the third.

Since the three broken pieces always add up to 1, this simplifies nicely:

largest piece < 1/2

Why? If the longest piece has length L, the other two pieces sum to 1 - L. For a triangle, we need:

1 - L > L
1 > 2L
L < 1/2

So the problem becomes:

  • Pick two random break points in [0,1]
  • Compute the 3 piece lengths
  • Ask whether the longest one is less than 1/2

3) Geometric probability in the unit square

Every pair of break points (x,y) is equally likely, so probability = area of success region.

first break x second break y 0 1/2 1 0 1/2 1

4) Area calculation

The successful region consists of two congruent right triangles.

Each triangle has legs:

1/2 and 1/2
Area of one triangle:

(1/2) × (1/2) × (1/2) = 1/8

So total successful area is:

2 × (1/8) = 1/4
Probability = 1/4 = 25%

5) Final answer

The probability that the three pieces can form a triangle is

1/4

equivalently, 25%