lordjeebus

I believe the answer is 1/4. I will publish my findings shortly.

EDIT: my theory.

so, you're going to get three segments. let's call them x, y, and z.

because the original line had length 1, x+y+z=1.

To form a triangle, no one side can be longer than the other two sides combined. Hence, x+y>z, y+z>x, x+z>y.

Combing the equations x+y+z=1 and x+y>z, we find that x+y = 1-z > z; 1 > 2z; so z<.5. Similarly, y<.5 and x<.5 to form a triangle.

What are the chances of randomly picking x, y, and z, so that all three are less than .5?

The set of all x, y, and z that add up to 1 form a plane. We will only consider the section of that plane in which x, y, and z are positive. This section of the plane is shaped like an equilateral triangle.

By connecting the midpoints of that equilateral triangle, a second triangle can be formed. Within this triangle, and only within this triangle, are the conditions x+y+z=1, x<.5, y<.5, and z<.5 met.

The larger triangle is the set of ways the line could be cut twice. The smaller one is the set of ways that could form a triangle.

The ratio of the smaller triangle to the larger one is 1:4. Hence, given two random cuts of the line, the chance of being able to form a triangle is 1/4.

If there is interest I'll make some illustrations to make things a bit clearer. I'm sure you can solve this without thinking so geometrically, but I find this way easier to understand.