[mturmon]

This post has an opportunity to find the error bars on its estimate of π. The estimate came out as 3.14616.

The number of in-the-circle counts is S ~ binomial(N,p) where N=1e5 and p=π/4, which we'll take to be 0.75. Then

  var(S) = p(1-p)N ≤ N/4

and the standard deviation of S is the square root of this, or about ∆=137.

(If we weren't willing to use a guess for p in the variance above, we could use the universal bound of N/4.)

The estimate of π is (S/N) * 4, so the sdev on the estimate of π is (∆/N)*4 = 0.0055.

So the estimate turned out to be a little less than one standard deviation higher than π. Which is satisfying.

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The point being that, for just a little extra work, you get not only a Monte Carlo estimate, but also a bound on its error. Sometimes the bound is just as important as the estimate.