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It appears each approach depends on the way random chords are selected. Each approach yields different probabilities of selecting a chord with a certain propert

It appears each approach depends on the way random chords are selected. Each approach yields different probabilities of selecting a chord with a certain property. But my intuition suggests that constructing a proof for all chords should be relatively simple, and indeed 1/3 of them hold the property that they are longer than √3.

sopooneo 11y ago
You may be right, but how do you define all cords?
- BrandonY 11y ago
  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.
  - amalcon 11y ago
    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.
    - dnautics 11y ago
      The probability density would not be different.
      - amalcon 11y ago
        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.
  - sopooneo 11y ago
    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.
    - DanielStraight 11y ago
      > 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.
  - Retra 11y ago
    The ninja story takes into account the symmetries of the circle, so it basically pre-divides your result.
  - sampo 11y ago
    > 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?
kmill 11y ago
When you say that 1/3 of them have a certain property, how are you counting that? Are you enumerating all chords with and without the property, then dividing? :-) This is exactly the issue: when dealing with infinities and probabilities, you need to choose a measure. Depending on how that measure is chosen, you might get different answers.
analog31 11y ago
I would put it in this way: Merely saying "random" is insufficient information. To solve the problem, you have to know from what distribution chords are drawn. It could be equal probability per unit angle, per unit distance along the circle, or something else altogether. Similar difficulties emerge when someone gives you a problem that starts with a "random real number."
- x3n0ph3n3 11y ago
  Exactly, most of the real numbers have an infinite number of digits!
  - JadeNB 11y ago
    All real numbers have an infinite number of digits. For the numbers that you probably meant to exclude, all but finitely many of those infinitely many digits are 0. (This is almost just nitpicking, but in another sense it's important: viewing an apparently terminating decimal expansion as actually 0-padded lets one say correctly that "a real number is rational if and only if its decimal expansion eventually repeats", full stop, rather than having to add the qualification "… or is eventually 0.")
    - x3n0ph3n3 11y ago
      Padding doesn't count...
      - JadeNB 11y ago
        Well, to be sure, one can say that it does count or that it doesn't—the axioms of mathematics are only as humans make them. I was arguing why it might make sense to have it count. Another reason that it might be a good reason to have it count: by my definition, I can say "the decimal expansion of a number eventually repeats if and only if the binary expansion of that number eventually repeats." This becomes considerably more awkward if you don't consider a terminating expansion as consisting of eventually repeating 0's; for example, 1/5 has a terminating decimal expansion 0.2 but an immediately-repeating binary expansion 0.repeat(0011).
sampo 11y ago
A point can by described by two numbers (the coordinates). If we are selecting a "random point in a square", everybody's intuition agrees that this means selecting the x coordinate and the y coordinate from a flat distribution. In a similar manner, a chord can by described by a point and a direction. – The first (1/2) method selects a point from a flat distribution along x-axis, and defines that the direction is always vertical. (Edit: this is how I interpreted the animated gif, but Wikipedia says that actually from a flat distribution along a random radius, and then the direction is perpendicular to the radius.) – The second (1/3) method selects a point from a flat distribution on the perimeter, and then actually selects another point also from the same distribution, to define the direction by connecting these points. – The third (1/4) method selects a point from a flat distribution on the circle area, and the defines that the direction is always perpendicular to the radius. The whole point of the paradox is to demonstrate that there is no single universally agreed upon way to define a flat distribution for the set of all chords. And when we start to try to define one, by breaking the definition of a cord into components, and sampling these components from a flat distribution (or just selecting one direction, because symmetry says they are all equal), we get different results depending on how we broke the concept of cord into components.
- phkahler 11y ago
  Interestingly, if you use the second method and chose a point somewhere on the circle, and then select a random direction (instead of another point on the circle) you'll get the same 1/3 answer.

sopooneo 11y ago

You may be right, but how do you define all cords?

BrandonY 11y ago
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.
- amalcon 11y ago
  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.
  - dnautics 11y ago
    The probability density would not be different.
    - amalcon 11y ago
      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.
- sopooneo 11y ago
  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.
  - DanielStraight 11y ago
    > 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.
- Retra 11y ago
  The ninja story takes into account the symmetries of the circle, so it basically pre-divides your result.
- sampo 11y ago
  > 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?

kmill 11y ago

When you say that 1/3 of them have a certain property, how are you counting that? Are you enumerating all chords with and without the property, then dividing? :-) This is exactly the issue: when dealing with infinities and probabilities, you need to choose a measure. Depending on how that measure is chosen, you might get different answers.

analog31 11y ago

I would put it in this way: Merely saying "random" is insufficient information. To solve the problem, you have to know from what distribution chords are drawn. It could be equal probability per unit angle, per unit distance along the circle, or something else altogether. Similar difficulties emerge when someone gives you a problem that starts with a "random real number."

x3n0ph3n3 11y ago
Exactly, most of the real numbers have an infinite number of digits!
- JadeNB 11y ago
  All real numbers have an infinite number of digits. For the numbers that you probably meant to exclude, all but finitely many of those infinitely many digits are 0. (This is almost just nitpicking, but in another sense it's important: viewing an apparently terminating decimal expansion as actually 0-padded lets one say correctly that "a real number is rational if and only if its decimal expansion eventually repeats", full stop, rather than having to add the qualification "… or is eventually 0.")
  - x3n0ph3n3 11y ago
    Padding doesn't count...
    - JadeNB 11y ago
      Well, to be sure, one can say that it does count or that it doesn't—the axioms of mathematics are only as humans make them. I was arguing why it might make sense to have it count. Another reason that it might be a good reason to have it count: by my definition, I can say "the decimal expansion of a number eventually repeats if and only if the binary expansion of that number eventually repeats." This becomes considerably more awkward if you don't consider a terminating expansion as consisting of eventually repeating 0's; for example, 1/5 has a terminating decimal expansion 0.2 but an immediately-repeating binary expansion 0.repeat(0011).

sampo 11y ago

A point can by described by two numbers (the coordinates). If we are selecting a "random point in a square", everybody's intuition agrees that this means selecting the x coordinate and the y coordinate from a flat distribution. In a similar manner, a chord can by described by a point and a direction. – The first (1/2) method selects a point from a flat distribution along x-axis, and defines that the direction is always vertical. (Edit: this is how I interpreted the animated gif, but Wikipedia says that actually from a flat distribution along a random radius, and then the direction is perpendicular to the radius.) – The second (1/3) method selects a point from a flat distribution on the perimeter, and then actually selects another point also from the same distribution, to define the direction by connecting these points. – The third (1/4) method selects a point from a flat distribution on the circle area, and the defines that the direction is always perpendicular to the radius. The whole point of the paradox is to demonstrate that there is no single universally agreed upon way to define a flat distribution for the set of all chords. And when we start to try to define one, by breaking the definition of a cord into components, and sampling these components from a flat distribution (or just selecting one direction, because symmetry says they are all equal), we get different results depending on how we broke the concept of cord into components.

phkahler 11y ago
Interestingly, if you use the second method and chose a point somewhere on the circle, and then select a random direction (instead of another point on the circle) you'll get the same 1/3 answer.