Solution
Break the Longer Piece
Show the problem again
A stick of length 1 is broken at a point chosen uniformly at random. The longer of the two pieces is then broken at a point chosen uniformly at random along its length. What is the probability that the three resulting pieces can form a triangle? Give an exact closed form, or a decimal rounded to 4 places.
Worked solution
The answer is 2 ln 2 − 1 ≈ 0.3863. The longer piece has length L, uniform on [1/2, 1], and the short piece 1 − L is automatically less than 1/2. Breaking the long piece at a, the pieces are a, L − a, and 1 − L. A triangle needs every piece below 1/2, so a < 1/2 and L − a < 1/2, i.e. a ∈ (L − 1/2, 1/2), an interval of length 1 − L out of L. Hence P = 2 ∫_{1/2}^1 (1 − L)/L dL = 2(ln 2 − 1/2) = 2 ln 2 − 1. This beats the 1/4 for two independent breaks: always cutting the longer piece avoids the most common failure, a piece longer than 1/2.
Source: Classic geometric-probability exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.