Unable to find error in finding area of inequality

Question

I was trying to find the probability of breaking a unit stick at two points to form a triangle which is 1/4 by graphing some inequalities. First I let the first point be x, and the second be y (on a unit stick). I let x be some arbitrary point between 0 and 1/2 and y be some point between 1/2 and 1. Then, I also graphed y-x < 1/2 as the distance between x and y should not exceed 1/2. Upon graphing the inequalities: $ 0<x<\frac{1}{2} $ , $ \frac{1}{2} < y < 1 $, and $ y-x < \frac{1}{2}$, I got the intersection of these regions to be a triangle of area 1/8, where did I go wrong in my reasoning?

Answer 1

I should have considered the possibility that y could have also been the first point, thus I have the other inequalities:

$0<y<\frac{1}{2}$

$\frac{1}{2}<x<1$

and $y-x < \frac{1}{2}$ becomes $|y-x| < \frac{1}{2}$

I then get two 1/8 triangles giving 1/4.