A rope of 1m is divided into three pieces by two random points. Find the average length of the largest segment.
This question has been answered here: Average length of the longest segment
Someone had a solution:
let the cuts be at $X, Y$, with $Y \gt X$:
Then each piece is equally likely to be the longest, and the expected length of the longest piece doesn't depend on which piece we choose. Then we can calculate $\mathop{\mathbb{E}}(X|X \text{ is the longest piece} )$.
We have the three inequalities: $$X \gt Y-X \implies Y < 2X$$ $$X \gt 1-Y \implies Y > 1-X$$ and, from our setup, $$Y \gt X$$ These can be represented by the following diagram:
Then the area satisfying our inequalities is the two triangles A and B. So we wish to find the expected value of $X$ within this area.
The expected value of $X$ in A is $\bar{X}_A = \frac{1}{2}-\frac{1}{3}(\frac{1}{2}-\frac{1}{3}) = \frac{8}{18}$.
The expected value of $X$ in B is $\bar{X}_B = \frac{1}{2}+\frac{1}{3}(\frac{1}{2}) = \frac{4}{6} = \frac{12}{18}$
The area of A is $A_A = \frac{1}{2} \times \frac{1}{2}\times (\frac{1}{2}-\frac{1}{3}) = \frac{1}{24}$.
The area of B is $A_B = \frac{1}{2} \times \frac{1}{2}\times \frac{1}{2} = \frac{1}{8} = \frac{3}{24} = 3 A_A$.
So $\mathop{\mathbb{E}}(X|X \text{ is the longest piece} ) = \frac{\tfrac{8}{18} + 3\left(\tfrac{12}{18}\right)}{4} = \frac{11}{18}$
I understand everything but I got lost when he said calculate expected value of $X$ within triangles. How do you exactly find the expected value of $X$ over the triangle $A$ and $B$? How did he come up with $\mathop{\mathbb{E}}(X|X \text{ is the longest piece} )$?
