The Longest Broken Piece

A stick of length $1$ is broken at two points chosen independently and uniformly along its length, giving three pieces. What is the expected length of the longest piece? What about the shortest and the middle one?

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#Three exchangeable pieces

Two uniform cuts split $[0,1]$ into three pieces whose lengths are exchangeable and sum to $1$ , so each piece on its own averages $\tfrac13$ . The order statistics, shortest to longest, split that $1$ unevenly.

#The shortest piece directly

All three pieces exceed $t$ only when the two cuts avoid the two end-zones of width $t$ , which by a geometric argument has probability $(1 - 3t)^2$ for $t \le \tfrac13$ . Hence

\E[\text{shortest}] = \int_0^{1/3} (1 - 3t)^2\, dt = \left[-\frac{(1-3t)^3}{9}\right]_0^{1/3} = \frac19. \tag{1}

#The order-statistic formula

For $n$ uniform spacings the $j$ -th smallest has expected length

\E[L_{(j)}] = \frac{1}{n}\sum_{i = n-j+1}^{n} \frac{1}{i}. \tag{2}

With $n = 3$ this gives

\E[\text{shortest}] = \tfrac13 \cdot \tfrac13 = \tfrac19, \qquad \E[\text{middle}] = \tfrac13\big(\tfrac13 + \tfrac12\big) = \tfrac{5}{18}, \qquad \E[\text{longest}] = \tfrac13\big(\tfrac13 + \tfrac12 + 1\big) = \tfrac{11}{18}. \tag{3}

Averaged over all breaks, the longest piece claims eleven eighteenths of the stick, the middle five eighteenths, and the shortest a ninth. The longest alone is more than half the stick, far from an even three-way split.

#Result

The longest piece averages $\tfrac{11}{18} \approx 0.61$ , the middle $\tfrac{5}{18} \approx 0.28$ , and the shortest $\tfrac19 \approx 0.11$ , and the three sum to $1$ as they must. The longest piece alone is more than half the stick on average.