The longest of three pieces

Solution. Let the break points be at positions $x, y \in [0,1]$, independent uniform. The three pieces have lengths $\min(x,y)$, $|x-y|$, and $1 - \max(x,y)$. The longest piece exceeds $\tfrac12$ iff some piece exceeds $\tfrac12$: $$\begin{align*} \mathbb{P}(\text{longest} > \tfrac12) = \mathbb{P}\big( &\min(x,y) > \tfrac12 \;\text{or} \\ &|x-y| > \tfrac12 \;\text{or} \\ &1 - \max(x,y) > \tfrac12 \big). \end{align*}$$ The three events are mutually exclusive: if $\min(x,y) > \tfrac12$ then both $x, y > \tfrac12$, forcing $|x-y| < \tfrac12$ and $\max(x,y) > \tfrac12$; similarly if $\max(x,y) < \tfrac12$ then $|x-y| < \tfrac12$. So we may add the three probabilities directly.

Each event corresponds to a region of the unit square:

• $\{\min(x,y) > \tfrac12\} = [\tfrac12, 1]^2$, area $\tfrac14$,

• $\{\max(x,y) < \tfrac12\} = [0, \tfrac12]^2$, area $\tfrac14$,

• $\{|x-y| > \tfrac12\}$ is the union of two corner triangles (upper-left and lower-right), each of area $\tfrac18$, total area $\tfrac14$.

Summing, $$\boxed{\mathbb{P}(\text{longest} > \tfrac12) = \tfrac14 + \tfrac14 + \tfrac14 = \tfrac34.} \qedhere$$

import numpy as np
x, y = np.random.rand(2, 10**7)
pieces = np.stack([np.minimum(x,y), np.abs(x-y), 1-np.maximum(x,y)])
p = (pieces.max(axis=0) > 0.5).mean()
print(f"sim: {p:.5f}   exact: 0.75")
# sim: 0.74991   exact: 0.75