Non-geometric way to calculate expected value of breaks?

Question

In "50 Challenging Problems in Probability", question #43 is the following:

"A bar is broken at random in two places. Find the average size of the smallest, of the middle-sized, and of the largest pieces."

The author gives what seems like a complicated geometric way of calculating the probabilities. He arrives at the solutions 1/9, 5/18, and 11/18. Is there a simpler, non-geometric way of calculating these probabilities?

Answer 1

Here's a sketch. Pick points $0\le x\le y\le 1$. Now let's find the expected value of $x$ given that the interval $[0,x]$ is shortest. That is, we look at the region satisfying \begin{align*} x&\le y \\ x&\le y-x \\ x&\le 1-y \end{align*} You can draw your own pictures. We then get a triangle with $0\le x\le 1/3$, $2x\le y\le 1-x$. This triangle has area $1/6$ (by inspection) or by double-integration, and $$\int_0^{1/3}\int_{2x}^{1-x} x\,dy\,dx = \frac1{54},$$ so the expected value of $x$ is $\dfrac{\frac1{54}}{\frac16} = \dfrac19$.

Next case. Suppose the interval $[0,x]$ has middle length. This corresponds to two regions: \begin{align*} x&\le y \\ y-x&\le x \\ x&\le 1-y \end{align*} OR \begin{align*} x&\le y \\ x&\le y-x \\ 1-y&\le x \end{align*} Interestingly, these both have area $1/12$ and for each of these we get $\displaystyle\iint x\,dA = \dfrac5{216}$. So the expected value of $x$ in the middle case is $\dfrac{\frac5{216}}{\frac1{12}} = \dfrac5{18}$.

The last case follows by subtraction.

Answer 2

The general solution has been discussed in a paper ``On the lengths of the pieces of a stick broken at random'', and some information is also there in Feller's book, vol. 2.

Also see here.

For the $k^{th}$ smallest piece indicated by $S_{(k)}$, the expected length is given by

\begin{align*} \mathbb{E}\left(n\, S_{(k)}\right) &= \mathbb{E}\left(X_{(k)}\right) \\ &= \sum_{i=0}^{k-1}\frac{1}{n-i} \end{align*} which simplifies to the formula given in the book of 50 challenging problems.

For the question asked, $n=3$ and calculate the sums for $k=1,2,3$, which we find to be \begin{align*} \mathbb{E}\left(S_{(1)}\right) &= \frac{1}{9}\\ \mathbb{E}\left(S_{(2)}\right) &= \frac{5}{18}\\ \mathbb{E}\left(S_{(3)}\right) &= \frac{11}{18} \end{align*}