1. Buffon’s Needle Problem Grant Weller Math 402

2. Georges-Louis Leclerc, Comte de Buffon French naturalist, mathematician, biologist, cosmologist, and author 1707-1788 Wrote a 44 volume encyclopedia describing the natural world One of the first to argue for the concept of evolution Influenced Charles Darwin

3. Georges-Louis Leclerc, Comte de Buffon Introduced differential and integral calculus into probability theory His “needle problem” is one of the most famous in the field of probability Introduced these concepts in his paper Sur le jeu de franc-carreau

4. Buffon’s Needle Problem Suppose that you drop a needle on ruled paper. What is the probability that the needle comes to lie in a position where it crosses one of the lines?

5. Buffon’s Needle Problem Answer: it depends on the length of the needle! For simplicity, assume that the length of the needle l is less than the distance d between the lines on the paper Then the probability is equal to (2/π)( l / d ) This means that you can use this experiment to get an approximation of π!

6. Finding π If you have a needle of shorter length than the distance between the lines, you can approximate π with this experiment. If you drop a needle P times and it comes to cross a line N times, you should eventually get π ≈ (2 l N)/( d P). This experiment is one of the most famous in probability theory!

7. Lazzarini’s Experiment In 1901 he allegedly built a machine to do this experiment He used a stick with ( l / d )=5/6 Dropped the stick 3408 times Found that it came to cross a line 1808 times Approximated π ≈ 2(5/6)(3408/1808) = (355/113) = 3.1415929… Correct to six digits of π, too good to be true!

8. Lazzarini’s Experiment Lazzarini knew that 355/113 was a great approximation of π He chose 5/6 for the length ratio because he knew he would be able to get 355/113 that way If this experiment is “successful,” we hope for P = 113N/213 successes So Lazzarini just did 213 trials at a time until he got a value that satisfied this ratio exactly!

9. Theorems If a short needle, that is, one with l ≤ d, is dropped on paper that is ruled with equally spaced lines of distance, then the probability that the needle comes to lie in a position where it crosses one of the lines is exactly p =(2/π)( l / d ). If l>d, this probability is p =1+(2/π)(( l / d )(1-√(1- d 2 / l 2 ))-arcsin( d / l ))

10. Proof Can be solved by integrals First, E. Barbier’s proof (1860): Consider a needle of length l ≤ d, and let E( l ) be the expected number of crossings produced by dropping the needle. Let l = x + y (break the length of the needle into two pieces Then we can get E( x + y ) = E( x )+E( y ) (linearity of expectation)

11. Proof ctd. Furthermore, we can show that E(cx) = cE(x) for all cєR, because for x≥0, as the length x of the needle increases, the expected number of crossings increases proportionately. We also know that c=E(1), the probability of getting one crossing from the needle. Consider needles that aren’t straight. For example a polygonal needle

12. Proof ctd. The number of crossings produced by this needle is the sum of the number of crossings produced by its straight pieces. If the total length of the needle is l, we again have E = cl (again by linearity of expectation) In other words, it is not important whether the needle is straight or even if the pieces are joined together rigidly or flexibly!

13. Proof ctd. Now consider a needle that is a perfect circle, call it C, with diameter d. This needle has length d π This needle always produces two intersections

14. Proof ctd. We can now use polygons to approximate the circle. Let’s draw an inscribed polygon P n and a circumscribed polygon P n. PnPn PnPn

15. Proof ctd. Remember that for the polygonal needles the expected number of crossings is just cl, or the constant times the length of the needle. Also, if a line intersects P n, it will intersect C, and if a line intersects C, it will hit P n. Thus E(P n ) ≤ 2 ≤ E(P n ) And cl (P n ) ≤ 2 ≤ cl (P n ) PnPn PnPn

16. Proof ctd. Since both the polygons approximate C for n→∞, we know that lim n→∞ l (P n ) = d π = lim n→∞ l (P n ) Thus for n→∞, we have cd π ≤ 2 ≤ cd π This gives us c = (2/π)(1/ d )!!! Thus the probability of a needle of length l crossing a line is cl = (2/π)( l / d ).

17. A much quicker proof! We could have done this proof with integral calculus! Consider the “slope” of the needle. Let it drop at an angle α from the horizontal α falls in the range 0 to π/2 The height of this needle is then l sin α, and the probability that it crosses a line of distance d is ( l sin α)/ d. α

18. A much quicker proof ctd. Thus the probability of an arbitrary needle crossing a line can be found by averaging this probability over the possible angles α.

19. What about for a long needle? For a long needle, as long as it falls in a position where the “height” l sin α is less than the distance between the lines d, the probability is still ( l sin α)/ d. This occurs when 0 ≤ α ≤ arcsin (d/l) If α is larger than this, the needle must cross a line, so the probability is 1 α

20. Long needle probability Thus for l ≥ d, we just have a longer integral: As you would expect, this formula yields 2/π for l = d, and goes to 1 as l →∞

21. Probability for any needle If we let the distance between the lines on the paper be 1, this is what the probability function looks like for needles of increasing length:

22. References “Buffon’s Needle Problem”. Chapter 21, Proofs from the Book. http://en.wikipedia.org/wiki/Georges- Louis_Leclerc%2C_Comte_de_Buffon http://en.wikipedia.org/wiki/Buffon's_needle http://mathworld.wolfram.com/BuffonsNeedl eProblem.html