### Buffon's Needle, the Easy Way

Buffon's needle is a popular probability problem. Rule lines on the floor a distance d apart and toss a needle of length l<d onto it. What is the probability that the needle crosses a line? A solution is described at wikipedia but it involves a double integral and some trigonometry. Nowhere does it mention that there is a less familiar but much simpler proof, though if you follow the links you'll find it. In addition, the usual solution involves π but gives little intuition as to why π appears. The simpler proof reveals that it appears naturally as a ratio of the circumference of a circle to its diameter. I've known this problem since I was a kid and yet I hadn't seen the simpler proof until a friend sold me his copy of

So instead of solving Buffon's needle problem we'll solve what appears to be a harder problem: when thrown, what is the expectation of the number of times a rigid curved (in a plane) wire length l (no restriction on l) crosses one of our ruled lines d apart? Here's an example of one of these 'noodles'. It crosses the ruled lines three times:

Expectation is linear in the sense that E(A+B) = E(A)+E(B). So if we imagine the wire divided up into N very short segments of length l/N the expectation for the whole wire must be the sum of the expectations for all of the little pieces. If the wire is well behaved, for N large enough the segments are close to identical straight line segments. Here's a zoomed up view of a piece of our noodle:

For a small straight line segment the expectation must simply be a function of the length of the segment. The expectation for the whole wire is the expectation for one segment multiplied by the number of segments. In other words, the expectation is proportional to the length of the wire and we can write E(l)=kl for some constant k.

Now we know it's proportional to the length, we need to find the constant of proportionality, k. We need to 'calibrate' by thinking of a noodle shape where we know in advance exactly how many times it will cross the ruled lines. The following picture gives the solution:

A circle of diameter d will almost always cross the lines in two places. The length of this wire is πd so E(πd)=2 and k=2/πd.

The expected number of crossings for a wire of length l is 2l/πd. A needle of length l<d can intersect only zero or one times. So the expected value is in fact the probability of intersecting a line. The solution is 2l/πd.

No integrals needed.

The expected number of crossings is an example of an invariant measure, something I've talked about before. There are only a certain number of functions of a noodle that are additive and invariant under rotations and just knowing these facts is almost enough to pin down the solution.

Now I can leave you with a puzzle to solve. In the UK, a 50p coin is a 7 sided curvilinear polygon of constant width. Being constant width means a vending machine can consistently measure its width no matter how the coin is oriented in its plane. Can you use a variation of the argument above to compute the circumference of a 50p coin as a function of its width?

*Introduction to Geometric Probability*for $5 a few days ago.So instead of solving Buffon's needle problem we'll solve what appears to be a harder problem: when thrown, what is the expectation of the number of times a rigid curved (in a plane) wire length l (no restriction on l) crosses one of our ruled lines d apart? Here's an example of one of these 'noodles'. It crosses the ruled lines three times:

Expectation is linear in the sense that E(A+B) = E(A)+E(B). So if we imagine the wire divided up into N very short segments of length l/N the expectation for the whole wire must be the sum of the expectations for all of the little pieces. If the wire is well behaved, for N large enough the segments are close to identical straight line segments. Here's a zoomed up view of a piece of our noodle:

For a small straight line segment the expectation must simply be a function of the length of the segment. The expectation for the whole wire is the expectation for one segment multiplied by the number of segments. In other words, the expectation is proportional to the length of the wire and we can write E(l)=kl for some constant k.

Now we know it's proportional to the length, we need to find the constant of proportionality, k. We need to 'calibrate' by thinking of a noodle shape where we know in advance exactly how many times it will cross the ruled lines. The following picture gives the solution:

A circle of diameter d will almost always cross the lines in two places. The length of this wire is πd so E(πd)=2 and k=2/πd.

The expected number of crossings for a wire of length l is 2l/πd. A needle of length l<d can intersect only zero or one times. So the expected value is in fact the probability of intersecting a line. The solution is 2l/πd.

No integrals needed.

The expected number of crossings is an example of an invariant measure, something I've talked about before. There are only a certain number of functions of a noodle that are additive and invariant under rotations and just knowing these facts is almost enough to pin down the solution.

### Puzzle

Now I can leave you with a puzzle to solve. In the UK, a 50p coin is a 7 sided curvilinear polygon of constant width. Being constant width means a vending machine can consistently measure its width no matter how the coin is oriented in its plane. Can you use a variation of the argument above to compute the circumference of a 50p coin as a function of its width?