Stick Broken in Three: EV of Middle Piece
What is the expected length of the middle piece if a 1 m stick is broken at two independent uniform random points?
Answer
Let \(X\) be the first breaking point and \(Y\) the second, each \(\mathrm{Uniform}(0,1)\). By symmetry it suffices to compute the contribution when \(X<Y\) and then double.
Case \(X<Y\): The three pieces have lengths
\[
X,\quad Y - X,\quad 1 - Y,
\]
so the middle piece is \(Y - X\). Its expected value in this region can be computed as
\[
\int_{y=0}^{1}\int_{x=0}^{y}(y - x)\,dx\,dy
= \int_{0}^{1}\frac{y^2}{2}\,dy
= \frac{1}{6}.
\]
By symmetry,
\[
E\bigl[\lvert Y - X\rvert\bigr]
= E\bigl[(Y - X)\mathbf{1}_{\{X<Y\}}\bigr]
+ E\bigl[(X - Y)\mathbf{1}_{\{Y<X\}}\bigr]
= \tfrac{1}{6} + \tfrac{1}{6}
= \tfrac{1}{3}.
\]
Notes and comments
Comment 1: Note that here we are computing the expectation of the piece that is “middle” in position (i.e., the segment between the two breakpoints), which is different from asking for the expected length of the piece whose length ranks in the middle when all three lengths are ordered.