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.
 A: 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 $3<\pi<4$, 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
$$\sigma(M)=\sqrt{Np(1-p)}<\frac{\sqrt{3N}}4.$$
We approximate
$$
\pi\approx\frac{4M}N.
$$
This has SD $\sigma(\pi)<4\sigma(M)/N=\sqrt{3/N}$. 
For 99% confidence we want $3\sigma(\pi)<10^{-8}$ or equivalently $\sqrt{N}>3\sqrt{3}\cdot 10^8$. 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?
A: The statistical analysis given in the other answer seems fine.
But the discussion of integrals is unnecessary, I think. The standard Monte-Carlo technique for estimating $\pi$ is to generate $N$ random points in a square centered at the origin -- say the square $[-1,1] \times [-1,1]$. After generating each point $(x,y)$, you can check to see if $x^2 + y^2 < 1$, which tells you whether that point lies in the unit circle. You keep track of the fraction $k$ of random points that lie within the unit circle. As $N$ gets larger, this fraction $k$ gives increasingly accurate estimates of the ratio $\text{(area of circle)} / \text{(area of square)}$. In other words $k \approx \pi/4$, when $N$ is large, so $\pi \approx 4k$.
The statistical analysis given in the other answer tells you how large $N$ must be in order to achieve the desired accuracy with a desired probability.
There is a fairly good explanation on this web page.
And another one on this page.
A: 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 $\sqrt{n}$
A: Basically, it is about the standard deviation of the Binomial distribution (a single independent experiment is called Bernoulli experiment, with a mean of p and the variance of p-p^2). For B(n,p), the variance is np(1-p), the standard deviation is  $$ \sqrt{p(1-p)/n} $$
In our case of Monte Carlo simulation, p = pi/4
Lastly, we need to time 4 to get pi. So $4*\sqrt{p*(1-p)/n}=1.642*\sqrt{1/n}<10^{-8}$, get n> 2.7e16
