This is a simulation that estimates Pi by randomly dropping sticks onto a set of horizontal lines spaced one stick length apart. It uses the fact that the probability of a randomly dropped stick intersecting one of the lines is equal to $\frac{\pi}{2}$. (NOTE: YOU NEED TO HAVE ADOBE FLASH INSTALLED TO VIEW) This simulation represents the Buffon's needle problem. Buffon's Needle is one of the oldest problems in the field of geometrical probability. It was first stated in 1777. 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. The remarkable result is that the probability is directly related to the value of pi. The problem in more mathematical terms is: Let's take the simple case first. In this case, the length of the needle is one unit and the distance between the lines is also one unit. 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). Theta can vary from 0 to 180 degrees and is measured against a line parallel to the lines on the paper. The distance from the center to the closest line can never be more that half the distance between the lines. The image on the left illustrates that. The needle in the picture misses the line. The needle will hit the line if the closest distance to a line (D) is less than or equal to 1/2 times the sine of theta. That is, D ≤ (1/2)sin(θ). How often will this occur? In the graph below, we plot D along the ordinate and (1/2)sin(θ) along the abscissa. The values on or below the curve represent a hit (D ≤ (1/2)sin(θ)). Thus, the probability of a success it the ratio shaded area to the entire rectangle. What is this to value? The shaded portion is found with using the definite integral of (1/2)sin(θ) evaluated from zero to pi. The result is that the shaded portion has a value of 1. The value of the entire rectangle is (1/2)(π) or π/2. So, the probability of a hit is 1/(π/2) or 2/π. That's approximately .6366197. 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). Resources
90 % of all people would not even try this question, and of those who do, only a select few can boast they have mastered it! Give it a shot yourself! Reply your answer in the comment section!
Post by Jacob Cheek. Calculus Humor does not endorse the content submitted by viewers. The views are only of the original poster of content. Thank you for your understanding!

PollOriginsThis started as a way to express the admins' love of calculus and math in general. As result, this has turned into a gathering place for mathbased humor and weekly challenges. This work by Calculus Humor is licensed under a Creative Commons AttributionNonCommercialShareAlike 3.0 Unported License. Archives
March 2016
Categories
All
