Snap a stick, make a triangle

You pick up a stick and snap it at two random spots, giving you three pieces. Can you arrange them into a triangle?

What do you think?

What fraction of random breaks produce three pieces that form a triangle?

Try it. Break some sticks and see how many triangles you get:

Stick Breaker

After enough breaks, your success rate locks in around 25%. But why that number?

The triangle inequality

Three lengths $a$ , $b$ , $c$ form a triangle only when every side is shorter than the other two combined:

$a + b > c \quad\text{and}\quad a + c > b \quad\text{and}\quad b + c > a$

Since the full stick has length 1, we know $a + b + c = 1$ . So $a + b > c$ becomes $1 - c > c$ , meaning $c < 1/2$ . The same logic applies to $a$ and $b$ .

A triangle forms only when no single piece is at least half the stick. Every piece must be shorter than 1/2.

Drag the sliders below to see this in action. When any piece hits 50%, the triangle collapses:

Triangle Inequality Explorer

Drag any slider to reshape the pieces. A triangle forms only when every piece is shorter than 1/2. Watch what happens as one side crosses the threshold.

a = 33.0%

b = 34.0%

c = 33.0%

✓ Valid triangleAll sides < 50%

That's a clean condition. But what fraction of random breaks satisfy it?

Mapping breaks to points

Call the two break points $x$ and $y$ , both chosen uniformly on $[0,1]$ . We use $[0,1]$ because we can always rescale any stick to have length 1 without changing the proportions, and proportions are all that matter for the triangle inequality. "Uniformly" means each point is equally likely to land anywhere along the stick, with no spot favored over another.

Try it yourself. Click twice on the stick below to place your two break points:

Pick Your Break Points

Click anywhere on the stick to place the first break point.

Your break points map to a point in the unit square. The green diamond is the region where a triangle forms:

Your Point on the Unit Square

Place break points above to see them here.

And here's the triangle (or failed triangle) your break points produce:

Triangle Attempt

Place two break points above to see the triangle attempt.

Since $x$ can be anything in $[0,1]$ and $y$ can independently be anything in $[0,1]$ , the pair $(x, y)$ lives in the square $[0,1] \times [0,1]$ . Every point in that square corresponds to one way of breaking the stick, and because both choices are uniform and independent, every point in the square is equally likely. That is the unit square.

After sorting so the smaller value comes first, the three piece lengths are:

$a = \min(x,y), \quad b = |x - y|, \quad c = 1 - \max(x,y)$

The triangle condition (each piece $< 1/2$ ) carves out a region inside the unit square. Points in that region produce triangles. Points outside don't.

Triangle Region

Each random point (x, y) in the unit square represents two break points. The shaded region is where the three pieces form a triangle.

Drop enough points and you'll see the valid region is a diamond taking up exactly 1/4 of the square.

Why 1/4?

\text{Break points } x, y \sim \text{Uniform}(0,1)

Two independent uniform random variables define a point in the unit square.

Step 1 of 6

Competition connections

This problem (or variations of it) has appeared in:

IMO Shortlist problems on geometric probability
Putnam competition practice sets
AMC/AIME in the form "what is the probability three random lengths form a triangle"

The technique of mapping a probability question to an area calculation is called geometric probability. You'll see it again and again in competition math.

What do you think?

What if you break the stick into three pieces by choosing the lengths uniformly (not the break points)? How does this differ?

Geometric Probability

When outcomes live in a continuous space (a line segment, a square, etc.), the probability of an event equals the measure (length, area, volume) of the "favorable" region divided by the measure of the total region. This only works when the distribution is uniform over the space.

What changes with more pieces?

If you make $n - 1$ random cuts (producing $n$ pieces), the probability that the pieces form an $n$ -gon is $1 - n / 2^{n-1}$ . The same inclusion-exclusion logic applies: at most one piece can be $\ge 1/2$ (since they sum to 1), and by symmetry each piece has probability $(1/2)^{n-1}$ of being that long.

Pieces	Polygon	P(valid)
3	Triangle	1/4 = 25.0%
4	Quadrilateral	1/2 = 50.0%
5	Pentagon	11/16 = 68.75%

Use the explorer below to see how the probability climbs as you add more pieces, and run a simulation to verify:

n-Piece Polygon Probability

Pieces (n):5

Cut a stick at 4 random points → 5 pieces → form a pentagon?

P = 1 − 5 / 2⁴ = 11/16 = 68.75%

How probability grows with the number of pieces:

More pieces actually make polygon formation more likely, because it's harder for any single piece to dominate when you're dividing the stick into many parts. As $n \to \infty$ , the probability approaches 1.

A stick of length 10 is broken at two random points. What is the probability the longest piece is more than 5? (decimal like 0.75)

1/3

The takeaway

The broken stick problem introduces geometric probability: instead of counting discrete outcomes, you compute areas (or volumes) of regions in continuous spaces. Mapping a probability question to a geometry question is a common technique in competition math. When you see "choose uniformly at random from a continuous set," think shapes and areas.