Compute value of $\pi$ up to 8 digits I am quite lost on how approximate the value of $\pi$ up to 8 digits with a confidence of 99% using Monte Carlo. I think this requires a large number of trials but how can I know how many trials?

I know that a 99% confidence interval is 3 standard deviations away from the mean in a normal distribution. From the central limit theorem the standard deviation of the sample mean (or standard error) is proportional to the standard deviation of the population $\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}$

So I have something that relates the size of the sample (i.e. number of trials) with the standard deviation, but then I don't know how to proceed from here. How does the "8 digit precision" comes into play?

UPDATE

Ok I think I am close to understand it. From CLT we have $\displaystyle \sigma_{M} = \frac{\sigma}{\sqrt{N}}$ so in this case $\sigma = \sqrt{p(1-p)}$ therfore $\displaystyle \sigma_{M} = \frac{\sqrt{p(1-p)}}{\sqrt{N}}$

Then from the Bernoulli distribution, $\displaystyle \mu = p = \frac{\pi}{4}$ therefore

$$\sigma_{M}=\frac{\sqrt{\pi(4-\pi)}}{\sqrt{N}}$$ but what would be the value of $\sigma_{M}$? and then I have $\pi$ in the formula but is the thing I am trying to approximate so how does this work? and still missing the role of the 8 digit precision.
I may completely off the mark, but I guess that Monte Carlo here refers to approximating some integral by a random process, and that the integral must be chosen in a way that knowing its value allows us to calculate $\pi$.

Let's guess that the integral we are interested in is
$$
\frac\pi4=\int_0^1\sqrt{1-x^2}\,dx
$$
giving the probability that a random point in the square $[0,1]\times[0,1]$ (so $x$ and $y$ independent and uniformly distributed in $[0,1]$) is also in the unit disk.

Assume that we generate $N$ such points $(x,y)$ and record the number of successes (points in the unit disk) $M$. We know from crude estimates for $\pi$, say $3&lt;\pi&lt;4$, that the success rate $p$ of an individual point is between $3/4$ and $1$. Therefore the standard deviation of $M$ is bounded from above by
$$\sigma(M)=\sqrt{Np(1-p)}&lt;\frac{\sqrt{3N}}4.$$

We approximate
$$
\pi\approx\frac{4M}N.
$$
This has SD $\sigma(\pi)&lt;4\sigma(M)/N=\sqrt{3/N}$. 

For 99% confidence we want $3\sigma(\pi)&lt;10^{-8}$ or equivalently $\sqrt{N}&gt;3\sqrt{3}\cdot 10^8$. This suggests that generating $N\approx 27\cdot10^{16}$ random points on the unit square would do the trick :-)

After this gedankenexperiment I quite appreciate Machin's formula. Even the alternating series for $\arctan 1$ beats this.

So this leaves open the possibility that something completely different was wanted?
The usual trick is to use the approximation of 355/113, repeating multiples of 113, until the denominator is 355.  This version of $\pi=355/113$ is the usual implied value when eight digits is given.

Otherwise, the approach is roughly $\sqrt{n}$