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 "pL" 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 pL) or it intersects 0 lines (with probability 1-pL). Thus we may take the expectation to get the probability:
pL = 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

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