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: $$\frac{\int_{x=0}^{L/2}(\int_{y=L/2}^{x+L/2}(1)dy)dx}{\int_{x=0}^{L}(\int_{y=x}^L(1)dy)dx}=0.25$$

-
This question was asked on Math Overflow a while back. There are some good answers there: mathoverflow.net/questions/2014 – Mike Spivey Mar 11 '11 at 18:01
I like this question! – Asaf Karagila Mar 11 '11 at 18:09
Indeed, the accepted answer is very clever. Is my reasoning correct thought? I'm not sure, even thought I arrived at the same answer. – Leonardo Fontoura Mar 11 '11 at 18:11
Yes, I think your reasoning is correct. – joriki Mar 11 '11 at 18:27
@Asaf: I favorited this question solely because it features a samurai. – JCCyC Mar 13 '11 at 0:44

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.

-
Very intuitive answer, thank you! – Leonardo Fontoura Mar 12 '11 at 15:17
I've deleted my other comment; no need to have two links to the same question. Nice answer, by the way. – Mike Spivey Mar 12 '11 at 19:00

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 $$\int_{x=0}^{L/2}\int_{y=L/2}^{x+L/2}\frac{1}{L^2} dy\,dx + \int_{x=L/2}^{1}\int_{y=x-L/2}^{L/2}\frac{1}{L^2}dy\,dx = \int_{x=0}^{L/2}\frac{x}{L^2}dx + \int_{x=L/2}^{1}\frac{L-x}{L^2}dx = \frac{1}{4}$$

-