A needle on a ruled floor

Idea. Reduce the random drop to two independent uniform variables, then compute the area of the crossing region.

Step 1: coordinates. Let $y\in [0,\frac{1}{2}]$ be the distance from the centre of the needle to the nearest line and $\theta \in [0,\frac{\pi }{2}]$ the acute angle with the lines. By symmetry both are uniform and independent.

Step 2: the crossing condition. The half-needle on the side of the nearest line reaches a distance $\frac{1}{2}\sin \theta$ across the lines, so it crosses exactly when $y\le \frac{1}{2}\sin \theta$.

Step 3: integrate. The joint density is $\frac{1}{1/2}\cdot \frac{1}{\pi /2}=\frac{4}{\pi }$, so $$\begin{array}{rl}\mathbb{P}(\text{cross})& =\frac{4}{\pi }{\int }_0^{\pi /2}\frac{\sin \theta }{2}d\theta \\ & =\frac{4}{\pi }\cdot \frac{1}{2}=\frac{2}{\pi }\approx \mathrm{0.637.}\end{array}$$

A way to remember it. The expected number of crossings of a needle is proportional to its length, whatever its shape. A circle of diameter $1$ always crosses exactly twice and has length $\pi$, so a straight needle of length $1$ crosses $\frac{2}{\pi }$ times on average. A needle of length $1$ crosses at most once, so this is also the probability.