My 2019 Mathematics A To Z: Buffon’s Needle


Today’s A To Z term was suggested by Peter Mander. Mander authors CarnotCycle, which when I first joined WordPress was one of the few blogs discussing thermodynamics in any detail. When I last checked it still was, which is a shame. Thermodynamics is a fascinating field. It’s as deeply weird and counter-intuitive and important as quantum mechanics. Yet its principles are as familiar as a mug of warm tea on a chilly day. Mander writes at a more technical level than I usually do. But if you’re comfortable with calculus, or if you’re comfortable nodding at a line and agreeing that he wouldn’t fib to you about a thing like calculus, it’s worth reading.

Cartoony banner illustration of a coati, a raccoon-like animal, flying a kite in the clear autumn sky. A skywriting plane has written 'MATHEMATIC A TO Z'; the kite, with the letter 'S' on it to make the word 'MATHEMATICS'.
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Buffon’s Needle.

I’ve written of my fondness for boredom. A bored mind is not one lacking stimulation. It is one stimulated by anything, however petty. And in petty things we can find great surprises.

I do not know what caused Georges-Louis Leclerc, Comte de Buffon, to discover the needle problem named for him. It seems like something born of a bored but active mind. Buffon had an active mind: he was one of Europe’s most important naturalists of the 1700s. He also worked in mathematics, and astronomy, and optics. It shows what one can do with an engaged mind and a large inheritance from one’s childless uncle who’s the tax farmer for all Sicily.

The problem, though. Imagine dropping a needle on a floor that has equally spaced parallel lines. What is the probability that the needle will land on any of the lines? It could occur to anyone with a wood floor who’s dropped a thing. (There is a similar problem which would occur to anyone with a tile floor.) They have only to be ready to ask the question. Buffon did this in 1733. He had it solved by 1777. We, with several centuries’ insight into probability and calculus, need less than 44 years to solve the question.

Let me use L as the length of the needle. And d as the spacing of the parallel lines. If the needle’s length is less than the spacing then this is an easy formula to write, and not too hard to calculate. The probability, P, of the needle crossing some line is:

P = \frac{2}{\pi}\frac{L}{d}

I won’t derive it rigorously. You don’t need me for that. The interesting question is whether this formula makes sense. That L and d are in it? Yes, that makes sense. The length of the needle and the gap between lines have to be in there. More, the probability has to have the ratio between the two. There’s different ways to argue this. Dimensional analysis convinces me, at least. Probability is a pure number. L is a measurement of length; d is a measurement of length. To get a pure number starting with L and d means one of them has to divide into the other. That L is in the numerator and d the denominator makes sense. A tiny needle has a tiny chance of crossing a line. A large needle has a large chance. That \frac{L}{d} is raised to the first power, rather than the second or third or such … well, that’s fair. A needle twice as long having twice the chance of crossing a line? That sounds more likely than a needle twice as long having four times the chance, or eight times the chance.

Does the 2 belong there? Hard to say. 2 seems like a harmless enough number. It appears in many respectable formulas. That π, though …

That π …

π comes to us from circles. We see it in calculations about circles and spheres all the time. We’re doing a problem with lines and line segments. What business does π have showing up?

We can find reasons. One way is to look at a similar problem. Imagine dropping a disc on these lines. What’s the chance the disc falls across some line? That’s the chance that the center of the disc is less than one radius from any of the lines. What if the disc has an equal chance of landing anywhere on the floor? Then it has a probability of \frac{L}{d} of crossing a line. If the radius is smaller than the distance between lines, anyway. If the radius is larger than that, the probability is 1.

Now draw a diameter line on this disc. What’s the chance that this diameter line crosses this floor line? That depends on a couple things. Whether the center of the disc is near enough a floor line. And what angle the diameter line makes with respect to the floor lines. If the diameter line is parallel the floor line there’s almost no chance. If the diameter line is perpendicular to the floor line there’s the best possible chance. But that angle might be anything.

Let me call that angle θ. The diameter line crosses the floor line if the diameter times the sine of θ is less than half the distance between floor lines. … Oh. Sine. Sine and cosine and all the trigonometry functions we get from studying circles, and how to draw triangles within circles. And this diameter-line problem looks the same as the needle problem. So that’s where π comes from.

I’m being figurative. I don’t think one can make a rigorous declaration that the π in the probability formula “comes from” this sine, any more than you can declare that the square-ness of a shape comes from any one side. But it gives a reason to believe that π belongs in the probability.

If the needle’s longer than the gap between floor lines, if L > d , there’s still a probability that the needle crosses at least one line. It never becomes certain. No matter how long the needle is it could fall parallel to all the floor lines and miss them all. The probability is instead:

P = \frac{2}{\pi}\left(\frac{L}{d} - \sqrt{\left(\frac{L}{d}\right)^2 - 1} + \sec^{-1}\left(\frac{L}{d}\right)\right)

Here \sec^{-1} is the world-famous arcsecant function. That is, it’s whatever angle has as its secant the number \frac{L}{d} . I don’t mean to insult you. I’m being kind to the person reading this first thing in the morning. I’m not going to try justifying this formula. You can play with numbers, though. You’ll see that if \frac{L}{d} is a little bit bigger than 1, the probability is a little more than what you get if \frac{L}{d} is a little smaller than 1. This is reassuring.

The exciting thing is arithmetic, though. Use the probability of a needle crossing a line, for short needles. You can re-write it as this:

\pi = 2\frac{L}{d}\frac{1}{P}

L and d you can find by measuring needles and the lines. P you can estimate. Drop a needle many times over. Count how many times you drop it, and how many times it crosses a line. P is roughly the number of crossings divided by the number of needle drops. Doing this gives you a way to estimate π. This gives you something to talk about on Pi Day.

It’s a rubbish way to find π. It’s a lot of work, plus you have to sweep needles off the floor. Well, you can do it in simulation and avoid the risk of stepping on an overlooked needle. But it takes a lot of needle-drops to get good results. To be certain you’ve calculated the first two decimal points correctly requires 3,380,000 needle-drops. Yes, yes. You could get lucky and happen to hit on an estimate of 3.14 for π with fewer needle-drops. But if you were sincerely trying to calculate the digits of π this way? If you did not know what they were? You would need the three and a third million tries to be confident you had the number correct.

So this result is, as a practical matter, useless. It’s a heady concept, though. We think casually of randomness as … randomness. Unpredictability. Sometimes we will speak of the Law of Large Numbers. This is several theorems in probability. They all point to the same result. That if some event has (say) a probability of one-third of happening, then given 30 million chances, it will happen quite close to 10 million times.

This π result is another casting of the Law of Large Numbers, and of the apparent paradox that true unpredictability is itself predictable. There is no way to predict whether any one dropped needle will cross any line. It doesn’t even matter whether any one needle crosses any line. An enormous number of needles, tossed without fear or favor, will fall in ways that embed π. The same π you get from comparing the circumference of a circle to its diameter. The same π you get from looking at the arc-cosine of a negative one.

I suppose we could use this also to calculate the value of 2, but that somehow seems to touch lesser majesties.


Thank you again for reading. All of the Fall 2019 A To Z posts should be at this link. This year’s and all past A To Z sequences should be at this link. I’ve made my picks for next week’s topics, and am fooling myself into thinking I have a rough outline for them already. But I’m still open for suggestions for the letters E through H and appreciate suggestions.

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Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.

2 thoughts on “My 2019 Mathematics A To Z: Buffon’s Needle”

  1. I enjoyed reading this; it made me think anew about the deep connexion between probability and pi. As a student I was fascinated (and baffled) by pi’s appearance in the formula for the normal distribution curve, which describes variations that appear in nature, like the heights of students in a college class.

    Liked by 1 person

    1. Thank you, and I’m glad that you enjoyed.

      Pi in the normal distribution curve is something I haven’t got a good reason-to-think-that-makes-sense. It may be I need to go back to a derivation of the normal distribution and think things out anew.

      Like

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