The Broken Stick

The problem
break a stick at two random points01what is P(the three pieces form a triangle)?

A stick of length 11 is broken at two points chosen uniformly at random and independently. The stick is now in three pieces.

What is the probability that the three pieces can be assembled into a triangle?

Tempting (but wrong)
most guesses: ½ or ⅓0.200.300.50biggest piece ≥ ½ → no triangle0.100.550.35biggest piece ≥ ½ → no triangle

The common guesses:

  • "12\tfrac{1}{2}" — feels like triangles are easy enough to form that it should work about half the time.
  • "13\tfrac{1}{3}" — three pieces, one of three orderings, vaguely 13\tfrac{1}{3}.

Both ignore the triangle inequality: three lengths a,b,ca, b, c form a triangle iff each is strictly less than the sum of the other two. With a+b+c=1a + b + c = 1, this collapses to: every piece must be <12< \tfrac{1}{2}. That's a stronger condition than intuition admits.