A stick of length 1 is broken at two points chosen at random and independently along its length…

Step 1. The condition. The pieces form a triangle when each is less than $\dfrac{1}{2}$. If the longest were at least $\dfrac{1}{2}$, it would be at least as long as the other two together.

Step 2. The unit square. Take $x < y$ (the other half of the square is the mirror image). The pieces are $x$, $y - x$ and $1 - y$. We need $x < \dfrac{1}{2}$, $y > \dfrac{1}{2}$ and $y - x < \dfrac{1}{2}$.

Step 3. The area. In the half of the square where $x < y$, the region described is a triangle with corners $\left(0, \dfrac{1}{2}\right)$, $\left(\dfrac{1}{2}, \dfrac{1}{2}\right)$ and $\left(\dfrac{1}{2}, 1\right)$. Its area is $\dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{8}$. The mirror half gives $\dfrac{1}{8}$ again, so the total area is $\dfrac{1}{4}$.

Step 4. The longest piece. The complement of a triangle is "the longest piece is at least $\dfrac{1}{2}$", so its probability is $1 - \dfrac{1}{4} = \dfrac{3}{4}$.

Step 5. The simulation. The usual error is $\sqrt{\dfrac{0.25 \times 0.75}{40\,000}} \approx 0.002$, so an estimate between $0.246$ and $0.254$ is expected.