pr.probability - 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?

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!