This is the third 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/ ) and Bertrand's Random Chord Paradox 2 (http://scratch.mit.edu/projects/20392511/) 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. In Paradox 2, the probability is 1/2. 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 3. Compare to Paradox 1 and 2.
This project first picks a random point in the circle. This point is taken as the midpoint of the chord through that point. The chord is drawn and then determined to be either longer or shorter than the side of the 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, Paradox 2, and Paradox 3 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 tree different, CORRECT, answers for the same problem. That's why the problem is a paradox!
