# Estimating pi with a Gaussian darts player

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?


A straightforward version of the dart experiment when using a Gaussian distribution is when the covariance matrix fits the circle, which corresponds to a $$\mathcal N_2(0,\sigma^2 I)$$ distribution since $$\mathbb P(||X||^2\le 1)=\mathbb P(\sigma^{-2}||X||^2\le \sigma^{-2})=\overbrace{\mathbb P(Z\le \sigma^{-2})}^{Z\sim\chi^2_2}=1-e^{-\sigma^{-2}/2}$$ Therefore the proportion of realisations inside the unit circle $$\mathcal C$$ does not involve $$\pi$$.

To estimate $$\pi$$ an importance sampling version could be used instead $$\pi = \int_{\mathcal C} 1\,\text{d}x=\int_{\mathcal C} \frac{\varphi(x)}{\varphi(x)}\,\text{d}x=\mathbb E[\varphi(X)^{-1}\mathbb I_{\mathcal C}(X)]$$ hence $$\pi\approx \frac{1}{n}\sum_{i=1}^n \varphi(X_i)^{-1}\mathbb I_{\mathcal C}(X_i)\qquad X_i\stackrel{\text{iid}}{\sim}\mathcal N_2(0,\Sigma)$$

The key to calculating $$\pi$$ this way is being able to calculate $$P[\text{dart in shape}]$$ both by counting darts $$(\frac{\text{total darts in shape}}{\text{total darts thrown}})$$, and also with an analytical expression that somehow involves $$\pi$$. That lets you substitute one into the other to solve for $$\pi$$.

When the darts are thrown with equal likelihood of landing anywhere in the 2x2 area, you have the analytical expression:

$$P_{\text{uniform}}[\text{dart in unit circle}] = \frac{\text{area of circle}}{\text{area of 2x2 square}} = \frac{\pi r^2}{4} = \frac{\pi}{4}$$

So you can calculate $$\pi$$ as:

$$\pi = 4 \cdot \frac{\text{total darts in circle}}{\text{total darts thrown}}$$

But in general, the analytical expression is of the form:

$$P[\text{dart in unit circle}] = \frac{\text{probability mass in shape}}{\text{probability mass in 2x2 square}}.$$

If your darts land according to a Gaussian distribution, then that expression is just the integral of the Gaussian distribution's PDF (which conveniently involves $$\pi$$) over the shape you choose, divided by its integral over the 2x2 square.