From Buffon's Needle to Buffon's Noodle

# From Buffon's Needle to Buffon's Noodle Source: [https://mbmccoy.dev/posts/buffons-noodle/](https://mbmccoy.dev/posts/buffons-noodle/) 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 result[1](https://mbmccoy.dev/posts/buffons-noodle/#fn:1)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 random[2](https://mbmccoy.dev/posts/buffons-noodle/#fn:2)\. 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 determine[3](https://mbmccoy.dev/posts/buffons-noodle/#fn:3)\. 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 arbitrary[4](https://mbmccoy.dev/posts/buffons-noodle/#fn:4)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\. Get new posts by email\. No spam, just posts\.