Buffon’s Needle
Todd Savage
Buffon's Needle Problem
Buffon's needle problem asks to find the probability that a needle of length ‘l’ will land on a line, given a floor with
equally spaced parallel lines a distance apart. The problem was first posed by the French naturalist Buffon in 1733
(Buffon 1733, pp. 43-45), and reproduced with solution by Buffon in 1777 (Buffon 1777, pp. 100-104).
Monte Carlo Method
Monte Carlo methods are a class of computational algorithms for simulating the behavior of
various physical and mathematical systems. They are distinguished from other
simulation methods (such as molecular dynamics) by being stochastic, that is non
deterministic in some manner - usually by using random numbers (or more often
pseudo-random numbers) - as opposed to deterministic algorithms. A classic use is for
the evaluation of definite integrals, particularly multidimensional integrals with
complicated boundary conditions.
The Simplest Case
Let's take the simple case first. In this case, the length of the needle is one unit and the distance between the lines is
also one unit. There are two variables, the angle at which the needle falls (theta) and the distance from the center of
the needle to the closest line (D). Theta can vary from 0 to 180 degrees and is measured against a line parallel to the
lines on the paper. The distance from the center to the closest line can never be more that half the distance between
the lines. The graph below depicts this situation.
The needle in the picture misses the line. The needle will hit the line if the closest distance to a line (D) is less
than or equal to 1/2 times the sine of theta. That is, D <= (1/2)sin(theta).
The Simplest Case (cont’d)
In the graph below, we plot D along the ordinate and (1/2)sine(theta) along the abscissa. The values on or below the
curve represent a hit (D <= (1/2)sin(theta)). Thus, the probability of a success it the ratio shaded area to the entire
rectangle.
The shaded portion is found using the definite integral of (1/2)sin(theta) evaluated from zero to pi. The result is that the
shaded portion has a value of 1. The value of the entire rectangle is (1/2)(pi) or pi/2. So, the probability of a hit is
1/(pi/2) or 2/pi. That's approximately 0.6366197.
To calculate pi from the needle drops, simply take the number of drops and multiply it by two, then divide by the
number of hits, or
2(total drops)/(number of hits) = pi (approximately).
Mathematica Notebook
Buffon’s Needle Java Applet
Buffon's Needle
Questions
Experimental Probability = (# of hits)/(total # of drops)
On the other hand, we have (approximately):
Theoretical Probability = 2/Pi
We can use this fact to conclude that:
2(total # of drops)/(# of hits)= Pi (approximately)
1. After 1,000 drops, how close would you expect to be to pi?
2. After 264 drops, the estimate of pi is 3.142857. This estimate is correct to within 2/1000 of the book value of pi.
Will the next drop:
A. make the estimate more accurate?
B. make the estimate less accurate?
C. make it more or less accurate depending on whether it's a hit or miss? or
D. impossible to say.
3. What about the next 10 drops?
References
http://www.mste.uiuc.edu/reese/buffon/buffon.html
http://www.answers.com/topic/buffon-s-needle
http://www.ms.uky.edu/~mai/java/stat/buff.html
http://en.wikipedia.org/wiki/Monte_Carlo_method
http://mathworld.wolfram.com/BuffonsNeedleProblem.html
http://scienceworld.wolfram.com/biography/Buffon.html