TITLE: expected length of broken stick
QUESTION [9 upvotes]: You break a stick of unit length in two.
You then subsequently break the biggest of the resulting two sides in two, thus obtaining three pieces.
What is the expected length of the smallest of the three?
(Each breaking of a stick is assumed to be at a random point in that stick, uniformly distributed.)

REPLY [10 votes]: If the shorter stick after the first break has length $S$ then $S$ uniformly distributed on $[0,\frac{L}{2}]$.
If $S=s$ then the shorter part of the longer stick will have length $T$ uniformly distributed on $[0,\frac{L-s}{2}]$.  You will certainly have $T \lt s$ if $\frac{L}{3} \lt s \le \frac{L}{2}$.  
So the expectation of the minimum of $S$ and $T$ is 
$$\int_{s=\frac{L}{3}}^\frac{L}{2} \int_{t=0}^\frac{L-s}{2} t \frac{2}{L-s} \frac{2}{L} \,dt\,ds + \int_{s=0}^\frac{L}{3} \int_{t=0}^s t \frac{2}{L-s} \frac{2}{L} \,dt\,ds + \int_{s=0}^\frac{L}{3} \int_{t=s}^\frac{L-s}{2} s \frac{2}{L-s} \frac{2}{L} \,dt\,ds $$    $$= \frac{7 L}{144}  +2L\log\left( \frac{3}{2}\right)-\frac{7L}{9} +\frac{5L}{3} -4L\log\left( \frac{3}{2}\right) $$ $$= L\left(\frac{15}{16} -2\log\left( \frac{3}{2}\right) \right)$$ the same as did's answer