random? chord

Originally posted by PBE6
Argh!! I think you're right. I just tried using a random midpoint selected 2 ways (one with polar variables and one with Cartesian variables), and got different answers for each. The polar one gave me 0.5 and the Cartesian one gave me 0.25. As you say, "uniform randomness" needs some explanation.

Apparently this is a well-known puzzle (unknown by me until ...[text shortened]... is no reason to disregard any particular type of random distribution in this solution?

PBE6 : I'm glad you are investigating the answers that I was too lazy to do!!!

As I said I remember hearing 3 'proofs' which maybe leads to mtthw's cnclusion that there are an infinite number of solutions based upon how you generate your 'random' chord.


btw: I said it would lead to arguments! 😉

Originally posted by wolfgang59
PBE6 : I'm glad you are investigating the answers that I was too lazy to do!!!

As I said I remember hearing 3 'proofs' which maybe leads to mtthw's cnclusion that there are an infinite number of solutions based upon how you generate your 'random' chord.


btw: I said it would lead to arguments! 😉

Even more interesting - if we do a physical experiment with straws, it turns out that the experimental probability is 1/2! Of course, as the article mentions, no one said that the puzzle was meant to be a model of a physical process, but if you're running a gambling ring it's nice to know the physical answer. 😉

http://en.wikipedia.org/wiki/Bertrand%27s_paradox_%28probability%29

Originally posted by PBE6


http://www.cut-the-knot.org/bertrand.shtml

I still feel that the sweep method generates the proper sample space (the analytical solution for the sample space, not randomly generated) because it generates every possible chord drawn from a fixed point on the circumference without double counting, and every rotation of this construction is symmetrical so they do not impact the weighting.

I checked the link and watched the demos.

The sweep method you describe is the only one that gives an unbiased answer imo.

The 2nd demo showed an inherent bias toward drawing lines away from the centre resulting in a 1/4 probability and the 3rd demo showed a bias toward drawing toward the centre resulting in 1/2 probability. For this reason they should be discounted.

The sweep is the only one that would make a true representation without any bias.

Random in the 2nd and 3rd appears to be a biased random generator whereas as the sweep is unbiased.