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 fact? - MathOverflow

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 fact?

2009-10-23T02:28:05Z

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?

Answer by Kevin Costello for 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 fact?

2009-10-23T02:43:55Z

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.

Answer by Jason Dyer for 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 fact?

2009-10-23T02:59:03Z

Is the argument you remember along the lines of: picking three points on a circle, what is the probability they lie in the same semicircle?

The problem is discussed here:

http://godplaysdice.blogspot.com/2007/10/probabilities-on-circle.html

Answer by Ilya Nikokoshev for 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 fact?

2009-10-23T03:57:55Z

Yes, here's a nice and beautiful argument!

First you should draw a picture of axes a and b. You're asked to select uniformly a point in the square [0,1]x[0,1]. Now because of the symmetry (sic!) it's equivalent to choosing the points a and b uniformly in the triangle cut from the square by b > a.

So you're actually uniformly selecting a point inside triangle defined by lines a>=0, b<=1, 'b>=a'.

Now let's find the conditions to be able to make a triangle of short sticks. We should have a + (1-b) > b-a, b-a + (1-b) > a and b > 1 - b which indeed, as you say, boils down to

b > 1/2,  a < 1/2,  b-a < 1/2

It remains to note that those lines create inside the big triangle a small triangle which is similar to big but with all lengths 1/2 of the big, so this small triangle has area of exactly 1/4 of original!