Expected Longest Piece From Two Random Cuts

Expectation · Medium · Free problem

You have a stick of unit length. You pick two points independently and uniformly at random on $[0,1]$ and cut the stick at those points, producing three pieces.

Let $U, V \sim \text{Unif}(0,1)$ independently, and define the ordered cut points $X = \min(U,V)$ and $Y = \max(U,V)$. The three piece lengths are $L_1 = X$, $L_2 = Y - X$, and $L_3 = 1 - Y$.

Compute $E[\max(L_1, L_2, L_3)]$.

Hints

  1. The three pieces are exchangeable -- by symmetry, you only need to compute the contribution from one of them being the largest and multiply by 3.
  2. The ordered cut points $(X, Y)$ have joint density $f(x,y) = 2$ on the simplex $\{0 \le x \le y \le 1\}$. Partition this simplex into regions based on which piece is longest.
  3. In the region where $L_1 = x$ is longest, the constraints are $x \ge y/2$ and $x \ge 1-y$. Split the integral at $y = 2/3$ (where $y/2 = 1 - y$) and evaluate each part.

Worked Solution

How to Think About It: Two random cuts split a unit stick into three pieces, and you want the expected length of the longest one. First observation: by symmetry, the three pieces are exchangeable -- they have the same marginal distribution (each has mean

/3$). So the longest piece must average more than
/3$, but how much more? The max of three things that sum to 1 is at least
/3$ (pigeonhole), and at most 1 (degenerate case). Intuitively, the random variation pushes the longest piece well above
/3$. This is a clean geometric integration problem over the simplex.

Quick Estimate: The three pieces follow a symmetric $\text{Dirichlet}(1,1,1)$ distribution (i.e., uniform on the 2-simplex). Each piece has mean

/3$ and variance
/18$, so standard deviation $\approx 0.236$. The max of three such pieces should be roughly the mean plus a bit more than one standard deviation:
/3 + 0.24 \approx 0.57$ to $0.63$. A more refined heuristic: for $n+1$ exchangeable pieces summing to 1, $E[\max] = H_{n+1}/(n+1)$ where $H_k$ is the $k$-th harmonic number. For 3 pieces: $H_3/3 = (1 + 1/2 + 1/3)/3 = (11/6)/3 = 11/18 \approx 0.611$. That is our target.

Approach: We integrate $\max(x, y-x, 1-y)$ over the simplex $0 \le x \le y \le 1$ with density 2 (the joint density of the order statistics of two uniform random variables).

Formal Solution:

The joint density of $(X, Y)$ on $\{0 \le x \le y \le 1\}$ is $f(x,y) = 2$, so:

$E[\max(L_1, L_2, L_3)] = 2 \int_0^1 \int_0^y \max(x,\; y-x,\; 1-y)\, dx\, dy$

We partition the simplex into three regions based on which piece is longest:

  • Region $R_1$: $L_1 \ge L_2$ and $L_1 \ge L_3$, i.e., $x \ge y - x$ and $x \ge 1 - y$, so $x \ge y/2$ and $x \ge 1 - y$.
  • Region $R_2$: $L_2 \ge L_1$ and $L_2 \ge L_3$, i.e., $y - x \ge x$ and $y - x \ge 1 - y$, so $x \le y/2$ and $y \ge (1+x)/2$.
  • Region $R_3$: $L_3 \ge L_1$ and $L_3 \ge L_2$, i.e.,
    - y \ge x$ and
    - y \ge y - x$, so $y \le 1 - x$ and $y \le (1+x)/2$.

By the exchangeability of the three pieces under the Dirichlet distribution, all three integrals give the same value. So:

$E[\max] = 3 \times 2 \int\!\!\int_{R_1} x\, dx\, dy$

We compute the integral over $R_1$. The constraints are $0 \le x \le y \le 1$, $x \ge y/2$, and $x \ge 1 - y$. Since $x \ge y/2$ and $x \le y$, we have $y/2 \le x \le y$, so $y \ge 0$. Also, $x \ge 1 - y$ combined with $x \le y$ requires $y \ge 1 - y$, i.e., $y \ge 1/2$. But we also need $x \ge \max(y/2, 1-y)$.

  • When $y/2 \ge 1 - y$, i.e., $y \ge 2/3$: lower bound on $x$ is $y/2$.
  • When $y/2 < 1 - y$, i.e.,
    /2 \le y < 2/3$: lower bound on $x$ is
    - y$.

For

/2 \le y < 2/3$:

$I_a = \int_{1/2}^{2/3} \int_{1-y}^{y} x\, dx\, dy = \int_{1/2}^{2/3} \frac{y^2 - (1-y)^2}{2}\, dy = \int_{1/2}^{2/3} \frac{2y - 1}{2}\, dy$

$= \frac{1}{2}\left[y^2 - y\right]_{1/2}^{2/3} = \frac{1}{2}\left[\left(\frac{4}{9} - \frac{2}{3}\right) - \left(\frac{1}{4} - \frac{1}{2}\right)\right] = \frac{1}{2}\left[-\frac{2}{9} + \frac{1}{4}\right] = \frac{1}{2} \cdot \frac{1}{36} = \frac{1}{72}$

For

/3 \le y \le 1$:

$I_b = \int_{2/3}^{1} \int_{y/2}^{y} x\, dx\, dy = \int_{2/3}^{1} \frac{y^2 - y^2/4}{2}\, dy = \int_{2/3}^{1} \frac{3y^2}{8}\, dy$

$= \frac{3}{8}\left[\frac{y^3}{3}\right]_{2/3}^{1} = \frac{1}{8}\left(1 - \frac{8}{27}\right) = \frac{1}{8} \cdot \frac{19}{27} = \frac{19}{216}$

So:

$E[\max] = 6(I_a + I_b) = 6\left(\frac{1}{72} + \frac{19}{216}\right) = 6\left(\frac{3}{216} + \frac{19}{216}\right) = 6 \cdot \frac{22}{216} = \frac{132}{216} = \frac{11}{18}$

Answer:

$E[\max(L_1, L_2, L_3)] = \frac{11}{18} \approx 0.611$

This is a special case of the general formula: for $n$ i.i.d. uniform cuts producing $n+1$ pieces, $E[\max] = H_{n+1}/(n+1)$, where $H_k = \sum_{j=1}^{k} 1/j$ is the $k$-th harmonic number.

Intuition

This problem is a gateway to the rich theory of spacings -- the gaps between order statistics of uniform random variables. Two uniform cuts produce three pieces whose joint distribution is Dirichlet(1,1,1), which is just the uniform distribution on the 2-simplex. Because the pieces are exchangeable, the expected maximum reduces to three times the expected value of one piece restricted to the event that it is the largest. The answer

1/18$ comes from a clean geometric integration, but it is also a special case of the general harmonic formula $E[\max] = H_{n+1}/(n+1)$ for $n$ cuts.

The deeper lesson is about how much "concentration" you get (or don't get) from random partitions. With three pieces averaging

/3$ each, the longest piece averages about $0.611$ -- nearly twice the mean. This reflects the high variability of uniform spacings. In practice, this kind of calculation appears whenever you model random fragmentation: breaking a time interval into trading periods, splitting a portfolio across random allocations, or analyzing the longest gap between events in a Poisson process. The harmonic number structure is a signature of these problems and connects directly to coupon collector, record values, and other classical probability results.

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