TITLE: If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a nice proof of this?
QUESTION [80 upvotes]: There is a standard problem in elementary probability that goes as follows.  Consider a stick of length 1.  Pick two points uniformly at random on the stick, and break the stick at those points.  What is the probability that the three segments obtained in this way form a triangle?
Of course this is the probability that no one of the short sticks is longer than 1/2.  This probability turns out to be 1/4.  See, for example, problem 5 in these homework solutions.
It feels like there should be a nice symmetry-based argument for this answer, but I can't figure it out.  I remember seeing once a solution to this problem where the two endpoints of the interval were joined to form a circle, but I can't reconstruct it.  Can anybody help?

REPLY [6 votes]: Here's how I explained it in my blog post a few years ago, as a step in solving a related problem:
If we break the stick at two random points, $x$ and $y$, the three resulting pieces will have lengths $x$, $(y - x)$, and $(1 - y)$ if $x$ is to the left of $y$

and $y$, $(x - y)$, and $(1 - x)$ if $x$ is to the right of $y$

The three pieces form a triangle if none of the pieces is greater than half the length of the stick. In other words, if
$(y > 1/2) AND (x < 1/2) AND (y - x) < 1/2$
when point $x$ is to the left of point $y$ (first image above), and
$(x > 1/2) AND (y < 1/2) AND (x - y) < 1/2$
when $x$ is to the right of $y$ (second image above). If we plot all six of these inequalities we get the area that represents the proportion of triangles formed from our broken stick.

The shaded regions representing the conditions that form a triangle add up to 1/4 of the total area, or a 0.25 probability of forming a triangle.