![]() The situation in which the needle is longer than the distance between the lines leads to a more complicated result. The situation in which the distance between the lines is greater than the length of the needle is an extension of the above explanation and the probability of a hit is 2(L)/(K)π where L is the length of the needle and K is the distance between the lines. A good discussion of these can be found in Schroeder, 1974. There are two other possibilities for the relationship between the length of the needles and the distance between the lines. To calculate pi from the needle drops, simply take the number of drops and multiply it by two, then divide by the number of hits, or 2(total drops)/(number of hits) = π (approximately). So, the probability of a hit is 1/(π/2) or 2/π. The value of the entire rectangle is (1/2)(π) or π/2. The result is that the shaded portion has a value of 1. The shaded portion is found with using the definite integral of (1/2)sin(θ) evaluated from zero to pi. ![]() The distance from the center to the closest line can never be more that half the distance between the lines. Theta can vary from 0 to 180 degrees and is measured against a line parallel to the lines on the paper. There are two variables, the angle at which the needle falls (θ) and the distance from the center of the needle to the closest line (D). In this case, the length of the needle is one unit and the distance between the lines is also one unit. ![]() This page will present an analytical solution to the problem along with a JavaScript applet for simulating the needle drop in the simplest case scenario in which the length of the needle is the same as the distance between the lines. The remarkable result is that the probability is directly related to the value of pi. It involves dropping a needle on a lined sheet of paper and determining the probability of the needle crossing one of the lines on the page. Buffon's Needle is one of the oldest problems in the field of geometrical probability.
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