Laplace’s extension of Buffon

Solution. Let $p$ be the probability that the needle crosses a horizontal line, and $q$ the probability it crosses a vertical line. Buffon’s calculation (Chapter 53) gives $p = q = \tfrac{2 \ell}{\pi d}$.

We want the probability of any crossing. By inclusion–exclusion, $$\mathbb{P}(\text{any crossing}) = p + q - \mathbb{P}(\text{both}).$$

To compute $\mathbb{P}(\text{both})$, parametrise as in Chapter 53: let $\theta \in [0, \pi/2]$ be the angle with the horizontal, uniform, and let $(u, v)$ be the location of the needle’s centre inside a single tile, uniform in $[0, d]^2$. The needle crosses a horizontal line iff $\min(v, d - v) \leq \tfrac{\ell}{2} \sin\theta$, an event of $v$-probability $\tfrac{\ell \sin\theta}{d}$ (since this uses both sides of the tile). Similarly it crosses a vertical line with $u$-probability $\tfrac{\ell \cos\theta}{d}$. Conditioned on $\theta$, these are independent events, so $$\mathbb{P}(\text{both} \mid \theta) = \frac{\ell^2 \sin\theta \cos\theta}{d^2} = \frac{\ell^2 \sin(2\theta)}{2 d^2}.$$ Averaging over $\theta$ uniform on $[0, \pi/2]$: $$\begin{align*} \mathbb{P}(\text{both}) &= \frac{2}{\pi} \int_0^{\pi/2} \frac{\ell^2 \sin(2\theta)}{2 d^2} \, d\theta \\ &= \frac{\ell^2}{\pi d^2} \int_0^{\pi/2} \sin(2 \theta) \, d\theta = \frac{\ell^2}{\pi d^2}. \end{align*}$$

Therefore $$\mathbb{P}(\text{any crossing}) = \frac{2 \ell}{\pi d} + \frac{2 \ell}{\pi d} - \frac{\ell^2}{\pi d^2} = \frac{4 \ell}{\pi d} - \frac{\ell^2}{\pi d^2},$$ which simplifies to $$\boxed{\mathbb{P}(\text{any crossing}) = \frac{\ell (4 d - \ell)}{\pi d^2}.} \qedhere$$