Broken Stick Triangle Cut Geometry

worked-example

Let the stick have length 1. We choose two points, x and y, uniformly and independently from the interval [0, 1]. For convenience, let c_1 be the smaller cut and c_2 be the larger cut. The three pieces have lengths c_1, c_2 - c_1, and 1 - c_2. To form a nondegenerate triangle, the strict triangle inequality must hold: the sum of any two sides must be strictly greater than the third side. Because the three sides sum to 1, this is equivalent to stating that the maximum piece length must be strictly less than 1/2.

Let's calculate the probability that the maximum piece length is less than or equal to a threshold q, using the exact CDF formula: CDF(q) = \sum_(j=0)³ (-1)^j C(3, j) max(1 - jq, 0)². Let's evaluate this for q = 1/2:

1. For j = 0: (-1)^0 C(3, 0) max(1 - 0, 0)² = 1 · 1 · 1² = 1
2. For j = 1: (-1)^1 C(3, 1) max(1 - 1/2, 0)² = -1 · 3 · (1/2)² = -3/4
3. For j = 2: (-1)² C(3, 2) max(1 - 2(1/2), 0)² = 1 · 3 · 0² = 0
4. For j = 3: (-1)³ C(3, 3) max(1 - 3(1/2), 0)² = -1 · 1 · 0² = 0

Summing these gives the CDF at q = 1/2, which is exactly the probability of forming a triangle (excluding degenerate equality cases): 1 - 3/4 = 1/4.

experiment

Use the interactive model to visualize the ordered-cut domain and see which cut combinations succeed or fail.

1. Open the interactive module.
2. Change the First cut and Second cut to different values. Watch how the point (x, y) moves in the 2D configuration space.
3. Observe the Maximum-piece threshold control. Adjusting it will show you the exact Maximum-piece CDF computed using the alternating sum formula.
4. Notice that any point falling inside the inner green triangular region produces a valid triangle (where the longest piece is strictly less than 1/2), while points outside fail the triangle inequality.

limitations

This model assumes the two cuts are made simultaneously and independently on the original unit interval. A different protocol—for example, breaking the stick once uniformly, then uniformly selecting one of the two pieces to break again—yields a different geometric distribution and a different probability. Furthermore, this model treats equality (a piece of exactly length 1/2) as a degenerate triangle with zero area, which fails the strict triangle inequality.

references

The geometry of the broken stick problem is widely covered in probability literature as a classic example of continuous uniform distributions mapped to 2D event spaces.

• UC Berkeley CS70 Lecture 36 Slides - Covers continuous probability, joint distributions, and the geometric interpretation of the broken stick problem.