Break the longer piece: does it still make a triangle?

Question

Break a stick of length 1 at a uniformly random point. Then take the longer of the two pieces and break it at a uniformly random point along its length, giving three pieces total. What is the probability the three pieces can form a triangle?

Accepted Answer

Three lengths form a triangle iff each is below 	frac12. Let the first break give a longer piece of length M = \max(X, 1-X), where X \sim U(0,1); since M \ge 	frac12, the untouched short piece 1 - M \le 	frac12 is never the problem. The longer piece is then split at U \sim U(0,1) into MU and M(1-U), and we need both below 	frac12: MU < 	frac12 \ 	ext{and}\ M(1-U) < 	frac12 \quad\Longleftrightarrow\quad 1 - 	frac{1}{2M} < U < 	frac{1}{2M}. That interval has length 	frac{1}{2M} - \left(1 - 	frac{1}{2M}ight) = 	frac1M - 1 = 	frac{1-M}{M}, so P(	ext{triangle} \mid M) = \frac{1 - M…

Break the longer piece: does it still make a triangle?

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