The problem
A stick of length 1 is broken at two points chosen uniformly at random. What is the probability the three pieces form a triangle?
The condition
Three lengths form a triangle exactly when each is less than the sum of the other two. Since they sum to 1, that reduces to a single clean condition:
No piece may be longer than 1/2.
The geometric solution
Let the two break points be x and y, uniform on [0,1]. The sample space is the unit square, area 1.
Assume x < y (the other case is symmetric). The three pieces are x, y - x, and 1 - y. All three must be under 1/2:
- x < 1/2
- y - x < 1/2
- 1 - y < 1/2, so y > 1/2
Shading that region in the triangle where x < y gives an area of 1/8. Doubling for the symmetric case x > y gives 1/4.
Drawing the square and shading it is the fastest way to present this, and interviewers usually accept the picture as the proof.
The sequential variant
Now break the stick once, then break the longer piece. The answer changes.
The first break is at x, uniform. Say x < 1/2, so the longer piece has length 1 - x and is broken uniformly within it. Working through the triangle condition and integrating gives
ln 2 - 1/2 ≈ 0.193
Lower than 1/4, which makes sense: always breaking the longer piece tends to produce one dominant piece more often than two independent breaks do.
Interviewers ask both versions back to back. Candidates who answer 1/4 to the second without noticing the change are the target.
The general lesson
For continuous uniform problems, sketching the sample space as a region and computing an area beats integrating almost every time - it is faster, less error-prone, and easier to explain out loud.
Related: symmetry arguments, and practise in probability.