How To Intuitively Understand Buffon’s Needle Problem

Buffon’s needle problem is a direct extension of the Franc-Carreau or the fair-square game. If you haven’t already, check out my original essay on this topic. I had covered the story of what motivated Georges-Louis LeClerc, Comte de Buffon to come up with and solve these problems.

In his attempt to make it into the Royal Academy of Sciences in Paris, Buffon started with the humble game of Franc-Carreau. This game combined the notion of probability with geometry. Back then, this was a novel approach because mathematicians of the time believed these to be unrelated fields.

But Buffon did not stop there. He pressed on to a much more complex problem that established the foundations of geometric probability. His paper made him so famous that the problem he presented and solved is known until today as Buffon’s needle problem in mathematical circles.

In this essay, we will look at the essence of Buffon’s needle problem. After establishing the context, we will proceed to solving it intuitively using simple mathematical concepts (not the way Buffon solved it). This approach enables us to appreciate the beauty and profoundness of the problem. Without any further ado, let us begin.

This essay is supported by Generatebg

What is Buffon’s Needle Problem?

Say that we have access to a needle that is ‘L’ units long. Further, say that we have constructed a bunch of equidistant lanes on a piece of paper as shown in the illustration below. The width between each lane is exactly equal to ‘L’ units.

Given this arrangement, we plan to just throw the needle randomly on to the paper with the lanes. As we do this, Buffon asks the following question:

“What is the probability that the needle crosses one of lane boundaries?”

This is basically Buffon’s needle problem statement. In order to answer this question, we could use a similar approach to the Franc-Carreau, but we run into a fundamental problem.

With the Franc-Carreau, we computed the probability that a circular coin lands squarely inside a square. We did not realise it back then, but a circle makes things significantly easier because it is symmetric around both the axes of a plane.

However, a needle is not symmetric. We need to factor in the direction of the needle in addition to its position. So, how shall we proceed?

---

What Does Intuition Say About Buffon’s Needle Problem?

Since the needle is ‘L’ units long, and the width between any two lane-boundaries is also ‘L’ units, we have just two extreme situations:

1. Case 1: Where the needle is parallel to the lane boundaries.

2. Case 2: Where the needle is perpendicular to the lane boundaries.

For case 1, the probability that the needle would cross a lane boundary is zero, since parallel lines do not ever meet. For case 2, the probability that the needle would cross a lane boundary is 1, because the lane width and needle length are equal.

So, the answer to Buffon’s question has to be somewhere in the middle. One approach would be to just weight the two extreme cases to arrive at an expected value of 0.5. That is, we could say that the needle crosses a lane boundary half on half the number of all throws.

However, in reality, this is a big underestimation, because the needle crosses a lane boundary more often than not. In fact, Buffon calculated the probability to be 2/π, which translates to approximately 64% of the time. But isn’t it strange that π shows up in a problem where we have no circles? Make a note of this; we will eventually answer this question.

Buffon originally used this problem to flex his math prowess, but we have no such lofty ambitions in this essay. Instead, we are going to cover an ingeniously intuitive solution proposed by Joseph-Émile Barbier.

Solving Buffon’s Needle Problem by Making Things More Complex

The very first thing that Barbier did was to rephrase Buffon’s question into a more complex question.

“Hang on a second! Why make a complex question even more complex? Why not make it simpler?”

If you are thinking along these lines, I empathise with you. In mathematics and logic, simplifying helps in some scenarios, but in other scenarios, important detail gets lost. In this case, we could choose to simplify, but the simplified version of this problem happens to be the Franc-Carreau.

In the Franc-Carreau, we work with a circle which is rotationally invariant, whereas we have a pin that is indeed rotationally variant. So, simplification does not help our cause.

Instead of doing this, what Barbier chose to do was as counter-intuitive back then as it is today. However, today, the concept has become standardized and is popularly known as making use of the linearity of expectation. Linearity of expectation states the following:

E(x + y) = E(x) + E(y), where E() denotes the expectation operator or the expectation function.

What is special about linearity of expectation is that it applies not just to independent random variables, but also dependent ones! This is precisely what makes it so counterintuitive.

There are certain conditions that these dependent variables have to meet though, which don’t concern us in this essay (I could go into stuff like Homoscedasticity, but it would be a long detour).

---

Buffon’s Needle Problem Rephrased

Coming back to Buffon’s needle problem, Barbier rephrased the question as follows:

“What is the expected number of lane boundaries that the needle crosses?”

If ‘p’ is the expected probability that a needle crosses a boundary, then, the expected number of crossed lane boundaries is calculated as follows:

Expected number of crossed lane boundaries = Sum of [(each possible number of crossed boundaries) * (probability of observing that number)]

= [(1 — p)*0] + (p*1) = p

So, the expected number of crossed lane boundaries is the same as the probability that the needle crosses a lane boundary (what Buffon’s question aims to answer).

How can this be? It might not be obvious, but this is an instantiation of the linearity of expectation: E(x + y) = E(x) + E(y).

Barbier’s Approach to Solving the Problem

If what you’ve seen so far is counterintuitive, the next bit is even more counterintuitive. Barbier imagined a situation where he threw a needle that was twice as long as the original needle (that is, the new needle is ‘2L’ units long).

“What is the probability that this needle crosses a lane boundary?”

Intuition says that it must be greater than the probability of a needle half its size. Barbier did even better. He imagined that the needle was made of two needles of length ‘L’ glued at the centre.

Here’s the counterintuitive bit: Apply linearity of expectation to these two needles!

Probability of the 2L needle crossing a boundary = E(Needle 1 + Needle 2) = E(Needle 1) + E(Needle 2)

E(Needle 1) or E(Needle 2) is none other than what we calculated previously as ‘p’. Therefore:

Probability of the 2L needle crossing a boundary = p + p = 2p

Extending Linearity of Expectation to More Needles

By the same logic we have used so far, we could compute the probability that a needle of any length crosses a boundary. Let us say that we have a needle of arbitrary length ‘nL’. Remember that ‘L’ is the width between any two lanes. If this is the case:

Probability of the nL needle crossing a boundary = np

Here’s the next counterintuitive bit: what we have figured out so far is true even if the needle is bent in different shapes as shown below:

The probability for the triangle on the left turns out to be 3p + 3p + 2p = 8p. Similarly, the probability of the for the irregular geometry on the right is p + p + p + p + p + p = 6p.

In fact, with our model, we could construct any polygon and compute its probability of crossing a lane boundary. With this insight, we are finally ready to solve Buffon’s needle problem.

---

How to Solve Buffon’s Needle Problem

What if we construct a geometry that looks like a circle with a diameter of ‘L’ units? How do we do this? Simple: We string together thousands and thousands of tiny needles to make a polygon that approximates a circle.

As a result, we end up with a rotationally invariant circle. In other words, we have regained the symmetry we had lost with the Franc-Carreau. With a circle, the problem becomes ridiculously simple.

Since the circle has a radius of ‘½ L’ units, its circumference is simply ‘πL’ units. Consequently, when we apply linearity of expectation:

Probability of the πL needle crossing a boundary = πp

Let us hold this result at the back of our minds. Since the circle has a diameter of ‘L’, no matter where it falls, it crosses a boundary exactly at two points (since it is a loop). Therefore:

Expected number of crossed lane boundaries for the circle = 2

Applying linearity of expectation to the last two equations, we arrive at the following result:

πp = 2

Dividing by π on both sides:

p = 2/π

There we go. We have solved Buffon’s needle problem (with Barbier’s help)! The probability is approximately 64%.

---

Reference: Joseph-Émile Barbier.

If you’d like to get notified when interesting content gets published here, consider subscribing.

Further reading that might interest you: How To Really Debunk Astrology Using The Barnum Effect and How To Intuitively Understand Euler’s Identity?

If you would like to support me as an author, consider contributing on Patreon.