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![Page 1: Buffon’s Needle Problem Grant Weller Math 402. Georges-Louis Leclerc, Comte de Buffon French naturalist, mathematician, biologist, cosmologist, and author.](https://reader031.fdocuments.net/reader031/viewer/2022032204/56649e445503460f94b38545/html5/thumbnails/1.jpg)
Buffon’s Needle Problem
Grant WellerMath 402
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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
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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
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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?
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Buffon’s Needle Problem
• Answer: it depends on the length of the needle!
• For simplicity, assume that the length of the needle is less than the distance between the lines on the paper
• Then the probability is equal to (2/π)(/)• This means that you can use this experiment
to get an approximation of π!
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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 ≈ (2N)/(P).
• This experiment is one of the most famous in probability theory!
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Lazzarini’s Experiment
• In 1901 he allegedly built a machine to do this experiment
• He used a stick with (/)=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!
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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!
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Theorems
• If a short needle, that is, one with ≤, 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 =(2/π)(/).
• If l>d, this probability is =1+(2/π)((/)(1-√(1-2/2))-arcsin(/))
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Proof
• Can be solved by integrals• First, E. Barbier’s proof (1860):• Consider a needle of length ≤, and let E() be
the expected number of crossings produced by dropping the needle.
• Let = + (break the length of the needle into two pieces
• Then we can get E(+) = E()+E() (linearity of expectation)
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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
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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 , we again have E = (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!
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Proof ctd.
• Now consider a needle that is a perfect circle, call it C, with diameter .
• This needle has length π
• This needle always produces two intersections
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Proof ctd.
• We can now use polygons to approximate the circle.
• Let’s draw an inscribed polygon Pn and a circumscribed polygon Pn.
PnPn
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Proof ctd.
• Remember that for the polygonal needles the expected number of crossings is just , or the constant times the length of the needle.
• Also, if a line intersects Pn, it will intersect C, and if a line intersects C, it will hit Pn.
• Thus E(Pn) ≤ 2 ≤ E(Pn)
• And(Pn) ≤ 2 ≤ (Pn)
PnPn
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Proof ctd.
• Since both the polygons approximate C for n→∞, we know that
lim n→∞ (Pn) = π = lim n→∞ (Pn)
• Thus for n→∞, we have π ≤ 2 ≤ π
• This gives us = (2/π)(1/)!!!• Thus the probability of a needle of length
crossing a line is = (2/π)(/).
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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
sin α, and the probability that itcrosses a line of distance is (sin α)/.
α
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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.
d
l
d
ld
d
lp
2)cos(
2sin
02/
1 2/0
2/
0
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What about for a long needle?
• For a long needle, as long as it falls in a position where the “height” sin α is less than the distance between the lines , the probability is still (sin α)/.
• This occurs when 0 ≤ α ≤ arcsin (d/l)• If α is larger than this, the needle
must cross a line, so the probability is 1α
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Long needle probability
• Thus for ≥, we just have a longer integral:
• As you would expect, this formula yields 2/π for =, and goes to 1 as→∞
))arcsin()11((2
1
)))arcsin(2(]cos[(
2
)1sin
(2
2
2
)/arcsin(0
2/
)/arcsin(
)/arcsin(
0
l
d
l
d
d
l
l
d
d
l
ddd
lp
ld
ld
ld
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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:
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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/
BuffonsNeedleProblem.html