One of my all time favorite problems is that of Buffon’s Needle. The problem itself is clearly unique and intriguing, and I will show you a solution that is equally astounding, and that requires no calculus or advanced mathematics whatsoever. The traditional way to solve this problem is using integral geometry, but I can affirmatively say this way is much more savvy.

The premise of the problem is that on the floor you have lines 1 cm apart, and you drop a needle 1 cm long onto the floor. The task is to find the probability the needle cuts or touches a line. An interesting aspect of this is that it blends the physical world into a math problem, without any physics as such. It brings probability which is generally thought of as drawing balls out of a bag into a more creative and visual setting. I would suggest you try it out before reading further to see the solution.

Assumption is one of the first things one must do in any problem. So let us assume that the probability is P. This P can also be thought of the expected number of lines that the needles cuts for every drop when it is carried out a large number of times. In fact let us keep P as an expected value (for example if it is .6, it means that out of every 100 drops, the needles cut 60 lines).

What can we do now? What if the needle was 2 cm? In a way, it is like dropping two needles onto the floor, but the instance says that they are connected together. But what would the expected number of lines that it would cut relative to P? Well it would just be 2P, because it is simply as if two needles were dropped, but they are tied together by the same probability, which would be the same regardless of whether they were dropped separately in the long run. So the expected lines cut for every drop is 2P.

What if the length of the needle is ½ cm? Then the expected lines cut would be simply 1/2 P by the same reasoning. Let’s try something new now. What if the line was a curve. When we have a curve, if we keep breaking it down further and further, we will basically have a set of lines, and so a curve of length 1 cm is approximately 1000 small lines of length 1/1000 cm, which each have an expected value of 1/1000 P by the logic we used before. So added together, they give a total expected value of 1000 x 1/1000 P, which equals P. So whether a line is straight or not is irrelevant to the expected values.

Now we have that insight, consider a circle of diameter 1 cm. Think about how many lines it would cut. If it lands completely in the middle of two lines, it will touch both, giving 2, and if it lands anywhere else it will cut one line twice.

So we can see that the expected value of this circle is 2. But what is the length of the circumference? It’s **π** x Diameter, which is Pi. We saw that whether the needle is curved doesn’t matter, so the expected value is simply **π x **P using the logic that we used before. So:

**π** x P = 2

P = 2/**π**

** **And there it is. We have a super awesome answer using a method that involved nothing but imagination and circles. Who would have guessed that this question, which had nothing to do with circles, would have an answer in terms of **π **? Beautiful, right?