Circle Puzzle Answers

This puzzle ended up being much more popular that I anticipated. Aside from the comments on the blog, some coworkers and other friends that don't post got into it. The name of the puzzle is Bertrand's Paradox.

rpc did some of the math wrong, but did point out another solution. If you randomly pick the cord and then orient the circle so that the cord is vertical, then all cords between x=-1/2 and x=1/2 are longer then sqrt(3). Therefor, the probability is 1/2.

The third classic answer is that the midpoint of the cord uniquely defines the cord. So, we randomly pick a point inside the circle and construct a cord so that that point is the midpoint of that cord. Now, any point we pick inside the circle of radius 1/2 is longer than sqrt(3), and outside is shorter. Comparing the inside area of the circle to the area of the whole circle gives a probability of 1/4.

Now rpc did the most work on this, and some of his ideas were on track to a cool outcome. He and I did some discussions at climbing on Tuesday, and he cleaned up the reasoning so that we can now cover all probabilities on the interval (0,1). It's not [0,1], just (0,1) because I wouldn't let him do constructions at infinity. I've modified it a bit from what rpc came up with. He used two constructions to cover (0,1/2) and (1/2,1), and then we filled in 1/2 with the previously mentioned case. I figured out a way to do a single construction to cover (0,1).

Start with the vertical line at x=-1. Pick a point at (n,0) where n>1 (but finite). Draw a line tangent to the circle from (n,0) to (-1,y); y>0. Randomly pick a cord, and then orient the circle so that one end of the cord lies at the point (-1,0) and the other point is oriented so that its y coordinate is positive. Now create a 1-to-1 and onto mapping (bijective) of every cord to a point on the tangent line (n,0);(-1,y) via the cord's projection. Because it is bijective, it is equivalent to randomly pick a point on the tangent line and look at the probability that the associated cord is longer than sqrt(3).

If n is very close to 1, then y is very large, and most points on the tangent line are associated with cords of length less than sqrt(3). If n is very large, the y is very nearly 1, and most of the cords will have length more than sqrt(3). Given any probability desired over the inteval (0,1), a tangent line can be constructed that gives a bijective map with that probability.

It is not an ill defined problem. The problem is that there are an uncountably infinite number of cords on the circle. I can create an uncoutnably infinite set of mappings of cords to the interval [0,1] such that I can get the probability to be whatever I want.

I don't let n be equal to 1 or infinity, because then I can't map the cords to the inteval [0,1], which seems necessary to get any sense of the problem (which fraction of the interval maps cords of length more than sqrt(3)). But since n>1, and n is finite, I get the full range of probabilities from (0,1).

Super duper extra credit to rpc for inspiring this solution.