Math Competition Question on Forming a Triangle from a broken stick This is a question from a Math-Competition. As the test is multiple choice and the authors only publishe what the correct answer is, but no reason as to why, I would appreciate if someone can tell me why this is correct.

We randomly break a stick in $3$ pieces. What is the probability that you can build a triangle out of those three pieces.

The correct answer: $1/4$
Why is that true? I thought about using the triangle inequality $c<a+b$, but I am not getting anywhere with that...
 A: First, we have to know how we break the stick into three pieces.  There are (at least) two different ways to do it, and I believe they will give different answers.
Assume the sticks has unit length.  We can choose two points independently uniformly at random from $[0,1]$  Assuming the two points are $0<x<y<1$, what is the probability that $x,y-x,1-y$ satisfy the triangle inequality? Look at the three possibilities for the longest, figure out the region where the triangle inequality is satisfied for each, and compute the area, by integration or geometry.  The sum of the three areas is the probability.
The second interpretation is that we break the stick in two, uniformly at random, pick one of the two pieces, uniformly at random, and break it in two, uniformly at random.  Then we ask if the triangle inequality holds.  Note that if we pick up the shorter piece, we have no hope of succeeding.
I would guess that the first interpretation is the one the examiners were looking for, but there's no way to be sure.  It's a badly-phrased question, since it asks for a probability without specifying the probability space.
