Hacker News Reader

Comment by BrandonY | Hacker News Reader

Maybe I'm missing something, but this doesn't seem so hard. Every cord must start at some point on the circle and end at some point on the circle. If P is the set of all points on the circle, then the set of all cords is the set of all possibilities of (p1, p2), where p1 and p2 are in P.

The ninja story doesn't work here, since for any p1 I think there's exactly one p2 possible. The photographer story DOES work, since he can start at any point, and the mermaid can then be at any second point. I'm not sure about the dragon story, but I'm guessing it doesn't work either.

View on HN

What about lines across a square? They must also all hit two points on its perimeter, but randomly choosing p1, p2 in that case will show a clear bias towards the diagonals.

Of course a circle has no corners to complicate things like that, but it can still have biases. Just because we have found a way to uniquely identify cords does not mean it will lead to a uniform distribution.

> Just because we have found a way to uniquely identify cords does not mean it will lead to a uniform distribution.

This is the sentence that helped me understand the paradox.

Perhaps this analogy will help. Think about how technically picking a value at random from a normal distribution can result in any value at all, extending to infinity in both directions, but that doesn't mean all values are equally likely. Every real number can be uniquely expressed as a number of standard deviations from mean in a normal distribution, but that still has a bias.

> If P is the set of all points on the circle, then the set of all cords is the set of all possibilities of (p1, p2), where p1 and p2 are in P.

So you are in favour of method 2.

But method 3 says: Take a random point inside the circle, and use this as a center point of the cord.

What do you find wrong with this method?

All you need to add to make the ninja story work is to randomly rotate the pizza after it is cut (which clearly has no effect on the size of the resulting piece). It is then possible to select any chord of the circle -- though the probability density would be different than in the other two methods.

The probability density would not be different.

Rotating doesn't alter the probability density of the length function, and I suppose that's one reading of what I wrote, but it's not what I meant.

This method has different probability density from the random-center and random-endpoints methods. That's the whole point of the exercise: you get different probability density across the chord space depending on how you select from it.

The ninja story takes into account the symmetries of the circle, so it basically pre-divides your result.

Reply on news.ycombinator.com