Drop a needle of length $L$ onto a hardwood floor with floorboards of width $W$. On average, the needle crosses $2L / \pi W$ lines between floorboards, a classic result of Buffon. But that $\pi$ in the formula means there’s a circle hiding somewhere. The trick to finding it? Bend the needle into a noodle.
The usual approach to Buffon’s problem involves a double integral. Respectable, but this hides the circle at the heart of the solution, and frankly, I don’t love doing integrals. Instead, we’ll derive the result1 by going from a straight needle, to a curvy noodle, to a circle. All we need is some basic geometric reasoning and probability.
Let’s fix some notation. Add ruled lines on $\mathbb{R}^2$ spaced $W > 0$ apart, and choose a line segment of length $L>0$ at random2. Let $X_1$ be the number of ruled lines that this random line segment crosses. We want to compute $\mathbb{E}[X_1] =: f(L)$ as a function of $L$.
Now suppose we drop two needles with lengths $L_1$ and $L_2$, and let $X_1$ and $X_2$ be the number of lines that each needle crosses. By linearity of expectation,
$$ \mathbb{E}[X_1 + X_2] = \mathbb{E}[X_1] + \mathbb{E}[X_2] = f(L_1) + f(L_2). $$
Linearity of expectation requires no independence assumption. In particular, we could weld the two segments together, and the equation would continue to be true. Joining the two segments end-to-end gives $f(L_1 + L_2) = f(L_1) + f(L_2)$, which holds for all lengths $L_1$, $L_2$. Since $f$ is non-negative and increasing with $f(0) = 0$, we deduce that $f(L) = c L$ for some constant $c \ge 0$ that we need to determine3.
We can then “bend” the needle into an arbitrary polygonal line with $N$ segments, each of length $L/N$. With $X_i$ the number of crossings on the $i$th segment, we find
$$ \mathbb{E}[X_1 + \dotsb + X_N] = N f(L/N) = c L, $$
that is, the average number of lines that a polygonal line strikes depends linearly on its length. Taking a limit gives us Buffon’s noodle: throw an arbitrary4 curve onto the plane, and the average number of lines it intersects is proportional only to its length.
The special circle
Only the value of the constant $c$ remains. Consider a circle of radius $W/2$. With probability one, this circle crosses a single ruled line twice; the alternative, being tangent to two lines, occurs with probability zero. (Try it out in the widget above!) This means that
$$ \mathbb{E}[\text{\# intersections with $W/2$-circle}] = 2. $$
This means that $cL = 2$ for this special circle; since $L = \pi W$, we conclude that
$$ c = \frac{2}{\pi W} $$
which completes the proof.
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See Klain, Daniel A., and Gian-Carlo Rota. Introduction to geometric probability. Cambridge University Press, 1997. ↩︎
To avoid ambiguity and paradoxes, we need to define what “random” means. Here, we’ll assume that the line segment is drawn from the Haar measure, which basically means that our answers must be independent of the orientation of the floorboards. The precise definition of the Haar measure is that it is the unique rotation- and translation-invariant measure on the space of line segments. In practice, one draws a center uniformly over a bounded space (say, $[0,\,1]\times[0,\,1]$) and an angle $\theta$ uniformly over $[0,\,2\pi)$. ↩︎
We elide some of the reasoning here for clarity. In particular, to make this argument rigorous, we must show that $f$ is continuous. ↩︎
The curve must, of course, be finite length and approximated as the limit of polygonal curves. Such curves are called rectifiable. ↩︎