probability
I am having trouble coming up with a solution for this problem:
There is a stick of unit length. We break it into two parts.
Now, we pick the bigger one and break it into two parts.
I want to calculate the probability that the three pieces form a triangle.
The problem is from "Introduction to Probability, Charles M. Grinstead", Chapter 2.2, Exercise 13
The three triangle inequalities are
\begin{align} x + y &> 1-x-y \\ x + (1-x-y) &> y \\ y + (1-x-y) &> x \\ \end{align}
Your problem is that in picking the smaller number first from a uniform distribution, it's going to end up being bigger than it would if you had just picked two random numbers and taken the smaller one. (You'll end up with an average value of $1/2$ for the smaller instead of $1/3$ like you actually want.) Now when you pick $y$ on $[0, 1-x]$, you're making it smaller than it should be (ending up with average value of $1/4$). To understand this unequal distribution, we can substitute $y (1-x)$ for $y$ in the original inequalities and we'll see the proper distribution.
(Note that the $y$-axis of the graph doesn't really go from $0$ to $1$; instead the top represents the line $y=1-x$. I'm showing it as a square because that's how the probabilities you were calculating were being generated.) Now the probability you're measuring is the area of the strangely-shaped region on the left, which is $$\int_0^{1/2}\frac1{2-2x}-\frac{2x-1}{2x-2}\,dx=\ln 2-\frac12\approx0.19314$$ I believe that's the answer you calculated.
Let $f(x)$ be the probability that we eventually cut off at least $x$ from a stick of length $1$, with $1\ge x\ge1/2$. We can either succeed by immediately cutting off at least $x$, with probability $1-x$, or by leaving $t\ge x$ and then cutting off $x$ from a stick of length $t$. Thus we have
$$ f(x)=1-x+\int_x^1f(x/t)\mathrm dt\;. $$
Substituting $u=x/t$ yields
$$f(x)=1-x+x\int_x^1f(u)/u^2\mathrm du\;.\tag1$$
Then differentiating with respect to $x$ yields
$$f'(x)=-1-f(x)/x+\int_x^1f(u)/u^2\mathrm du\;,$$
and differentiating again yields
$$f''(x)=-f'(x)/x+f(x)/x^2-f(x)/x^2=-f'(x)/x\;,$$
so
$$ \frac{f''(x)}{f'(x)}=-\frac1x $$
and thus
$$ \begin{align} \log f'(x)&=-\log x +c\;,\\ f'(x)&=a/x\;,\\ f(x)&=a\log x+b\;. \end{align} $$
Now $f(1)=0$ yields $b=0$, and then substituting into $(1)$ yields $a=-1$, so $f(x)=-\log x$ and
$$f(1/2)=\log2\approx0.693\;.$$
Best Answer
Let $\lambda$ be the length of the bigger piece and we split it into two smaller pieces $\lambda\mu$ and $\lambda(1-\mu)$. It is clear $\lambda$ and $\mu$ are uniform random variables $\sim \mathcal{U}(\frac12,1)$ and $\mathcal{U}(0,1)$ respectively.
In order for the three pieces with lengths $\;1-\lambda, \lambda\mu, \lambda(1-\mu)\;$ to form a triangle, the necessary and sufficient conditions are the fulfillment of following three triangular inequalities: $$\begin{cases} \lambda \mu + \lambda (1-\mu) &\ge 1-\lambda\\ \lambda \mu + (1 - \lambda) &\ge \lambda (1-\mu)\\ \lambda (1-\mu) + (1-\lambda) &\ge \lambda \mu \end{cases} \quad\iff\quad \begin{cases} \lambda \ge \frac12\\ \mu \ge 1 - \frac{1}{2\lambda}\\ \frac{1}{2\lambda} \ge \mu \end{cases}$$ The first inequality is trivially satisfied because we are told to break the bigger piece.
The probability we seek is given by:
$$2\int_{1/2}^1 \int_{1-\frac{1}{2\lambda}}^{\frac{1}{2\lambda}} d\mu d\lambda = 2\int_{1/2}^1 \left(\frac{1}{\lambda} - 1 \right) d\lambda = 2\log 2 - 1 \approx 38.6294\%$$