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Sun, 05 Jan 2020 14:39:45 0500

A satisfying proof
https://ianfox.io/posts/favouriteproof/
Sun, 05 Jan 2020 14:39:45 0500
ian@ianfox.io (Ian Fox)
https://ianfox.io/posts/favouriteproof/
<p>In 1777, GeorgesLouis Leclerc, Comte de Buffon posed the following question:</p>
<p>Suppose we have a floor made of parallel wooden boards, each the same width, and we drop a needle onto the floor. What is the probability that the needle will lie across a line between two boards?</p>
<p>He solved it with some fancy calculus, which is a fine way of doing it, but not particularly satisfying to me. However in 1860 a man named JosephĂ‰mile Barbier came up with this super slick proof, which is without a doubt the coolest proof I’ve ever seen.</p>
<p>The trick is that rather than trying to calculate the probability directly, Barbier examined it through the lens of the expected number of lines the needle would cross.</p>
<p>Suppose we define our boards to be 1 unit of distance wide, and we find that dropping a needle of length 1 unit has a probability $$p$$ of crossing one of the lines. We will ignore the possibility of the needle being exactly along a line, or having one end each on precisely two lines, since the probability of those is 0.</p>
<p>If we drop one needle $$n_1$$, our expected number of crossings is simply $$p$$, the probability that the needle crosses a line.</p>
<p>If we drop a second needle $$n_2$$, then by linearity of expectation we find that our expected total number of crossings is</p>
<p>$$E(n_1+n_2)=E(n_1)+E(n_2)=2p$$</p>
<p>In general, the expected number of crossings when you drop $$m$$ needles is $$mp$$.</p>
<figure>
<img src="twoneedles.png"
alt="Two needles"/> <figcaption>
<p>Two needles on the floor</p>
</figcaption>
</figure>
<p>Similarly, if we drop a needle $$n$$ and break it into $$m$$ identical pieces $$n_1 \ldots n_m$$ we find that each piece must be responsible for the same expected number of crossings:</p>
<p>$$E(n) = \sum_{i=0}^{m}E(n_i) \implies E(n_i)=\frac{1}{p}$$</p>
<p>In short, we can drop a needle of any length $$\ell$$ we want and the expected number of crossings will be $$\ell p$$. This holds even if we drop some really bizarre needles:</p>
<figure>
<img src="strangeneedles.png"
alt="Strangely shaped needles"/> <figcaption>
<p>Some strangely shaped needles</p>
</figcaption>
</figure>
<p>Now suppose we take a needle and bend it into a circle of diameter 1. We know that the expected number of crossings must be $$\pi p$$ from before. But consider the ways the needle could fall on the floor:</p>
<figure>
<img src="circleneedles.png"
alt="Some needles bent into circles"/> <figcaption>
<p>Three needles, bent into circles</p>
</figcaption>
</figure>
<p>It always crosses in exactly two places! So $$\pi p=2$$ and we find that $$p=\frac{2}{\pi}$$. Any straight needle we drop on the floor will have probability $$\frac{2\ell}{\pi}$$ of crossing a line. And we didn’t do a single bit of calculus!</p>