Expected lengths of the three broken-stick pieces

Solution. The three pieces sum to $1$, so the three expectations sum to $1$. The pieces are exchangeable (invariant under relabelling), but their order statistics $S, M, L$ have different distributions and hence different means.

Step 1: density of the simplex. The pieces $(a, b, c)$ with $a + b + c = 1$ and $a, b, c \geq 0$ are uniform on the $2$-simplex with density $2$ (the simplex has area $1/2$).

Step 2: $\mathbb{E}[S]$ via tail integration. Compute $\mathbb{P}(S > t) = \mathbb{P}(\text{every piece} > t)$. The event requires $a > t$, $b > t$, and $1 - a - b > t$, which describes a smaller simplex of side $1 - 3t$. Thus $$\mathbb{P}(S > t) = (1 - 3t)^2, \qquad 0 \leq t \leq \tfrac13.$$ Therefore $$\mathbb{E}[S] = \int_0^{1/3} (1 - 3t)^2 \, dt = \left[ -\tfrac{(1-3t)^3}{9} \right]_0^{1/3} = \tfrac19.$$

Step 3: $\mathbb{E}[L]$ via inclusion–exclusion. The longest piece exceeds $t$ iff some piece does. The three events $\{a > t\}$, $\{b > t\}$, $\{c > t\}$ have pairwise and triple intersections computable from the simplex density. By inclusion–exclusion, $$\mathbb{P}(L > t) = 3 \mathbb{P}(a > t) - 3 \mathbb{P}(a > t, b > t) + \mathbb{P}(a > t, b > t, c > t).$$ Compute each term for $t \in [0, 1]$ (the pairwise and triple terms vanish when $t > 1/2$ and $t > 1/3$ respectively). After integration one obtains $$\mathbb{E}[L] = \frac{11}{18}.$$

Step 4: $\mathbb{E}[M]$ by subtraction. Since $S + M + L = 1$, $$\mathbb{E}[M] = 1 - \mathbb{E}[S] - \mathbb{E}[L] = 1 - \tfrac19 - \tfrac{11}{18} = \tfrac{5}{18}.$$

In summary, $$\boxed{\mathbb{E}[S] = \tfrac{1}{9}, \qquad \mathbb{E}[M] = \tfrac{5}{18}, \qquad \mathbb{E}[L] = \tfrac{11}{18}.} \qedhere$$

import numpy as np
x, y = np.random.rand(2, 10**7)
lo, hi = np.minimum(x, y), np.maximum(x, y)
pieces = np.stack([lo, hi - lo, 1 - hi])
sorted_pieces = np.sort(pieces, axis=0)
print(f"E[S] sim: {sorted_pieces[0].mean():.5f}  exact: {1/9:.5f}")
print(f"E[M] sim: {sorted_pieces[1].mean():.5f}  exact: {5/18:.5f}")
print(f"E[L] sim: {sorted_pieces[2].mean():.5f}  exact: {11/18:.5f}")
# E[S] sim: 0.11116  exact: 0.11111
# E[M] sim: 0.27782  exact: 0.27778
# E[L] sim: 0.61102  exact: 0.61111