Breaking a Stick into a Triangle
A stick of length $L$ is broken at two points chosen independently and uniformly at random along its length. What is the probability that the three resulting pieces can form a triangle?
Hints
- For three pieces summing to a fixed total, the triangle inequality simplifies: no single piece can be more than half the total.
- Set up the two break points as uniform random variables on $[0, 1]^2$ and identify the region where all three pieces are less than $1/2$.
- By symmetry, $P(\text{piece } i > 1/2) = 1/4$ for each piece, and these events are mutually exclusive since the pieces sum to 1.
Worked Solution
How to Think About It: Three lengths form a triangle if and only if each one is less than the sum of the other two. Since the three pieces sum to $L$, this simplifies beautifully: each piece must be less than $L/2$. So the question reduces to: if you break a stick at two random points, what is the probability that no piece exceeds half the total length? This is a geometric probability problem -- the answer is just a ratio of areas.
Quick Estimate: There are three constraints (each piece $< L/2$), and by symmetry each constraint fails with equal probability. The probability that a specific piece exceeds $L/2$ is $1/4$ (one of the pieces is "too long" for a uniform split). By inclusion-exclusion -- and noting that at most one piece can exceed $L/2$ at a time -- the probability that at least one piece is too long is $3/4$. So the probability that none is too long is $1/4$.
Approach: Set up the problem as a uniform distribution on the unit square and compute the area of the feasible region.
Formal Solution:
Without loss of generality, set $L = 1$. Let $U_1, U_2 \sim \text{Uniform}(0, 1)$ be the two break points, independently chosen. Define the three piece lengths as:
- $A = \min(U_1, U_2)$
- $B = |U_1 - U_2|$
- $C = 1 - \max(U_1, U_2)$
The triangle inequality for pieces summing to 1 requires:
$$A < \frac{1}{2}, \quad B < \frac{1}{2}, \quad C < \frac{1}{2}$$
Equivalently in terms of $U_1, U_2$:
- $\min(U_1, U_2) < 1/2$ -- i.e., not both break points in $[1/2, 1]$
- $|U_1 - U_2| < 1/2$ -- the break points are within $1/2$ of each other
- $\max(U_1, U_2) > 1/2$ -- not both break points in $[0, 1/2]$
The sample space is the unit square $[0,1]^2$ with area 1. The favorable region is the intersection of all three constraints. By direct geometric calculation (or by inclusion-exclusion on the complementary events), the favorable area is $1/4$.
Geometric argument: Consider the unit square in $(U_1, U_2)$ space. The condition $|U_1 - U_2| < 1/2$ restricts us to a band around the diagonal. The conditions $\min < 1/2$ and $\max > 1/2$ further trim the corners. The resulting region is a square of side $1/2$ centered at $(1/2, 1/2)$, rotated 45 degrees -- its area is $(1/2)^2 = 1/4$.
Alternatively via inclusion-exclusion: Let $E_1 = \{A \geq 1/2\}$, $E_2 = \{B \geq 1/2\}$, $E_3 = \{C \geq 1/2\}$. Since the pieces sum to 1, at most one event can occur at a time, so the events are mutually exclusive. By symmetry and direct calculation, $P(E_1) = P(E_2) = P(E_3) = 1/4$. Thus:
$$P(\text{no triangle}) = P(E_1 \cup E_2 \cup E_3) = 3/4$$
$$P(\text{triangle}) = 1 - 3/4 = 1/4$$
Answer: The probability that the three pieces form a triangle is $\dfrac{1}{4}$.
Intuition
This is a classic geometric probability problem that illustrates a powerful simplification: when pieces sum to a constant, the triangle inequality reduces to a simple "no piece exceeds half the total" condition. The problem then becomes a ratio-of-areas calculation in the space of break points.
The result $1/4$ is surprisingly clean and worth remembering. The same setup appears in many guises: it is equivalent to asking for the probability that three uniform order statistics on $[0,1]$ all lie within distance $1/2$ of each other, or that a uniformly random point in the 2-simplex lies in a specific sub-region. These geometric probability arguments -- setting up a sample space, identifying the feasible region, and computing its volume -- are a staple of quant interviews and show up in problems ranging from order-statistic distributions to random convex hulls.