A stick is broken in three pieces at random. Compute the probability that the pi

Question

A stick is broken in three pieces at random. Compute the probability that the pieces are sides of the triangle (that it, the sum of any two is bigger than the third). Consider the following three possibilities (a) selecting two iid uniforms on the stick (then the probability is 1/4, best seeing by thinking of the stick as the height of an equilateral triangle); (b) breaking the stick at a uniform point in two pieces, then taking one of the pieces with equal probability and breaking it again (then the probability is 2(ln 20.5)); (c) breaking the stick at a uniform point in two pieces, then taking one of the pieces with probability proportional to the length of the piece and breaking it again (then the probability is 1/4).

Explanation / Answer

The answer is 1/4 for a circular arena and 33/128 for a square one, so probably very close to that for an equilateral triangle.
Here's what seems like the sort of argument you're looking for (based off of a trick Wendel used to compute the probability the convex hull of a set of random points on a sphere contains the center of the sphere, which is really the same question in disguise):
Connect the endpoints of the stick into a circle. We now imagine we're cutting at three points instead of two. We can form a triangle if none of the resulting pieces is at least 1/2, i.e. if no semicircle contains all three of our cut points.
Now imagine our cut as being formed in two stages. In the first stage, we choose three pairs of antipodal points on the circle. In the second, we choose one point from each pair to cut at. The sets of three points lying in a semicircle (the nontriangles) correspond exactly to the sets of three consecutive points out of our six chosen points. This means that 6 out of the possible 8 selections in the second stage lead to a non-triangle, regardless of the pairs of points chosen in the first stage
Consider an equilateral triangle with altitude 1. It is not hard to show that if you choose a point randomly in this triangle, the distances to the three sides gives the same distribution of lengths that you obtain by breaking a stick at two random points. Now, the locus of points for which no distance is longer than 1/2 is the smaller equilateral triangle formed by joining the midpoints of the edges, which has area 1/4 that of the original triangle.

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