This is a fun demonstration of how Pi can be found seemingly out of nowhere. What I am about to show is a popularior simulation of a Monte Carlo method. Monteo Carlo methods are general computational algorithms that rely on repeated random sampling as a way to try to extract valueable information. For this demonstration, we are going to try to to get an estimation for the value of Pi using nothing but a dartboard.

We will begin by first showing the demonstration, and then going over some of the math behind it.

II. The Demonstration

To begin, we will imagine that in front of us we have a standard dartboard. For our purposes, we do not care about the points associated within the dartboard, we only care about the circular shape of it. Furthermore, we will assume that our dartboard is perfectly inscribed within a square.

From here, we are going to throw a bunch of darts at the board and hope it hits. Unfortunately for us, we have terrible coordination so the location where the dart dies is going to be completely random.

Let's throw 100 darts and mark the ones that land on the board in black, and those that miss the board completely in red.

Next, we will go ahead and divide the number of misses by the total number of throws and multiple that value by four for seemingly no reason. \[4~*~\frac{\#~of~Misses}{Total}\]
Given this is a relatively small number, we can go ahead and count the values ourselves. We see that we have 73 hits out of 100 attempts.
\[4~*~\frac{73}{100} = \boxed{2.92}\]
We will keep track of this approximation as we increase the amount of throws (n). As we increase N, we might begin to notice something astonishing.

III. Some Math

When we take a step back, we recall that to find this approximation, we are getting the proportion of attempts that landed within the circle and dividing it by the misses. If we assume that there were a large amount of attempts taken to the point where the entire board is covered, we can then think of the proportion as simply the area of the circle, divided by the area of the outer square.

\[Area~of~circle~=~ \pi r^2\] \[Area~of~square~=~ lw = (2r)^2\]
\[\frac{\pi r^2}{(2r)^2}\]
\[\frac{\pi r^2}{4r^2}\]
\[\frac{\pi}{4}\]