Probability of forming triangle from breaking a stick Given a stick and break it randomly at two places, what is the probability that you can form a triangle from the pieces?
Here is my attempt and the answer does not match, so I am confused what went wrong with this argument.
I first denote the two randomly chosen positions by $X$ and $Y$, and let $A=\max(X,Y)$, $B=\min(X,Y)$. We are interested in the probability of the event $\{A>\frac{1}{2}, B>A-\frac{1}{2}\}$. Thus, we want the joint distribution of $A$ and $B$. To compute that, I computed
$$F_{A,B}(w,z)=\mathbb{P}(A\leq w, B\leq z)=\mathbb{P}(A\leq w)-\mathbb{P}(A\leq w, B>z)=\mathbb{P}(X\leq w,Y\leq w)-\mathbb{P}(X\leq w, Y\leq w, X>z, Y>z)$$
Therefore, we have if $z\leq w$
$$F_{A,B}(w,z)=w^2-(w-z)^2$$
otherwise
$$F_{A,B}(w,z)=w^2$$
Then the joint density of $A$ and $B$ is
$$f_{A,B}(w,z)=\frac{\partial^2 F}{\partial w\partial z}(w,z)=2$$
if $z\leq w$ and $0$ otherwise.
Finally
$$\mathbb{P}(A>\frac{1}{2},B>A-\frac{1}{2})=\int_{\frac{1}{2}}^1\int_{w-\frac{1}{2}}^w2dzdw=\frac{1}{2}$$
The answer is $\frac{1}{4}$ instead, but I can't figure out what went wrong with this argument.
 A: The lengths of the sides of the stick will be $1-A, A-B$, and $B$. For the stick to form a valid triangle, the following three conditions must hold by the triangle inequality:
$$\begin{align*}
1-A + A-B>B \to B&<\frac{1}{2} \\
1-A + B>A-B \to B&>A-\frac{1}{2} \\
A-B+B > 1-A \to A&>\frac{1}{2}
\end{align*}$$
You did not include the first condition, $B<1/2$, which threw off your answer. Other than this, everything else was solved correctly; your answer would have been correct if only the second and third conditions were needed.
A: Here is my view of this problem, not a debug of your procedure(sorry)
Actually I figure this 'forming triangle' problem by 'forming triangles'.

Following your definition of X,Y, the joint choice of X,Y are uniform in the unit cell (density 1). The task remained is to find the cutting position of X and Y such that resulting 3 segments can form a triangle.
Consider case when Y > X (so three segments length are X, Y-X, 1-Y). Inserting the fact that sum of any two sides is greater than the third. The constraints are:  $X + (Y-X) > 1- Y ,  X + (1-Y) > Y-X$ and $Y-X + 1-Y > X$, resulting in $Y>0.5, X > Y - 0.5$ and $X < 0.5$, which the boundary of Area A1 in my figure. With similar argument, you can sketch Area A2.
The geometric area of A1 and A2 are 1/8 respectively, sum these two, that is how I get 1/4.
A: The following simulation in R looks at a million such randomly broken
sticks, finds the lengths of the three pieces, and finally finds the
length of the longest piece. If the longest has length less
than half, you can make a triangle. Answer: $0.250\pm 0.001.$
set.seed(2020)
mx = replicate(10^6, max(diff(c(0, sort(runif(2)), 1))))
mean(mx < .5)
[1] 0.250222       # aprx 1/4
2*sd(mx < .5)/1000
[1] 0.000866282    # aprx 95% marg of sim error

Note: A related but different problem breaks the stick once uniformly at random and then breaks the longer piece uniformly at random.
