I've a straight stick which has been broken into three random-length pieces.

What is the probability that the pieces can be put together to form a triangle?

If you can answer this at this point, please do.

If not, perhaps this will help: here are several methods to break the stick into the three random length pieces:

- I select, independently, and at random, two points from the points that range uniformly along the stick, then break the stick at these two points.
- I select one point, independently, and at random (again uniformly), and break the stick at this point. I then randomly (with even chances) select one of the two sticks and randomly select a point (again uniformly) along that stick, and break it at that point.
- I select one point, independently, and at random (again uniformly), and break the stick at this point. I then select the
**larger** stick, and randomly select a point (again uniformly) along that stick, and break it at that point.

If this clarifies the problem, please show how this affects your work.

<BLOCKQUOTE>

Let the points be `A` and `B`, so `0<=A<=1` and `0<=B<=1`. For the three line segments to form a triangle, the sum of the lengths of any two must be greater than or equal to the length of the third.
**Case 1:** If `0<=B<=A`, the three line segments have lengths `B`, `A-B`, and `1-A`. For these to form a triangle, the following three conditions must be true:

</BLOCKQUOTE>
<BLOCKQUOTE>

` B + (A-B) >= 1-A ==> A >= 1/2` ` (A-B)+(1-A) >= B ==> B <= 1/2` ` B + (1-A) >= A-B ==> A-B <= 1/2`
**Case 2:** If `A<=B<=1`, the same conditions must apply as in case 1, except with `A` substituted for `B` and `B` subsituted for `A`, that is:

` A + (B-A) >= 1-B ==> B >= 1/2`

` (1-B)+(B-A) >= A ==> A <= 1/2`

` A + (1-B) >= B-A ==> B-A <= 1/2`

Plotting these cases on a graph with horizontal axis for `A` and vertical axis for `B`, we get:

` 1 +--+--+`

` | /| |`

` |/t| |`

`1/2 +--+--+`

` | |t/|`

` | |/ |`

` 0 +-----+`

` 0 1/2 1`

The two triangles labeled `t` are the only regions of this unit square where `A` and `B` create three line segments that agree with the conditions for case 1 (the lower right triangle) or case 2 (the upper left triangle). They total 1/4 of the area of the unit square, so the probability of two randomly chosen points craeting line segments that form a triangle is `1/4`.

</BLOCKQUOTE>