A samurai cuts a piece of bamboo Suppose a samurai wants to try out his new sword and cuts a piece of bamboo twice, randomly, so now there are $3$ lenghts of bamboo. What is the probability of these 3 pieces being able to form a triangle?
I have never came across a continuous probability problem before, but I tried doing it anyway and got a result of 0.25 probability.
My solution: Let $L$ be the original lenght of the bamboo, $x$ be the place of the first cut and $y$ be the place of the second cut. Writing out all the 3 triangle inequalities, we come to the conclusion that no piece of bamboo can have more than $L/2$ lenght, then the probability we're looking for is:
$$
\frac{\int_{x=0}^{L/2}(\int_{y=L/2}^{x+L/2}(1)dy)dx}{\int_{x=0}^{L}(\int_{y=x}^L(1)dy)dx}=0.25
$$
 A: There is also a nice geometric-probability solution to the problem.  For simplicity, let $L=1$, with $x$ and $y$ as you describe.  The space of all possible values of $x$ and $y$ is the unit square $[0,1]\times[0,1]$, with each point being equally likely (as $x$ and $y$ are uniformly distributed).  In order for the three pieces to form a triangle, each piece must have length less than $\frac{1}{2}$, so:


*

*if $x<y$: $x<\frac{1}{2}$, $y-x<\frac{1}{2}$, and $1-y<\frac{1}{2}$;

*or if $x>y$: $y<\frac{1}{2}$, $x-y<\frac{1}{2}$, and $1-x<\frac{1}{2}$.


Graphing these in the unit square gives the shaded region shown below.

This region is $\frac{1}{4}$ of the total area of the square, so the probability is $\frac{1}{4}$.

It's worth noting this similar but slightly different question, which arose from a mis-written Monte Carlo simulation of this problem.
A: That works, assuming randomly means each point is uniformly distributed on $[0,L]$.  
A similar approach is to note that the density of $x$ and $y$ is each $\frac{1}{L}$ so the probability of a triangle is 
$$\int_{x=0}^{L/2}\int_{y=L/2}^{x+L/2}\frac{1}{L^2} dy\,dx + \int_{x=L/2}^{1}\int_{y=x-L/2}^{L/2}\frac{1}{L^2}dy\,dx = \int_{x=0}^{L/2}\frac{x}{L^2}dx + \int_{x=L/2}^{1}\frac{L-x}{L^2}dx = \frac{1}{4}$$
