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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?

flag	asked Oct 23 at 2:28 Michael Lugo 4,540●3●23

Kevin Costello · Accepted Answer · 2009-10-23 02:50:55Z UTC

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.

Jason Dyer · Answer 2 · 2009-10-23 02:59:03Z UTC

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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

answered Oct 23 at 2:59

Jason Dyer
1,195●2●15

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This would be the part where I, somewhat embarrassedly, point out that I wrote that. – Michael Lugo Oct 23 at 4:13

Ilya Nikokoshev · Answer 3 · 2009-10-23 15:59:55Z UTC

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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!

edited Oct 23 at 15:59

answered Oct 23 at 3:57

Ilya Nikokoshev
5,908●1●7●32

"First, the way you define the probability, because of the symmetry (sic!) it's equivalent to first choosing first point a uniformly and then second point b uniformly on [a, 1]." I don't think that this is the case. In the original problem, the probability that there was a piece longer than 1-epsilon decayed like C/epsilon^2 as epsilon tended to 0 (a necessary condition is that no cut lies in (epsilon, 1-epsilon). In your model, it decays like c/epsilon (a sufficient condition is to have A larger than 1-ep). – Kevin Costello Oct 23 at 4:35

Yes, you were right to point my mistake. The geometric picture is correct, but I referred to it incorrectly. Will edit. – Ilya Nikokoshev Oct 23 at 15:57

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Just rephrasing your argument: One can partition the 2-dim simplex, defined by x>=0, y>=0, z>=, x+y+z=1 into 4 identical triangles defined by adding the conditions: 1) x>=½ 2) y>=½ 3) z>=½ 4) x<=½ and y<=½ and z<=½ of which the 4th is the event that you can form a triangle. – Ori Gurel-Gurevich Nov 4 at 4:09

Great way to say it indeed. – Ilya Nikokoshev Nov 4 at 20:59

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?

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