Buffon’s Needle Problem
Grant Weller
Math 402
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
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
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?
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!
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 ≈ (2lN)/(dP).
• This experiment is one of the most famous in
probability theory!
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!
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!
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-d2/l2))-arcsin(d/l))
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)
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
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!
Proof ctd.
• Now consider a needle that is a perfect circle, call
it C, with diameter d.
• This needle has length dπ
• This needle always produces two intersections
Proof ctd.
• We can now use polygons to approximate the
circle.
• Let’s draw an inscribed polygon Pn and a
circumscribed polygon Pn.
Pn
Pn
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)
n
Pn P
Proof ctd.
• Since both the polygons approximate C for
n→∞, we know that
lim n→∞ l(Pn) = dπ = lim n→∞ l(Pn)
• Thus for n→∞, we have
cdπ ≤ 2 ≤ cdπ
• 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).
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
α
lsin α, and the probability that it
crosses a line of distance d is (lsin α)/d.
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.
1
p
 /20
 /2

0
l sin 
2 l
2 l
 /2
d 
( cos )0 
d
d
d
What about for a long needle?
• For a long needle, as long as it falls in a
position where the “height” lsin α is less than
the distance between the lines d, the
probability is still (lsin 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
Long needle probability
• Thus for l≥d, we just have a longer integral:
p
2

arcsin( d / l )
(

0
 /2
l sin 
d 
1d )

d
arcsin( d / l )
l

d
arcsin( d / l )

( [  cos ]0
 (  arcsin( )))
 d
2
l
2
l
d2
d
 1  ( (1  1  2 )  arcsin( ))
 d
l
l
2
• As you would expect, this formula yields 2/π for
l=d, and goes to 1 as l→∞
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:
References
• “Buffon’s Needle Problem”. Chapter 21,
Proofs from the Book.
• http://en.wikipedia.org/wiki/GeorgesLouis_Leclerc%2C_Comte_de_Buffon
• http://en.wikipedia.org/wiki/Buffon's_needle
• http://mathworld.wolfram.com/BuffonsNeedl
eProblem.html