TITLE: Probability of forming triangle from breaking a stick
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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.