Buffon's Needle
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The probability is equal to the ratio of the two areas in this case because each possible value of θ and ''d'' is equally probable. The probability of an intersection is | The probability is equal to the ratio of the two areas in this case because each possible value of θ and ''d'' is equally probable. The probability of an intersection is | ||
| - | <math>P_{hit} = \cfrac{ \frac{1}{2} }{\frac{\pi}{4}} = \frac {2}{\pi} = .6366197 | + | <math>P_{hit} = \cfrac{ \frac{1}{2} }{\frac{\pi}{4}} = \frac {2}{\pi} = .6366197\ldots.\qquad \mbox{and thus} \qquad \pi = \frac 2{P_{hit}}.</math> |
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| + | Thus, in order to approximate <math>\pi</math>, it remains only to approximate <math>P_{hit}</math>. | ||
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| + | [[User:Smaurer1|Smaurer1]] 18:10, 14 June 2010 (UTC)<font color=DarkOrange> | ||
| + | Following my general desire to separate the probability aspects of this page from the approximation aspects of this page, I think this probability section should end with the conclusion that pi can be expressed in terms of the probability. Then the next section can concentrate on the approximation. The middle two equations of that section would have to be changed. | ||
| + | </font> | ||
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| - | The Buffon’s needle problem has been generalized so that the probability of an intersection can be calculated for a needle of any length and paper with any spacing. For a needle shorter than the distance between the lines, it can be shown by a similar argument to the case where ''d'' = 1 and ''l'' = 1 that the probability of a intersection is <math> \frac {2*l}{\pi*d} </math>. Note that this agrees with the normal case, where ''l'' =1 and ''d'' =1, so these variables disappear and the probability is <math> \frac {2}{\pi} </math>. | + | The Buffon’s needle problem has been generalized so that the probability of an intersection can be calculated for a needle of any length and paper with any spacing. For a needle shorter than the distance between the lines, it can be shown by a similar argument to the case where ''d'' = 1 and ''l'' = 1 that the probability of a intersection is <math> \frac {2*l}{\pi*d} </math>. Note that this expression agrees with the normal case, where ''l'' =1 and ''d'' =1, so these variables disappear and the probability is <math> \frac {2}{\pi} </math>. |
Revision as of 14:10, 14 June 2010
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Contents |
Basic Description
The method was first used to approximate π by Georges-Louis Leclerc, the Comte de Buffon, in 1777. Buffon was a mathematician, and he wondered about the probability that a needle would lie across a line between two wooden strips on his floor. To test his question, he apparently threw bread sticks across his shoulder and counted when they crossed a line.Calculating the probability of an intersection for the Buffon's Needle problem was the first solution to a problem of geometric probability.Smaurer1 17:45, 14 June 2010 (UTC) Do a mouseover here, something like "a probability problem is geometric if the probability can be computed as the ratio of two areas or volumes." The solution can be used to design a method for approximating the number π. Smaurer1 17:45, 14 June 2010 (UTC) I cut out "in the case where is needle is not greater" because any case can be used to approximate pi; it's just that the work is harder in these other cases.
Subsequent mathematicians have used this method with needles instead of bread sticks, or with computer simulations. We will show that when the distance between the lines is equal the length of the needle, an approximation of π can be calculated using the equation

A More Mathematical Explanation
Will the Needle Intersect a Line?
To prove that the Buffon's Needle experiment will give an approximation of π, we can consider which positions of the needle will cause an intersection. Since the needle drops are random, there is no reason why the needle should be more likely to intersect one line than another. As a result, we can simplify our proof by focusing on a particular strip of the paper bounded by two horizontal lines.
The variable θ is the acute angle made by the needle and an imaginary line parallel to the ones on the paper. Smaurer1 17:49, 14 June 2010 (UTC) Since we are considering the case where the interline distance equals l, we might as well take that common distance to be 1 unit. Finally, d is the distance between the center of the needle and the nearest line. Also, there is no reason why the needle is more likely to fall at a certain angle or distance, so we can consider all values of θ and d equally probable. (Smaurer1 17:49, 14 June 2010 (UTC) good addition)
We can extend line segments from the center and tip of the needle to meet at a right angle. A needle will cut a line if the green arrow, d, is shorter than the leg opposite θ. More precisely, it will intersect when
See case 1, where the needle falls at a relatively small angle with respect to the lines. Because of the small angle, the center of the needle would have to fall very close to a line in order for the needle to intersect the line. In case 2, the needle intersects even though the center of the needle is far from both lines because the angle is so large.
The Probability of an Intersection
In order to show that the Buffon's experiment gives an approximation for π, we need to show that there is a relationship between the probability of an intersection and the value of π. If we graph the possible values of θ along the X axis and d along the Y, we have the sample space for the trials. (Smaurer1 17:49, 14 June 2010 (UTC) the mouseover is a good addition.)In the diagram below, the sample space is contained by the dashed lines.
Each point on the graph represents some combination of an angle and a distance that a needle might occupy.
There will be an intersection if
, which is represented by the blue region. The area under this curve represents all the combinations of distances and angles that will cause the needle to intersect a line. Since each of these combinations is equally likely, the probability is proportional to the area – that's what makes this a geometric probability problem. The area under the blue curve, which is equal to 1/2 in this case, can found by evaluating the integral
Then, the area of the sample space can be found by multiplying the length of the rectangle by the height.
The probability is equal to the ratio of the two areas in this case because each possible value of θ and d is equally probable. The probability of an intersection is
Thus, in order to approximate
, it remains only to approximate
.
Smaurer1 18:10, 14 June 2010 (UTC) Following my general desire to separate the probability aspects of this page from the approximation aspects of this page, I think this probability section should end with the conclusion that pi can be expressed in terms of the probability. Then the next section can concentrate on the approximation. The middle two equations of that section would have to be changed.
Using Random Samples to Approximate Pi
The original goal of the Buffon's needle method, approximating π, can be achieved by using probability to solve for π. If a large number of trials is conducted, the proportion of times a needle intersects a line will be close to the probability of an intersection. That is, the number of line hits divided by the number of drops will equal approximately the probability of hitting the line.
Also, recall from above that
So
Therefore, we can solve for π:
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Why It's Interesting
Teaching Materials
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References
[1] http://www.maa.org/mathland/mathtrek_5_15_00.html
[2] http://mste.illinois.edu/reese/buffon/bufjava.html
[3] http://www.absoluteastronomy.com/topics/Monte_Carlo_method
[4] The Number Pi. Eymard, Lafon, and Wilson.
[5] Monte Carlo Methods Volume I: Basics. Kalos and Whitlock.
[6] Heart of Mathematics. Burger and Starbird
[7] http://math.tntech.edu/techreports/TR_2001_4.pdf
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.

involving repeatedly dropping needles on a sheet of lined paper and observing how often the needle intersects a line.




. Note that this expression agrees with the normal case, where l =1 and d =1, so these variables disappear and the probability is
.




