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 each fails with the same probability. Take $L = 1$. The left piece is too long exactly when both break points land in the right half, probability $\tfrac12 \cdot \tfrac12 = \tfrac14$. The right piece is the mirror image, also $\tfrac14$. The middle piece is too long when the two break points are more than $\tfrac12$ apart, which is two corner triangles of the unit square with legs $\tfrac12$, area $\tfrac18$ each, again $\tfrac14$. Since at most one piece can exceed half the total, these events never overlap, so the probability that some piece is too long is $3 \cdot \tfrac14 = \tfrac34$. So the probability that none is too long is $\tfrac14$.
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. Split by which break point is smaller. When $U_1 < U_2$, the conditions become $U_1 < 1/2 < U_2$ with $U_2 - U_1 < 1/2$: the triangle with vertices $(0, 1/2)$, $(1/2, 1/2)$, $(1/2, 1)$, area $1/8$. The case $U_2 < U_1$ gives the mirror triangle with vertices $(1/2, 0)$, $(1/2, 1/2)$, $(1, 1/2)$, another $1/8$. The favorable region is these two triangles meeting at $(1/2, 1/2)$, total area $2 \cdot 1/8 = 1/4$. (Not a single connected region: a point like $(0.55, 0.55)$ lies in the diagonal band but fails $\min < 1/2$.)
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. Each has probability $1/4$: $P(E_1) = P(U_1 \geq 1/2,\ U_2 \geq 1/2) = \tfrac14$, $P(E_3) = P(U_1 \leq 1/2,\ U_2 \leq 1/2) = \tfrac14$, and $E_2 = \{|U_1 - U_2| \geq 1/2\}$ is the two corner triangles $\{U_2 \geq U_1 + 1/2\}$ and $\{U_1 \geq U_2 + 1/2\}$, each with legs $1/2$ and area $1/8$, so $P(E_2) = \tfrac14$. 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 uniformly random points on a circle form a triangle containing the center (all three arcs shorter than half the circumference), or that a uniformly random point in the 2-simplex lies in its middle sub-triangle (the one joining the edge midpoints). 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.