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Buffon's Needle

Calculus II Honors Project
by

Sarah Mayer

on 7 December 2012

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Transcript of Buffon's Needle

Buffon's Needle Problem Sarah Mayer Calculus II Honors Project Fall 2012 Proposed by Georges-Louis Leclerc, Comte de Buffon in 1773 Solution published in 1777 Classic problem in geometric probability Can be used to experimentally approximate π If you drop a needle on a wood floor with equally spaced cracks, what is the probability that the needle will cross one of the cracks? Solution: Assume L ≤ D
Measure y from bottom of needle to the closest line above
Needle crosses crack if y ≤ d → if y ≤ Lsinθ D d y θ d=Lsinθ
0<θ<π
0<y<D Graphical Representation: The probability of a crossing is equal to the area under the
graph of y=Lsinθ divided by the total area of the graph
-Total area comes from the restrictions 0<θ<π and 0<y<D

Area under the curve: Lsinθdθ = -Lcosθ 0 π 0 π =-L(cosπ - cos0) = -L(-2) = 2L Probability of a crossing = (2L)/(πD) The probability of crossing a crack is based on two independent events: 1) The needle falls close enough to a crack that it is in range to cross the crack One way to think of this is to consider the midpoint of the needle... if the distance between the midpoint and the nearest crack is less than half the length of the needle, it is possible for the needle to cross (if it is oriented correctly, which brings us to the second event...) 2) The needle is oriented such that it actually crosses the crack ie, if the needle pivots around its midpoint (which is within range), is it at such an angle that it goes over the crack? To start, first let L equal the length of the needle
and let D equal the distance between cracks The angle of the needle only goes up to pi because that accounts for all possible orientations; after pi radians, the needle is taking the same positions, just with its "head" and "tail" in opposite directions, which doesn't matter for this problem! Each axis represents one of the two independent events: position of the needle and orientation of the needle A more visual way to approach the problem: Again, the probability of crossing a crack is based on two independent events: P1) the needle falling close enough to a line that it could cross, and P2) the needle falling at such an angle that it does cross. In order for a crossing to be possible, the distance from the midpoint of the needle to the nearest line must be less than L/2 (otherwise even a needle perpendicular to the line cannot reach). Since the midpoint could land within L/2 units from either of the lines it falls between, and potentially result in a crossing, the area that would allow for a crossing is (L/2 + L/2)/D
Therefore, P1=L/D (Again, we assume here that the L<D) The figure above shows x as the position of the midpoint. The grey area then represents the orientations of the needle that result in a crossing. The probability of a crossing is equal to all those orientations divided by all possible orientations. For simplicity, we will only integrate the left grey region and divide by π (instead of doing both grey areas over 2π), and the grey area will be equal to 2θ. From the triangle formed by the needle and the axes, we see that cosθ=x, and x can go from 0 to 1.
Represented as a function of x: θ=arccos(x). The integral of θ=arccos(x) from 0 to 1, then multiplied by 2, represents all orientations that result in a crossing (the left grey region) 0 1 arccos(x)dx = (x)arccos(x) - 1-x 2 1 0 =[(1)arccos(1)-0]-[(0)arccos(0)-1] = 1 Orientations represented by grey region
divided by possible orientations = 2/π x 2=2 Therefore, P2=2/π Probability of a crossing=P1 x P2 = L/D x 2/π
=(2L)/(πD) So far we've only considered cases where L<D...
What happens if L>D? If the needle is longer than the distance between cracks, then it is possible for the needle to cross more than one crack at once. However, the question is only if it will cross a single crack or not, so we will keep the same restrictions as before: y (the distance from the bottom of the needle to the crack above it) must be less than D, and θ is between 0 and π. This case is easiest to consider with the graphical approach: taking the area under the curve divided by the area of the graph. But, since the function y=Lsinθ can be greater than our boundary D (a consequence of L being greater than D), we cannot simply integrate, as part of the curve goes out of the region bounded by Dπ (see figure at left). To find the area under the curve AND within possible D and π values, we must integrate from 0 to A and from (π-A) to π, then add that to the area of the rectangle between A and (π-A). Since y=Lsinθ is symmetric, the integral from 0 to A equals the integral from (π-A) to π. Thus, the area between O and A, and (π-A) and π = 2 Lsinθdθ 0 A First, we must find the value of A in terms of θ. Since A is the intersection of Lsinθ and D, we set those terms equal and get:
Lsinθ=D → sinθ=D/L → θ=arcsin(D/L) 0 arcsin(D/L) 2L sinθdθ = 2L[-cos(arcsin(D/L)) + cos0] θ D L L -D 2 2 =2L[-( )/L + 1] = 2(L- ) L -D 2 2 L -D 2 2 D[π-2arcsin(D/L)] + 2(L- ) L -D 2 2 Then, to find the area of the rectangle, we know the height is D and the width is π-2A = π-2arcsin(D/L)
so the area = D[π-2arcsin(D/L)].
Thus, the total area under the curve and within the boundaries is and the probability of a crossing is D[π-2arcsin(D/L)] + 2(L- ) L -D 2 2 πD Since the number of crossings over uniformly-spaced cracks is proportional to the length of what is dropped, the formula P=(2L)/(Dπ) can be applied to shapes besides just straight lines.

This concept relies on the idea that different regions of the needle independently contribute to the probability of a crossing. Imagine one needle of length L compared to another needle of length L/2. The probability of the shorter needle crossing is half that of the longer needle crossing. If you tossed two short needles at the same time, the probability of a crossing occurring equals their P values added together - which equals the P value for a single needle twice as long! The leap we must make here is that, even if the two needles were attached (in a straight line OR at any random angle), their probabilities add together and equal the P value for a single needle of the shorter needles' combined length.

So, you can apply P=(2L)/(Dπ) along any segment of a needle, and the probability of that segment crossing is simply proportional to its length. If you apply this concept to infinitely short segments of a non-linear form, and add up the probabilities for each segment, the probability of the curve making a crossing is proportional to its length - even if it's not a straight line! We find this to be true in the case of a circle of diameter D dropped over cracks D units apart. The circumference of the circle (its "length" for our purposes) equals πD, meaning that its length is greater than the distance between the cracks. In this case, P should be thought of as the average number of crossings per drop.

No matter how the circle lands, it will always have exactly two crossings: (P=2) The circle either falls on top of a line, or exactly between two lines; in either case, there are exactly two intersections If we were to plug the length L=πD into our formula (2L)/(πD), we predict P=[2(πD)]/(πD)=2 ... just as we observe! An application of the solution to Buffon's Needle Problem: If P=(2L)/(πD), then we can rearrange the equation to get π=(2L/(DP).
P can be experimentally found by repeatedly dropping a needle and dividing the number of crosses by the number of tosses A "Monte Carlo" method for approximating π (Monte Carlo methods use random numbers and repeated sampling to make a computation) Once you have P, plug it into the formula with L and D to find π ~I experimentally determined my value of P by tossing a pencil (my "needle") across the room toward lines I taped on the carpet.
~I made D equal to the length of my pencil so that π=2/P
~To avoid the "crack" having a width itself, I made tape and carpet regions, so the "crack" is the transition between the two. The photo at left shows a crossing. Out of 200 tosses, I got 129 crosses
P=0.645
Approximation of pi=2/0.645=3.10 The greatest source of error in my experiment is that my tosses were not entirely random... I stood several feet away and had to aim in the general direction of the tape, so the pencil tended to land toward the center of the area. This may have increased the number of crosses, making my approximation less than the actual value of π. A better approximation could be achieved with a computer simulation. That said, I have pretty bad aim, so my tosses were still pretty random! Tossing objects to approximate π... perhaps a new party game? Sources:
Bogomolny, Alex. "Math Surprises: An Example." Math Surprises. N.p., Aug. 2001. Web. 06 Dec. 2012. <http://www.cut-the-knot.org/ctk/August2001.shtml>.

"The Buffon Needle Problem." The Buffon Needle Problem Lecture Note. N.p., n.d. Web. 06 Dec. 2012.
"Buffon's Needle Problem." Buffon's Needle Problem. N.p., n.d. Web. 06 Dec. 2012. <http://webspace.ship.edu/deensley/mathdl/stats/Buffon.html>.

"Buffon's Needle." Wikipedia. Wikimedia Foundation, 11 Dec. 2012. Web. 06 Dec. 2012.

"How to Calculate Pi by Throwing Frozen Hot Dogs." WikiHow. N.p., n.d. Web. 06 Dec. 2012. <http://www.wikihow.com/Calculate-Pi-by-Throwing-Frozen-Hot-Dogs>.

Kunkel, Paul. "Buffon's Needle." Buffon's Needle. N.p., n.d. Web. 06 Dec. 2012. <http://whistleralley.com/buffon/buffon.htm>.

Kunkel, Paul. "What If D<L?." N.p., n.d. Web. 06 Dec. 2012. <http://whistleralley.com/buffon/solve/solution.htm>.

Peterson, K. M. A Problem in Geometric Probability: Buffon's Needle Problem. Lynchburg, VA: Lynchburg College, n.d. PPT.

Reese, George. "Buffon's Needle." Buffon's Needle. N.p., n.d. Web. 06 Dec. 2012. <http://mste.illinois.edu/reese/buffon/buffon.html>.
To make calculations easier, assume the needle is 2 units long. This means x must be less than 1 since the midpoint must fall within L/2 units from the nearest crack for a crossing to occur.
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