# Break a stick into two pieces, and break the shorter one again, what's the pdf of the shortest pieces.

Break a stick of unit length at a uniformly chosen random point. Then take the shorter of the two pieces and break it into two again at a uniformly chosen random point. Let X denote the shortest of the final three pieces. Find the probability density function of X.

I let the shorter one on the first break be x, which should be between (0, $$1/2$$), so the pdf should be $$f(x)=2, 0 then the conditional probability of $$f(x|y)$$ should be (since it is another uniform distribution) $$f(x|y)=\frac{1}{y/2-0}=\frac{2}{y}$$ by applying conditional probability formular $$f_X(x)=\int_0^{1/2}f_{X|Y}(x|y)f_Y{y}dy=\int_0^{1/2}\frac{2}{y}\times2dy=4ln(y)|^{1/2}_0$$

I stopped here since it is impossible. Can someone help?

$$p(y) = 8 - 32 y\quad {\rm for}\ 0\leq y \leq 1/4$$
and $$0$$ otherwise.
After the first break the short leg is uniformly likely between $$0 \leftrightarrow 1/2$$, as shown by the range of the abscissa in the graph.
After this short piece is broken, it longer remaining part is uniformly distributed in $$1/4 \leftrightarrow 1/2$$. The other (shorter) part obeys a triangle distribution between $$0 \leftrightarrow 1/4$$. So just write down a triangle distribution that goes between $$0 \leftrightarrow 1/4$$.
The last integral should be taken from $$x$$ to $$1/2$$, since $$f(x \mid y)$$ is zero when $$x > y$$.