This is the second of three Bertrand's Chord Paradoxes. Set the Picks slider to 1 and then click on the green flag to see how the random chord is selected. Do this several times before running 1000 picks. In this project, the chord is randomly selected using a method different from that used in the project Bertrand's Random Chord Paradox 1 ( http://scratch.mit.edu/projects/20157686/ ). In Bertrand's Paradox 1, the probability that the chord length is greater than the length of the side of the triangle can be shown to be, using basic geometry, 1/3. Run the Monte Carlo simulation for 1000 picks and you should be able to form a conjecture as to what value the probability of the chord being greater in length than the side is for Paradox 2. Compare to Paradox 1.
This project first draws a random radius. Next, a random point is picked along the radius. A chord through this point, perpendicular to the radius is then drawn. The length of this chord is then compared top the length of the side of the equilateral triangle. If the length of this chord is greater than the length of the side of the triangle, then a counter is incremented by 1. Even though Bertrand's Paradox 1 and Paradox 2 are the same problem, that is, draw a random chord in a circle and compare its length to the length of a side of an inscribed equilateral triangle, be prepared to find two different, CORRECT, answers for the same problem. That's why the problem is a paradox!
