Rod cut at two points, expected length of largest part This question is similar to this but I cant find what goes wrong in my answer. The full statement of problem is this:

If a 1 meter rope is cut at two uniformly randomly chosen points (to give three pieces), what is the average length of the largest piece?

What I am doing:
|-------|------|------|
0       x      y      1

I assume rod as of being length 1 unit and divide it at distance $x$ from left and another cut is at distance $y$ from left ($y > x$). Then I take two cases:

*

*If $x \ge \frac{1}{2}$ then wherever we cut $y$ we always get $x$ as largest. So expectation is
$$\int_{1/2}^{1}x(1-x) dx = \frac{1}{12}$$


*If cut is made $\frac{1}{3} \le x \le \frac{1}{2}$ then $y$ may lie only in region $1-x$ to $2x$ for leftmost part to be largest. Then we have
$$\int_{1/3}^{1/2} x(3x-1) dx = \frac{1}{54}$$
So i thought expected length to be $1/12 + 1/54 = \frac{11}{108}$ but this is very wrong correct result is $11/18$
 A: Another approach: multivariable. Let the cuts be $x,y$ from the left, $y>x$. Then the pieces are $x,\,y-x,\,1-y$. The desired expected value is then $2$ times
$$\frac12 E=\iint\limits_{x>y-x\\x>1-y} x\,\mathrm{d}x\,\mathrm{d}y+
\iint\limits_{y-x>x\\y-x>1-y} (y-x)\,\mathrm{d}x\,\mathrm{d}y+
\iint\limits_{1-y>y-x\\1-y>x} (1-y)\,\mathrm{d}x\,\mathrm{d}y.$$

$$\frac12 E=
\left(\int\limits_{1/3}^1 \int\limits_{1-x}^{2x}
-\int\limits_{1/2}^1 \int\limits_{1}^{2x}
-\int\limits_{1/2}^1 \int\limits_{1-x}^{x}\right) x\,\mathrm{d}y\,\mathrm{d}x+\\
\left(\int\limits_{0}^{1/2}\int\limits_{2x}^{1}-
\int\limits_{0}^{1/3}\int\limits_{2x}^{(x+1)/2}\right) (y-x)\,\mathrm{d}y\,\mathrm{d}x+\\
\left(\int\limits_{0}^{1/2}\int\limits_{x}^{1-x}-
\int\limits_{0}^{1/3}\int\limits_{(1+x)/2}^{1-x}\right) (1-y)\,\mathrm{d}y\,\mathrm{d}x$$
It appears that it suffices to take only the first integral, the other should be equal by symmetry.
$$\frac16 E=
\int\limits_{1/3}^1 (3x-1)x\,\mathrm{d}x-
\int\limits_{1/2}^1 (2x-1)x\,\mathrm{d}x-
\int\limits_{1/2}^1 (2x-1)x\,\mathrm{d}x=\\
\frac{14}{27}-2\cdot \frac{5}{24}=\frac{11}{108}.$$
