It's well known that if you draw a square of 2x2 units side, and inside you draw a circle with a radius 1, and then randomly throw darts/count raindrops/etc. etc. that fall within the square, you can estimate pi as ~3.66 times as many darts/raindrops/etc. will fall inside the circle, as opposed to outside the circle but within the square (pi / (4-pi) to be exact).
For an explanation see this video
However, if I throw darts at a dartboard, I probably am (or at the very least can be thought of as) better modelled as a gaussian darts thrower, where both the x and y coordinates have a mean of 1 and a standard deviation of x.
Is there a way of intuiting what 'shape' should 'fit' inside the 2x2 square so that my dart throws can be used as a Monte Carlo method to estimate pi, given that for any throw the x and y coordinates are normally distributed around a mean of 1?
I'm not entirely sure such a shape should necessarily exist, but the question was raised to me and I had no answer.
Some basic R code below if it helps describe the problem
#do a lot of draws n <- 10000000 #create data frame of dart/raindrop coordinates df <- data.frame( x = runif(n), y = runif(n) ) #find which fall inside/outside the circle df$inside_circle <- ifelse(sqrt(df$x^2 + df$y^2) <= 1, 1, 0) #estimate pi length(which(df$inside_circle == 1)) / length(which(df$inside_circle == 0)) / (pi / (4 - pi)) #what if darts are thrown with a mean one and a sd of x? df <- data.frame( x = rnorm(n, 0.5, 0.12), y = rnorm(n, 0.5, 0.12) ) #how to estimate pi given these throws?