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Hi, what are the "best" values for lenght of needle (l) and distance between paralles (d) for an accurate approximation of pi? Does it have to be l-d-1.0 or ld? Thanx

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closed as not a real question by Henry Cohn, Andreas Blass, Andy Putman, Dan Petersen, Alain Valette Apr 4 '12 at 21:15

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Note that it is only the ratio $l/d$ that matters. – André Henriques Apr 4 '12 at 12:51
It's not clear what "Does it have to be l-d-1.0 or ld?" means. Overall, Buffon's needle is entertaining and beautiful, but it's a terrible way to compute $\pi$ (no matter how you set it up). – Henry Cohn Apr 4 '12 at 13:09
I suppose that you could realquestionify this by considering how the expected number of needle drops required to approximate $\pi$ to within a given amount depends on $l$ and $d$. I wondered about that once recently when discussing this with someone, but didn't get around to thinking about it – Ramsey Apr 4 '12 at 14:11

Indeed, the result of the experiment is strongly dependent on arithmetic properties of the parameters - the ratio $\rho$ of needle's lenght to the distance between parallel lines, and of course, the number $N$ of needles. There is a funny story about that. In 1901 the Italian mathematician Mario Lazzarini claimed to have obtained an experimental value of $\pi$ with 7 correct digits, using 3408 needles, and a ratio $\rho=5/6$ (a harmless choice, apparently). That was quite embarrassing, as it was quite clear he had cheated -after and before Lozzerini's experiment, everybody else, even with larger numbers of needles, never got a better result than 3.13 or 3.15., and today Lazzerini's experiment it is sometimes reported as a case of false. The point is a little more subtle: with the parameters he had taken, he was quite likely to obtain as approximantion the ratio 355/113 (the lucky number being 1808 intersections). My personal guess is that he somehow meant to mock the other scientists, physicists or naturalists, who performed Buffon's experiment ignoring the arithmetic axpect of the matter. One can easily do even more striking experiments, of course: with a convenient (irrational) choice of the ratio, and just 2 or 3 needles, you have a good chance to obtain an exact value of $\pi$ from Buffon's experiment...

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The naive estimator is biased. If there are $N$ trials and $i$ success, a Rao-Blackwellisation of the naive estimator gives the unbiased estimator $\frac{2l}{d}\left(\frac{n}{i}+\frac{1}{i}\right)$ (to be fair this hides an assumption for a uniform prior for the probability of crossing, which induces a weird prior on $\pi$).

One can look at the variance of the estimator conditional on obtaining one success. The strategy is then to set $l=d$. Intuitively this makes sense, we want the term in $1/i$ to be as small as possible. Though for $n$ small enough, the prior dominates and the optimum is actually achieved for a small value of $2l/(d\pi)$.

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I'm guessing that the longer the needle the better (meaning, highest $\ell/d$ ratio). It seems like you'll get significant experimental issues from the cases where the needle just barely crosses-or-doesn't-cross a line, and if your needle is long enough to cross lots of lines then this becomes less relevant. Probably you want to color the lines so every tenth and every hundredth are a different color, the better to accurately count them.

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example: which is the best conmbination for the most accurate approximation of Pi? 1. d = ℓ=1 2. d=1, ℓ=0.8 3. d=1, ℓ=0.6 4. d=1.2, ℓ=1.0 – spyros Apr 4 '12 at 12:23

The question is well defined for estimating $\frac 1 \pi$, but not for estimating $\pi$. If $l>d$, you need to evaluate $cos^{-1}$ which requires knowledge of $\pi$, otherwise the variance of the estimator decreases with$ \frac l d$, so you'd "practically" settle for $l=d$.

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