Timeline 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?

6 events

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Nov 4, 2009 at 20:59	comment	added	Ilya Nikokoshev		Great way to say it indeed.
Nov 4, 2009 at 4:09	comment	added	Ori Gurel-Gurevich		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.
Oct 23, 2009 at 15:59	history	edited	Ilya Nikokoshev	CC BY-SA 2.5	corrected
Oct 23, 2009 at 15:57	comment	added	Ilya Nikokoshev		Yes, you were right to point my mistake. The geometric picture is correct, but I referred to it incorrectly. Will edit.
Oct 23, 2009 at 4:35	comment	added	Kevin P. Costello		"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).
Oct 23, 2009 at 3:57	history	answered	Ilya Nikokoshev	CC BY-SA 2.5