Mistake in 'average size of smallest piece'

Question

This is related to the question

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.

I have already seen related questions: 1, 2.

My question is with regard to a way that I thought of:
Let the pieces have lengths $a$, $b$, $1-a-b$. Let the separating points be $A$ and $B$ respectively. Then, the probability distribution of all these lengths is the same.
WLOG, let the leftmost piece be the smallest, and rightmost piece be of largest length [middle one being in between, position- and length-wise]. This is one of the $3!$ possible cases. So, I calculate the mean length in this case, and multiply it by $3!$:

Edited based on further thought, and the comment by @lulu:

Let the length of the leftmost segment be $x\ (\le t)$. Then:

In the case we are considering (described above, as one of the $3!$ cases), the CDF is $$F(t,x)=Prob(a\le t, and\ the\ order\ smallest-middle-largest)=Prob(A\in[0,t)\ \cap\ B\in[2x,x+(1-x)/2))$$ $$=t \cdot \frac{1-3x}{2}$$

I am not sure on how to proceed, but replacing $x$ by its expected value in terms of $t$ doesn't seem to work:

$$F_1(t) = t \cdot \frac{1-3(t/2)}{2} = \frac{t(2-3t)}{4}$$ This is a monotonically increasing function in $[0,1/3]$.

Using this to compute the expectation as $$\int_0^{1/3} (t/2)\cdot \frac{d}{dt}F_1(t)\ dt=1/216$$ Multiplying this by $3!$ gives $1/36$, which is off by a factor of $4$.

Answer 1

Set the length of the bar at $1$. Break off a piece of length $x$. Of the remaning $1-x$ break off a piece of length $y$. This leaves a third part $z = 1-x-y$. We now have a probability space $0 < x < 1$, $0 < y < 1-x$ in which we can perform calculations.

In order to evaluate the average values of the shortest, middle and longest part we must first identify how these are defined in the triangular $(x,y)$ space. It turns out that the different regions are separated by three straight lines: $z = x$ (or $y = 1-2x$); $z=y$ (or $y=(1-x)/2$); and $y=x$. These three lines all pass through the point $(1/3, 1/3)$ and divide the original triangle into six smaller equal-sized triangles.

Each of these six smaller triangles have basically the same properties. So we can take one that is convenient. For example the region where $x < y < z$ corresponds to the trangle $0 < x <1/3$, $x < y < (1-x)/2$. We can now evaluate averages by performing integrations:

$$I = \int_0^{1/3}dx \int_{x}^{(1-x)/2}dy f(x,y)$$

We can take $f=1$ for normalization purposes. And we can take $f=x$, $f=y$, $f=z = (1-x-y)$ to calculate the average of the three lengths. Performing these double integrals is straightforward. The result is that the average shortest length is $2/18$, the average middle one is $5/18$ and the average longest length is $11/18$.