Building a triangle from random segments The problem
Let's consider the interval $[0, 1]$. We break this interval randomly in two parts. Then we choose the bigger part and again we break it randomly into two parts. At the end we have three parts of the interval.
Breaking points are distributed uniformly.
I am to find the probability of getting a triangle from these three pieces.
My attempt
First of all let's notice that a triangle can be build iff:


*

*$a+b>c$,

*$|a-b| < c$,


where $a, b, c$ are lengths of our triangle sights.
Let's denote the breaking points by $x$ and $y$.
Suppose $x > \frac{1}{2}$. We do have three intervals after breaking our segment: $[0, y], [y, x], [x, 1]$.
The first condition of triangle's existence is trivially fulfilled. The second one implies:
$$y > x - {1}{2} \text{  and  } y < \frac{1}{2}.$$
I calculated that:
$$f_{Y|X}(y|x) = \frac{f(x, y)}{\int \limits_{\infty}^{\infty}f_X(x) \mathbb{1}_{[0,x]}(y) \mbox{d}y}.$$
Where $f(x) = \mathbb{1}_{[0,1]}(x)$ and $f(x, y) = \mathbb{1}_{[0,1]}(x)\mathbb{1}_{[0,x]}(y)$.
I don't really know how the integration should look like. I would appreciate any tips or hints.
 A: Let $x$ and $1-x$  be the length of the two pieces after first split.
Relabel the two pieces if necessary, we can assume $x \sim \mathcal{U}(\frac12,1)$ and
the longer piece has length $x$.
Let $xy,x(1-y)$ be the length of pieces after we split the longer piece.
Relabel the two new pieces if necessary, we can assume $y \sim \mathcal{U}(\frac12,1)$.
We now have $3$ pieces of length $xy, x(1-y)$ and $1-x$. In order for them to form
a triangle, the conditions are 


*

*$xy + x(1-y) \ge 1-x \iff x \ge 1-x$

*$xy + 1 - x  \ge x(1-y)$

*$x(1-y) + 1-x \ge xy \iff xy \le \frac12$
The first two conditions are trivially satisfied. In order to satisfy the third condition.
The point $(x,y)$ needs to fall into a shape bounded by two lines $x = \frac12$, $y = \frac12$ and hyperbola $xy = \frac12$. The desired probability is $4$ times the area of this.
$$
4\int_{\frac12}^1 \int_{\frac12}^{\frac{1}{2x}} dy dx =
2\int_{\frac12}^1  \left( \frac{1}{x} - 1 \right) dx
= \log(4)-1 \approx 0.3862943611$$
As pointed out by M.Nestor in comment, a very similar question has been asked before.
However, our answer is different from there 
($\log 2 - \frac12 \approx 0.19314$) because we have been asked to break the longer piece after the first split. Since breaking the shorter piece 
"almost never" give you a valid triangle, our answer is exactly twice of the accepted answer in above question.
