Buffon

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Buffons Needle Problem

• Grant Weller
• Math 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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Buffons 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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Buffons 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/p)(l / d)
• This means that you can use this experiment to
  get an approximation of p!

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Finding p

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

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Lazzarinis 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 p 2(5/6)(3408/1808) (355/113)
  3.1415929
• Correct to six digits of p, too good to be true!

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Lazzarinis Experiment

• Lazzarini knew that 355/113 was a great
  approximation of p
• 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 ld, 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/p)(l/d).
• If lgtd, this probability is
• p1(2/p)((l/d)(1-v(1-d2/l2))-arcsin(d/l))

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Proof

• Can be solved by integrals
• First, E. Barbiers proof (1860)
• Consider a needle of length ld, and let E(l) be
  the expected number of crossings produced by
  dropping the needle.
• Let l xy (break the length of the needle into
  two pieces
• Then we can get E(xy) E(x)E(y) (linearity of
  expectation)

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Proof ctd.

• Furthermore, we can show that E(cx) cE(x) for
  all c?R, because for x0, as the length x of the
  needle increases, the expected number of
  crossings increases proportionately.
• We also know that cE(1), the probability of
  getting one crossing from the needle.
• Consider needles that arent 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 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!

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Proof ctd.

• Now consider a needle that is a perfect circle,
  call it C, with diameter d.
• This needle has length dp
• This needle always produces two intersections

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Proof ctd.

• We can now use polygons to approximate the
  circle.
• Lets draw an inscribed polygon Pn and a
  circumscribed polygon Pn.

Pn
Pn
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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 Pn, it will intersect
  C, and if a line intersects C, it will hit Pn.
• Thus E(Pn) 2 E(Pn)
• And cl(Pn) 2 cl(Pn)

Pn
Pn
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Proof ctd.

• Since both the polygons approximate C for n?8, we
  know that
• lim n?8 l(Pn) dp lim n?8 l(Pn)
• Thus for n?8, we have
• cdp 2 cdp
• This gives us c (2/p)(1/d)!!!
• Thus the probability of a needle of length l
  crossing a line is cl (2/p)(l/d).

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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 a from the horizontal
• a falls in the range 0 to p/2
• The height of this needle is then
• lsin a, and the probability that it
• crosses a line of distance d is (lsin a)/d.

a
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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 a.

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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 lsin a is less than
  the distance between the lines d, the probability
  is still (lsin a)/d.
• This occurs when 0 a arcsin (d/l)
• If a is larger than this, the needle
• must cross a line, so the probability is 1

a
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Long needle probability

• Thus for ld, we just have a longer integral
• As you would expect, this formula yields 2/p for
  ld, and goes to 1 as l?8

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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

• Buffons 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