Buffon's Needle

An oddity of the Universe first discovered by Georges-Louis Leclerc, Comte de Buffon.

It's a simple experiment, yet one of the most curious in all of mathematics.

Ingredients
1 sheet of lined paper (or anything with evenly spaced lines)
any number of needles or similar objects

Preparation
Cut the needles so that they are exactly the same length as the space between the lines on the paper.

Experiment
These needles can be in one of two states (binary, like a computer): they can either be crossing a line, or they can not. Record the state of each needle as you drop it.

The result? 2 x the number of drops, divided by the number of needles in the crossed state = pi

The more needles you drop, the closer the result gets to becoming pi.

And who said the world wasn't an interesting place?

It's not that curious, throw in a little calculus and everything makes sense. The root of the problem is that we need to find the probability that a needle will cross a line. For simplicity, we will assign the length of the needle and the distance between the lines a value of 1. As the needle is dropped, it can land at any angle. Assuming the lines run vertically, the maximum horizontal span of the needle, 1, occurs when the angle between the needle and the vertical is π/2 radians. The minimum span, essentially 0, occurs when the needle falls parallel to the lines, so that the angle between it and the vertical is 0.

(If you're not into geometry, π/2 radians is the same as 90 degrees.)

Applying some trigonometry, we can easily show that the horizontal span of the needle is given by sin(x) where x is the angle the needle makes with the vertical. The next step is to find the average span, that is, the average value of the function sin(x) as x varies between 0 and π/2 degrees. From calculus, this is defined as:

      π/2              /
     ∫    sinx dx     /  
      0              /   π/2 - 0
                    /

(Or if you don't like calculus, it is the area under the curve sin(x) divided by the length of the base of the curve, in this case π/2.)

Evaluating this integral, we get

        |π/2
 -cos(x)|     =   cos(0) - cos(π/2)   =   1
        |0

Dividing by π/2 - 0, we get 2/π. This is the average horizontal span of a needle dropped on the vertical lines. Without going into a proof, we can show by common sense that the probability that a needle will touch a line equals its average span divided by the distance between lines (which we defined as 1, so the probability is just 2/π). (Just think, as span approaches the distance between lines, this ratio approaches one, or maximum probability, so as span approaches the average span, the ratio goes to average probability.)

By definition, probability is the number of needles that cross a line (c) divided by the total number of needles (t) Therefore:

 P =   c/t = 2/π

Which can be rearranged to π = 2t/c
As with all probability problems, the more trials involved, the better the estimate becomes.

Note, I didnt mean to ruin the mystery of it...in my opinion the beautiful simplicity of this makes the world an even more interesting place.

Sewing without Calculus

Buffon's needle is deeply unsatisfying: a question with only a passing relationship to circles (the needle can fall in any orientation -- but there's still the small matter of lateral motion!) ends up connected to π. Why?

The standard proof -- above in devout's writeup, with integrals -- does little to explain the mysterious appearance of π. Calculation is often like that.

Geometric measure theory offers a "clean" proof, one that almost does explain where π comes from. And it goes like this.

1. We are interested in the probability "p_L" that a needle of length L, randomly dropped on a surface ruled with lines at interval 1 from each other, will intersect a line. Suppose L≤1. Then almost always the needle can intersect at most 1 line, so either it intersects 1 line (with probability p_L) or it intersects 0 lines (with probability 1-p_L). Thus we may take the expectation to get the probability:

   p_L = e(L) = E(number of intersections of a needle of length L with a line)

2. Suppose we connect 2 needles of lengths L and M at their edge (not necessarily in a straight line -- we're interested in a crooked needle). Since expectation is linear, the expected number of intersections of our crooked needle with a line is

   e(L+M) = e(L) + e(M)

   (Note that our notation e(L+M) ignores the shape of the needle, as the RHS does not depend on this shape!). Naturally, we can do this for any number of needles glued at their ends. So e is a linear function, and e(t)=ct for some constant c.
3. We wish to find c -- then we will be able to determine e(L) and, especially, e(1).
4. By using our favourite convergence theorem on expectation (e.g., the monotone convergence theorem) we see that for any curve with length L, if we drop it at random, the expected number of intersections with a line will be e(L)=cL.
5. OK, so let's pick a particularly easy curve. Take the circle of diameter 1. Its perimeter is π, so the expected number of intersections of a circle of diameter 1 with a surface ruled with lines at interval 1 is e(π)=cπ. On the other hand, no matter how we drop this circle, it always intersects exactly 2 lines! So e(π)=2, and therefore c=2/π, as required.

QED.

---

The proof above is based on what appears in the introduction to

• Daniel A Klain and Gian-Carlo Rota, Introduction to Geometric Probability, Lezioni Lincee, published Cambridge University Press 1997.

Naturally, that book goes considerably further -- and generally in more algebraic directions.