Broken stick probability question (variation) A stick is broken into two at random, then the longer half is broken again into two pieces at random. What is the probability that the three pieces make a triangle?
I've been stumped at this question for a while now, and cannot seem to find a convincing answer on the internet. The answers that I find are usually 0. 386 (which is 2ln2 - 1), but I have worked the answer out to be 0.27865. 
I let X be the event of the position of the first cut of a length of unit 1. Then X ~ Unif(0,1).
Then I let Y be the event of the position of the second cut. Finding the PDF of Y is slightly more tricky.
Y is Unif (a/(1-x)) for 0 < x < 0.5 and Unif (a/x) for 0.5 < x < 1
where the normalising constant a, I worked out to be 0.5/ln2.
Now the answers I have seen did not work out this normalising constant, and you will get 0.386 as the probability to create a triangle, but with the normalising constant, I get an answer of 0.27865...
Now, as with these probability questions, you can always make an experiment, and estimate a probability. I was too lazy to do this and found a simulation here:
http://www.sineofthetimes.org/the-broken-stick-puzzle/
I found the simulation to support the 0.38 answer... but then again, I don't know how they created the simulation.
What have I done wrong here? (Or am I right?) Thanks so much! 
 A: I think without loss of generality you can consider $X$~$U(\frac12, 1)$ and $Y$~$U(0,X)$. 
If you set $M=\max(Y,X-Y,1-X)$ (the longer side) then the three segments do not form a triangle if and only if $M$ is larger than the sum of the other two sides, that is $M> 1-M$, in other words whether $M> \frac12$.
Note that $P(1-X)>\frac 12=0$, so $$\{M>\frac 12\}=\{\max(Y,X-Y)>\frac 12\}$$
Now \begin{align*}P(M>\frac12\mid X=x)&=P(Y<x-\frac12)+P(\frac12<Y\leq x)\\&=2\frac{x-\frac12}{x}\\&=2-\frac1x\end{align*}
So that \begin{align*}P(M>\frac12)&=\int_{\frac12}^1P(M>\frac12\mid X=x)\cdot 2dx\\&=\int_{\frac12}^1 (4-\frac2x )dx\\&=\left[4x-2\ln x\right]_{\frac12}^1\\&=(4-2)+2\ln(\frac12)\\&=2(1-\ln 2)\end{align*}
This is the probability that the three segments do not form a triangle.
A: You do not need a normalising constant


*

*If $X=x < \frac12$ you have  $Y$ conditionally distributed uniformly on $[x,1]$ with density $\frac{1}{1-x}$ and you have a triangle if $Y \in \left(\frac 12, x+\frac12\right)$  which has conditional probability $\frac{x}{1-x}$ 

*If $X=x > \frac12$ you have  $Y$ conditionally distributed uniformly on $[0,x]$ with density $\frac{1}{x}$  and you have a triangle if $Y \in \left(x-\frac12, \frac12\right)$  which has conditional probability $\frac{1-x}{x}$

*So integrating over $x$ uniform on $[0,1]$, you get the overall probability of a triangle being $$\int_0^{1/2} \frac{x}{1-x} \, dx+ \int_{1/2}^1 \frac{1-x}{x} \, dx = 2\log_e(2)-1 \approx 0.386$$
The following simulation in R gives much the same result
set.seed(1)
cases <- 10^6
X <- runif(cases, min=0, max=1)
Y <- ifelse(X > 1/2, runif(cases, min=0, max=X), runif(cases, min=X, max=1))
triangle <- ((X > 1/2 & Y < 1/2) | (X < 1/2 & Y > 1/2)) & abs(X-Y) < 1/2
mean(triangle)
[1] 0.386273

