Broken stick, second break on the longer piece, Geometric Probability

Problem 7.1

A unit stick is broken at a uniform random point $X \in [0, 1]$ . Of the two resulting pieces, the longer one is then broken at a uniform random point along its length. What is the probability that the three resulting pieces form a triangle?

Solution. By symmetry between the two outcomes $X \leq \tfrac12$ and $X \geq \tfrac12$ , we may condition on $X \leq \tfrac12$ ; the other case is handled identically by reversing the stick. So take $X \leq \tfrac12$ , giving a shorter piece of length $X$ and a longer piece of length $1 - X$ . Now break the longer piece: let $Y \in [0, 1 - X]$ be uniform, giving two pieces of lengths $Y$ and $1 - X - Y$ .

The three lengths are $X$ , $Y$ , and $1 - X - Y$ . They form a triangle iff each is less than the sum of the other two, equivalently iff each is less than $\tfrac12$ : $X < \tfrac12, \quad Y < \tfrac12, \quad 1 - X - Y < \tfrac12.$ The first is automatic under our conditioning. The remaining two are: $Y < \tfrac12, \qquad Y > \tfrac12 - X.$ Given $X$ , $Y$ is uniform on $[0, 1 - X]$ , so $\mathbb{P}(\text{triangle} \mid X) = \frac{\tfrac12 - \left( \tfrac12 - X \right)}{1 - X} = \frac{X}{1 - X}.$ Integrating over $X \in [0, \tfrac12]$ with density $2$ (after conditioning on $X \leq \tfrac12$ ): $\mathbb{P}(\text{triangle}) = \int_0^{1/2} \frac{X}{1 - X} \cdot 2 \, dX = 2 \int_0^{1/2} \left( \frac{1}{1 - X} - 1 \right) dX.$ Evaluate: $= 2 \left[ -\ln(1 - X) - X \right]_0^{1/2} = 2 \left[ -\ln \tfrac12 - \tfrac12 \right] = 2 \ln 2 - 1.$ So $\boxed{\mathbb{P}(\text{triangle}) = 2 \ln 2 - 1 \approx 0.386.} \qedhere$

Compare with the ordinary two-cut broken stick of Chapter 2: breaking at two independent uniform points of the whole stick, the three pieces form a triangle iff no piece exceeds $\tfrac12$ , which happens with probability $1 - \tfrac34 = \tfrac14$ . Breaking the longer piece second gives $2 \ln 2 - 1 \approx 0.386$ , which is larger than $\tfrac14$ : this biased procedure increases the chance, by $(2 \ln 2 - 1) - \tfrac14 = 2 \ln 2 - \tfrac54 \approx 0.136$ . (The value $\tfrac12$ belongs instead to Chapter 1, where three lengths are sampled independently from $[0,1]$ , a different model.) The problem is a well-known variant in the Putnam / Eastern European olympiad tradition.

import numpy as np
x = np.random.rand(10**7)           # first break point
short = np.minimum(x, 1 - x)
longp = 1 - short                   # longer piece
y = np.random.rand(10**7) * longp    # second break on the longer piece
pieces = np.stack([short, y, longp - y])
triangle = (pieces < 0.5).all(axis=0)
print(f"sim: {triangle.mean():.5f}   exact: {2*np.log(2) - 1:.5f}")
# sim: 0.38640   exact: 0.38629