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.


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