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`Best' length of Buffon's needle

Mathematics Asked by Václav Mordvinov on January 5, 2021

Suppose we conduct Buffon’s needle experiment, randomly dropping a needle of length $Lin(0,1]$ between horizontal lines which are a distance of $1$ apart. It is well known that the probability that the needle intersects one of these lines equals $p=2L/pi$, so that we can estimate $pi$ by $2L/hat p=2Ln/m$, where $n$ is the size of the Monte Carlo experiment, and where $m$ is the number of intersections.

I am looking for the optimal length $L$ of the needle, making the estimate for $pi$ as accurate as possible. It seems reasonable that when an intersection is as likely as no intersection, so when $L=pi/4$, the estimation is most efficient (although I am not sure it this is correct, and if so, why).

More specifically, I’m looking for the variance of $2L/hat p$. This seems to come down to estimate the variance of the reciprocal of a binomial r.v., for which no nice expression exist. Probably there is some other approach I’m not seeing. Any help is much appreciated.

N.B.: this is just for theoretical purposes; I don’t mind to use $pi$ in the process of estimating $pi$ itself.

One Answer

You can also estimate $1/pi$ by $hat p/2L,$ whose variance is easier to calculate.

For large enough $R,$ the distribution of $hat p$ around $p$ will be concentrated near $p$ and very nearly symmetric near $p.$ When you suppose that a $95%$ confidence interval is $hat ppm1.96frac{2L}pileft(1-frac{2L}piright)/sqrt R,$ you are assuming that the distribution is symmetric within that interval around $p$.

The symmetric CI around $hat p$ will map to an asymmetric CI around $1/hat p$ when you use $1/hat p$ as an estimator for $pi,$ but this just means that the symmetric CI that you want around $1/hat p$ comes from an asymmetric CI around $hat p$. If your estimate of $pi$ has a decent amount of precision, the asymmetry of the desired CI around $hat p$ is not even very much. It seems reasonable to suppose that the value of $L$ that minimizes the asymmetric CI is the same as that which minimizes the symmetric CI. This may not be perfectly accurate, but neither is the use of a normal distribution in place of the actual binomial distribution of $hat p$.

Correct answer by David K on January 5, 2021

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