Asking the maths library

My computer tells me that \(\pi\) is approximatively

pi  
## [1] 3.141593

Buffon’s needle

Applying the method of Buffon’s needle, we get the approximation

set.seed(42)    
N = 100000  
x = runif(N)    
theta = pi/2*runif(N)   
2/(mean(x+sin(theta)>1))    
## [1] 3.14327

Using a surface fraction argument

A method that is easier to understand and does not make use of the \(\sin\) function is based on the fact that if \(X\sim U(0,1)\) and \(Y\sim U(0,1)\), then \(P[X^2+Y^2\leq 1] = \pi/4\) (see “Monte Carlo method” on Wikipedia). The following code uses this approach:

set.seed(42)    
N = 1000    
df = data.frame(X = runif(N), Y = runif(N)) 
df$Accept = (df$X**2 + df$Y**2 <=1) 
library(ggplot2)    
ggplot(df, aes(x=X,y=Y,color=Accept)) + geom_point(alpha=.2) + coord_fixed() + theme_bw()

It is then straightforward to obtain a (not really good) approximation to \(\pi\) by counting how many times, on average, \(X^2 + Y^2\) is smaller than 1:

4*mean(df$Accept)   
## [1] 3.156