Break a stick once, longer and shorter piece

Question

A stick of length 1 is broken at a single point chosen uniformly at random, producing two pieces. What are the expected lengths of the longer and shorter pieces? And what is the probability the longer piece is at least twice the shorter?

Accepted Answer

Let U \sim 	ext{Uniform}(0,1) be the break point. The pieces are U and 1 - U, so the longer piece is L{\max} = \max(U, 1 - U) \in [	frac12, 1] and the shorter is L{\min} = 1 - L{\max}. Expected longer piece. E[L{\max}] = \int0^1 \max(u, 1 - u)\,du = 2\int{1/2}^{1} u\,du = 2\left[\frac{u^2}{2}ight]{1/2}^{1} = 2\left(\frac12 - \frac18ight) = \frac34. (The factor 2 uses the symmetry of \max(u, 1-u) about u = 	frac12.) Expected shorter piece. Since the two pieces sum to 1, E[L{\min}] = 1 - E[L{\max}] = \frac14. Probability the longer is at least twice the shorter. We need L{\max} \ge 2 L{\…

Break a stick once, longer and shorter piece

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