14 ms·

Bertrand's Paradox

82 points by jcr 11y ago

jcr 11y ago
This will most likely be helpful as well: https://en.wikipedia.org/wiki/Bertrand_paradox_%28probability%29 https://en.wikipedia.org/wiki/Bertrand_paradox_%28probabilit...
- rchowe 11y ago
  I think that the most important part of the article is the following quote in the introduction: > probabilities may not be well defined if the mechanism or method that produces the random variable is not clearly defined. So comparing the answer of 1/2, where the chords are generated by moving along the radius in increments of dr, with the answer of 1/3, where the chords are generated based on the angle from the starting point. The problem doesn't specify how to pick random points, though I suspect that if you were to choose two random points on the radius of a circle by picking a reference "going around the edge" of the circle clockwise by some amount in [0, 2pi), we get the angular situation that produces the 1/3 answer.
  - dvanduzer 11y ago
    The '"maximum ignorance" principle' section from the Wikipedia article addresses most of what you are saying best.
  - trhway 11y ago
    >> probabilities may not be well defined if the mechanism or method that produces the random variable is not clearly defined. i don't agree. Because it is too vague and contradictory. If you defined probabilities it strongly implies that you defined Borel algebra of sets in the space. This is what meant by the "mechanism or method that produces the random variable". Without well defined Borel set (implicitly or explicitly) there is no probability at all, not just "well defined". To illustrate - lets take our example of random chord. One would think - take the set of all possible chords and assign equal probability to each. As the number of chords uncountable, one can't really do such assignment and still obtain an additive measure (one of the condition of something to be a probability). We're forced to go more coarse grained. Find or choose a set of subsets such that we can assign numbers to them such that there would be additivity when the subsets are combined and other properties making that assignment of numbers a "probability", ie. we need to define a Borel algebra. The way one builds such set of subsets, an algebra, is the actual definition of the probability and it is "the mechanism or method that produces the random variable". So, in the Wikipedia article, for example, taking random endpoints means that we're using, as the basis for the algebra, the open sets in the space x S1, while taking random midpoints means that we're building the algebra on the basis of the open sets in the space of the 2 dimensional sphere. It is pretty obvious that these 2 algebras are completely different and thus the probabilities they define are completely different, and one can't talk about probabilities without specifying the Borel algebra defining it.
- gburt 11y ago
  This is much better than the linked site.
transfire 11y ago
I came up with a third -- it seemed fairly obvious after some thought. I do not quite get the 1/2 and 1/4 explanations. (Unfortunately these gif analogies do not do such a great job of clarifying.) So I decided to ask good old "Reality". N = 1000 SQRT3 = Math.sqrt(3).to_r def point_on_circle x = (rand().to_r * 2) - 1 y = Math.sqrt(1 - x**2) y = -y if rand(2) == 0 return [x,y] end tally = 0 N.times do x1, y1 = point_on_circle x2, y2 = point_on_circle d = Math.sqrt( (x2 - x1)**2 + (y2 - y1)**2 ) tally += 1 if d > SQRT3 end puts "#{tally}/#{N}" Oddly, it gives an answer close to a third (yeah me!). But... it is always a bit high at approx 36%.
- x3n0ph3n3 11y ago
  That's because you used "random endpoints" chord selection (which has the probability of 1/3). Now try it with the "random radius" and "random midpoint" methods described here: https://en.wikipedia.org/wiki/Bertrand_paradox_%28probability%29#Bertrand.27s_formulation_of_the_problem https://en.wikipedia.org/wiki/Bertrand_paradox_%28probabilit... It's probably not wise to assume that you just easily solved a 100+ year old unsolved mathematical problem.
  - personjerry 11y ago
    Why do they say "unsolved" if the classical solution seems to deal with the "paradox"?
    - x3n0ph3n3 11y ago
      We have conjectures and approaches but no proven solution.
  - antimagic 11y ago
    I'm thinking though that if you talk a fairly large area surrounding a circle, and then randomly selected two points within it. Draw a line between the two points. If the line intersects the circle, calculate the length the length of it's chord. Bigger then sqrt(3) or less get put into bins, and the probability is calculated from there. Obviously this would work better if the area was an infinite plane, but I would expect to see the answer tending to a value as the size of the area increased. I don't believe that this approach makes any assumptions about the chords that it generates. Here's a ruby snippet that does just that (err, I think!) plane_size = 10000.0 chords = [] while chords.size < 100 p1 = {} p2 = {} p1[:x] = Random.rand(plane_size) - plane_size / 2 p1[:y] = Random.rand(plane_size) - plane_size / 2 p2[:x] = Random.rand(plane_size) - plane_size / 2 p2[:y] = Random.rand(plane_size) - plane_size / 2 puts p1 puts p2 #gradient 'm' m = (p2[:y] - p1[:y]) / (p2[:x] - p1[:x]) #y = mx + b # => b = y - mx b = p2[:y] - m * p2[:x] #x^2 + y^2 = 1 #sub in y from the line equation #x^2 + (mx + b) * (mx + b) = 1 #x^2 + m^2x^2 + 2bmx + b^2 = 1 #x^2(1 + m^2) + 2bmx + b^2 - 1 = 0 #apply the quadratic equation to find the roots #Ax^2 + Bx + C = 0 #=>x = (-B +- sqrt(b^2 - 4AC)) / 2A #in our case, A = (1 + m^2), B = 2bm, C = b^2 -1 #the line intersects if B^2 -4AC > 0 discriminator = (2 * b * m) * (2 * b * m) - 4 * (1 + m * m) * (b * b - 1) if discriminator > 0 then chord1 = {} chord2 = {} chord1[:x] = (-2 * b * m + Math.sqrt(discriminator) ) / (2 * (1 + m * m)) chord1[:y] = m * chord1[:x] + b chord2[:x] = (-2 * b * m - Math.sqrt(discriminator) ) / (2 * (1 + m * m)) chord2[:y] = m * chord2[:x] + b length = Math.sqrt((chord2[:x] - chord1[:x]) * (chord2[:x] - chord1[:x]) + (chord2[:y] - chord1[:y]) * (chord2[:y] - chord1[:y])) chords << length end end smaller = 0 chords.each {|c| smaller += 1 if c < Math.sqrt(3) } puts "The number of chords smaller than sqrt(3) = #{smaller/100.0}%" If I believe the result, the real world seems to feel that the answer is 0.5 Edit: Hah! Should have read just a few comments further, and I would have seen some other posts doing exactly the same thing. Good to see we ended up with the same answer :D
arexsutton 11y ago
The 1/4th answer has a typo. "The probability of being hit is the probability of being inside the area of fire divided by the total area" No. The probability of being hit is the probability of being inside the area of fire. The probability of being inside the area of fire is the area of fire divided by the entire area.
jamesrom 11y ago
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
sopooneo 11y ago
Years ago I posted a question on CraigsList (https://forums.craigslist.org/?ID=20211237 https://forums.craigslist.org/?ID=20211237) Science and Math discussion board, that I think may be nearly equivalent, and equivalently ambiguous in it's notion of random, to this post. It went on for quite some time, and was even reprised later (https://forums.craigslist.org/?areaID=4&ID=20432604 https://forums.craigslist.org/?areaID=4&ID=20432604) by someone else. Helpful people went back and forth with beautiful 3d renders to illustrate their solutions, though those links are sadly broken now. It is very gratifying to read this post, as I think it explains why we were never able to come to a satisfying answer.
deleted 11y ago
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dstyrb 11y ago
I don't agree with any of these solutions. In my opinion the circle should be positioned in an infinite plane, then random vectors should be drawn on the plane and ignored unless they intersect the circle. Then after making some large amount of intersecting vectors you calculate. I'm guessing that this essentially reduces to the random endpoints method. The other two methods don't make sense at all to me. Random chords is just obviously not a random sampling of possible vectors-- it's essentially the same as sampling a gaussian(x) distribution using an equal sampling in x and ignoring the fact that a gaussian is a PDF and you can't sample that way, you need to invert it into a CDF and take an equal sampling in _y_. Very basic I think... And random midpoint doesn't strike me as random either as it forces the orientations of the vectors to be non-random and systematically smaller than random because of the pi*r^2 area-dependence of concentric circles... But, what do I know.
- airza 11y ago
  i think that any probability measure you can define on your proposed solution assigns a probability of zero to actually hitting the circle, which might complicate things somewhat :)
  - dstyrb 11y ago
    It would assign a probability approaching zero as the plane approached infinity, yeah, I agree. But I think I have a way around that.
- arexsutton 11y ago
  I tried to script that together after the same idea occurred to me. What's interesting is the selection plane affects the outcome quite heavily. As the plane of random line selection increases in size, the probability of a random chord having length>sqrt3 seems to approach about 0.13 My math might be wrong; doesn't seem to be a floating point precision issue. from math import sqrt from random import random def randomPoints(width): return [ [random()*width-width/2, random()*width-width/2], [random()*width-width/2, random()*width-width/2] ] def chordCalc(points): def quadraticEquation(a,b,c): descriminant=(b**2 - 4*a*c) if descriminant <= 0: return None x1 = (-b+sqrt(descriminant))/(2*a) x2 = (-b-sqrt(descriminant))/(2*a) y1 = sqrt(1-x1**2) y2 = sqrt(1-x2**2) return [[x1, y1],[x2,y2]] m = (points[0][1] - points[1][1]) / (points[0][0] - points[1][0]) b = points[0][1] - m*points[0][0] qa = 1+m**2 qb = 2*b*m qc = b**2-1 return quadraticEquation(qa, qb, qc) def distance(points): return sqrt((points[0][1] - points[1][1])**2 + (points[0][0] - points[1][0])**2) chords=0 largeChords=0 sqrt3=sqrt(3) while True: if random() > 0.9999: print float(largeChords)/chords points = randomPoints(500) #adjust me down to 0.5 and see probability go to ~0.33 intersections=chordCalc(points) if intersections is None: continue chords += 1 if distance(intersections) > sqrt3: largeChords += 1
murbard2 11y ago
For extra credit compute the probability under Jeffrey's prior.
amalcon 11y ago
One could just as easily make an argument for a ratio of sqrt(3)/2. The chord is selected by randomly picking a length out of the range [0, 2], placing it arbitrarily on the circle, and then rotating it randomly between zero and one revolution[1]. Before we can select a random chord, we must first map the set of chords onto the set of real numbers (or, equivalently, tuples of real numbers). This is required in order to meaningfully apply a distribution to them. The problem is that there are multiple ways to do this. Which mapping we select has as much effect on the outcome as the probability distribution. Likewise, if we want to "count" all random chords, we need to do the same thing (map them onto the set of real numbers). We will get different "counts" depending on which mapping we select. [1]- Edit: This is not a very satisfying approach, but it does demonstrate the problem well. Being satisfying isn't a requirement for a mathematical construct.
andrew-lucker 11y ago
There should be infinitely many "solutions" to this if you are allowed to define the sampling method arbitrarily. Intuitively, one would think that random points are uniformly distributed, but that is not maintained in two of these solutions. If you randomly generate two points in [0,2pi) with a uniform distribution then you will quickly get the 1/3 probability empirically or provably by the limit. However, if you are allowed to derive the second point from the first (such as the cross section example) or create some other dependency, then you are given great liberty as to what the resulting "probability" will be. Sure you can even guarantee that every two points on the circle will be present "eventually" but infinity is very large and how you approach it will determine your eventual outcome. Here we could subsample preferred points first and actually derive any specific probability (0,1) and maybe even make it sound reasonable. This behaviour will even be maintained into the limit, but it is certainly not intuitive, at least to me.
- xyzzyz 11y ago
  The other two solutions are also results of generating points from uniform distribution: radius and point on radius result from uniformly drawing pair from [0, 2pi) x [0, 1) (radius, distance from center), while "random midpoint" draws from the uniform distribution on the open unit disc. There's no "dependency" at the drawing time. However, if one remembers the substitution formula for multiple variable integrals, this hints why the Jacobian multiplier is necessary in it: all three underlying spaces mentioned ([0,2pi)^2, [0,2pi)x[0,1], B^2\boundary) are diffeomorphic. They clearly don't have the same measure though -- the first one measures 4pi^2, second 2pi, and third pi. However, one can obtain one value from another by integrating Jacobian of appropriate diffeomorphism.
  - andrew-lucker 11y ago
    On most discrete samples for the 1/4 or 1/2 case, the probability of choosing any two points in the subset may be uniform, but the location of the permitted points is definitely not. If you add the requirement that all discrete point samples should be equidistant, then you fix yourself on 1/3 and there is really no way to break that conclusion. Sorry, in this case "uniform" is ambiguous and I didn't mean the usual definition related to probability, but rather to point spacing.
ihm 11y ago
I got 1/3, but with a "different" distribution. (I took symmetrized pairs rather than pairs. But I realized after I wrote this comment that folding a square in half preserves relative area so symmetrizing makes no difference.) I started off with a flat torus (two ordered points on the circle) and then quotiented out by (x, y) ~ (y, x) (since the chord corresponding to both is the same) to get a triangle as in the diagram [0]. By geometric intuition/trigonometry/etc it's clear that a chord has length < sqrt(3) iff the difference between the angles x and y is less than 2pi/3. The region corresponding to pairs with distance distance less than 2pi/3 is shaded in [0]. One adds it up and sees that the unshaded region is 1/3 of the whole. [0]: http://i.imgur.com/ow8Dkgd.png http://i.imgur.com/ow8Dkgd.png
reilly3000 11y ago
The average of the provable probabilities is the resultant probability. About 36%.
agnes49 11y ago
The Bertrand paradox is a problem within the classical interpretation of probability theory. http://www.trainingintambaram.in/web-designing-training-in-chennai.html http://www.trainingintambaram.in/web-designing-training-in-c...
aaron695 11y ago
You and another incredibly smart human have to chose an answer (of the three) in secret. You differ you die. Which one do you chose? To me this is the correct answer. I don't think the other two would even be in consideration. I also think aliens would agree, although I'm not sure that would be fair since the original question (on this website) is being asked to me, a human. [edit] Interesting this is also not the exact Bertrand paradox since they perhaps have added units to it to get around Edwin Jaynes solution *See Wiki
lordnacho 11y ago
IMO the most natural interpretation of "random" with no qualifiers is the one that gives you 1/3. Pick a point, pick another point, a third of the time the second point in within the far arc of the triangle. I'm not sure the ninja method gives you what I'd call a representative sample of possible chords. Looks like he only cuts vertically, so he can never give you a skew-slice. Similar with the dragon.
- chadzawistowski 11y ago
  It doesn't matter if he only cuts vertically in the animation. The pizza could be on a spinning wheel, and as long as it's a perfect circle, that wouldn't matter to the ninja's cuts. User antimagic thought it was most natural to draw lines by picking random (x,y) coordinates on a plane, then overlaying a circle. This "natural" choice corresponds to selection method #2, so they ended up getting the 1/2 answer.
deleted 11y ago
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flowerjeni 11y ago
It seems every methodology relies on upon the way arbitrary harmonies are chosen. Every methodology yields diverse probabilities of selecting a harmony with a certain property. http://www.trainingintambaram.in/ccna-training-in-chennai.html# http://www.trainingintambaram.in/ccna-training-in-chennai.ht...
deleted 11y ago
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82 points by jcr 11y ago

jcr 11y ago

This will most likely be helpful as well: https://en.wikipedia.org/wiki/Bertrand_paradox_%28probability%29 https://en.wikipedia.org/wiki/Bertrand_paradox_%28probabilit...

rchowe 11y ago
I think that the most important part of the article is the following quote in the introduction: > probabilities may not be well defined if the mechanism or method that produces the random variable is not clearly defined. So comparing the answer of 1/2, where the chords are generated by moving along the radius in increments of dr, with the answer of 1/3, where the chords are generated based on the angle from the starting point. The problem doesn't specify how to pick random points, though I suspect that if you were to choose two random points on the radius of a circle by picking a reference "going around the edge" of the circle clockwise by some amount in [0, 2pi), we get the angular situation that produces the 1/3 answer.
- dvanduzer 11y ago
  The '"maximum ignorance" principle' section from the Wikipedia article addresses most of what you are saying best.
- trhway 11y ago
  >> probabilities may not be well defined if the mechanism or method that produces the random variable is not clearly defined. i don't agree. Because it is too vague and contradictory. If you defined probabilities it strongly implies that you defined Borel algebra of sets in the space. This is what meant by the "mechanism or method that produces the random variable". Without well defined Borel set (implicitly or explicitly) there is no probability at all, not just "well defined". To illustrate - lets take our example of random chord. One would think - take the set of all possible chords and assign equal probability to each. As the number of chords uncountable, one can't really do such assignment and still obtain an additive measure (one of the condition of something to be a probability). We're forced to go more coarse grained. Find or choose a set of subsets such that we can assign numbers to them such that there would be additivity when the subsets are combined and other properties making that assignment of numbers a "probability", ie. we need to define a Borel algebra. The way one builds such set of subsets, an algebra, is the actual definition of the probability and it is "the mechanism or method that produces the random variable". So, in the Wikipedia article, for example, taking random endpoints means that we're using, as the basis for the algebra, the open sets in the space x S1, while taking random midpoints means that we're building the algebra on the basis of the open sets in the space of the 2 dimensional sphere. It is pretty obvious that these 2 algebras are completely different and thus the probabilities they define are completely different, and one can't talk about probabilities without specifying the Borel algebra defining it.
gburt 11y ago
This is much better than the linked site.

transfire 11y ago

I came up with a third -- it seemed fairly obvious after some thought. I do not quite get the 1/2 and 1/4 explanations. (Unfortunately these gif analogies do not do such a great job of clarifying.) So I decided to ask good old "Reality". N = 1000 SQRT3 = Math.sqrt(3).to_r def point_on_circle x = (rand().to_r * 2) - 1 y = Math.sqrt(1 - x**2) y = -y if rand(2) == 0 return [x,y] end tally = 0 N.times do x1, y1 = point_on_circle x2, y2 = point_on_circle d = Math.sqrt( (x2 - x1)**2 + (y2 - y1)**2 ) tally += 1 if d > SQRT3 end puts "#{tally}/#{N}" Oddly, it gives an answer close to a third (yeah me!). But... it is always a bit high at approx 36%.

x3n0ph3n3 11y ago
That's because you used "random endpoints" chord selection (which has the probability of 1/3). Now try it with the "random radius" and "random midpoint" methods described here: https://en.wikipedia.org/wiki/Bertrand_paradox_%28probability%29#Bertrand.27s_formulation_of_the_problem https://en.wikipedia.org/wiki/Bertrand_paradox_%28probabilit... It's probably not wise to assume that you just easily solved a 100+ year old unsolved mathematical problem.
- personjerry 11y ago
  Why do they say "unsolved" if the classical solution seems to deal with the "paradox"?
  - x3n0ph3n3 11y ago
    We have conjectures and approaches but no proven solution.
- antimagic 11y ago
  I'm thinking though that if you talk a fairly large area surrounding a circle, and then randomly selected two points within it. Draw a line between the two points. If the line intersects the circle, calculate the length the length of it's chord. Bigger then sqrt(3) or less get put into bins, and the probability is calculated from there. Obviously this would work better if the area was an infinite plane, but I would expect to see the answer tending to a value as the size of the area increased. I don't believe that this approach makes any assumptions about the chords that it generates. Here's a ruby snippet that does just that (err, I think!) plane_size = 10000.0 chords = [] while chords.size < 100 p1 = {} p2 = {} p1[:x] = Random.rand(plane_size) - plane_size / 2 p1[:y] = Random.rand(plane_size) - plane_size / 2 p2[:x] = Random.rand(plane_size) - plane_size / 2 p2[:y] = Random.rand(plane_size) - plane_size / 2 puts p1 puts p2 #gradient 'm' m = (p2[:y] - p1[:y]) / (p2[:x] - p1[:x]) #y = mx + b # => b = y - mx b = p2[:y] - m * p2[:x] #x^2 + y^2 = 1 #sub in y from the line equation #x^2 + (mx + b) * (mx + b) = 1 #x^2 + m^2x^2 + 2bmx + b^2 = 1 #x^2(1 + m^2) + 2bmx + b^2 - 1 = 0 #apply the quadratic equation to find the roots #Ax^2 + Bx + C = 0 #=>x = (-B +- sqrt(b^2 - 4AC)) / 2A #in our case, A = (1 + m^2), B = 2bm, C = b^2 -1 #the line intersects if B^2 -4AC > 0 discriminator = (2 * b * m) * (2 * b * m) - 4 * (1 + m * m) * (b * b - 1) if discriminator > 0 then chord1 = {} chord2 = {} chord1[:x] = (-2 * b * m + Math.sqrt(discriminator) ) / (2 * (1 + m * m)) chord1[:y] = m * chord1[:x] + b chord2[:x] = (-2 * b * m - Math.sqrt(discriminator) ) / (2 * (1 + m * m)) chord2[:y] = m * chord2[:x] + b length = Math.sqrt((chord2[:x] - chord1[:x]) * (chord2[:x] - chord1[:x]) + (chord2[:y] - chord1[:y]) * (chord2[:y] - chord1[:y])) chords << length end end smaller = 0 chords.each {|c| smaller += 1 if c < Math.sqrt(3) } puts "The number of chords smaller than sqrt(3) = #{smaller/100.0}%" If I believe the result, the real world seems to feel that the answer is 0.5 Edit: Hah! Should have read just a few comments further, and I would have seen some other posts doing exactly the same thing. Good to see we ended up with the same answer :D

arexsutton 11y ago

The 1/4th answer has a typo. "The probability of being hit is the probability of being inside the area of fire divided by the total area" No. The probability of being hit is the probability of being inside the area of fire. The probability of being inside the area of fire is the area of fire divided by the entire area.

jamesrom 11y ago

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

sopooneo 11y ago

Years ago I posted a question on CraigsList (https://forums.craigslist.org/?ID=20211237 https://forums.craigslist.org/?ID=20211237) Science and Math discussion board, that I think may be nearly equivalent, and equivalently ambiguous in it's notion of random, to this post. It went on for quite some time, and was even reprised later (https://forums.craigslist.org/?areaID=4&ID=20432604 https://forums.craigslist.org/?areaID=4&ID=20432604) by someone else. Helpful people went back and forth with beautiful 3d renders to illustrate their solutions, though those links are sadly broken now. It is very gratifying to read this post, as I think it explains why we were never able to come to a satisfying answer.

deleted 11y ago

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dstyrb 11y ago

I don't agree with any of these solutions. In my opinion the circle should be positioned in an infinite plane, then random vectors should be drawn on the plane and ignored unless they intersect the circle. Then after making some large amount of intersecting vectors you calculate. I'm guessing that this essentially reduces to the random endpoints method. The other two methods don't make sense at all to me. Random chords is just obviously not a random sampling of possible vectors-- it's essentially the same as sampling a gaussian(x) distribution using an equal sampling in x and ignoring the fact that a gaussian is a PDF and you can't sample that way, you need to invert it into a CDF and take an equal sampling in _y_. Very basic I think... And random midpoint doesn't strike me as random either as it forces the orientations of the vectors to be non-random and systematically smaller than random because of the pi*r^2 area-dependence of concentric circles... But, what do I know.

airza 11y ago
i think that any probability measure you can define on your proposed solution assigns a probability of zero to actually hitting the circle, which might complicate things somewhat :)
- dstyrb 11y ago
  It would assign a probability approaching zero as the plane approached infinity, yeah, I agree. But I think I have a way around that.
arexsutton 11y ago
I tried to script that together after the same idea occurred to me. What's interesting is the selection plane affects the outcome quite heavily. As the plane of random line selection increases in size, the probability of a random chord having length>sqrt3 seems to approach about 0.13 My math might be wrong; doesn't seem to be a floating point precision issue. from math import sqrt from random import random def randomPoints(width): return [ [random()*width-width/2, random()*width-width/2], [random()*width-width/2, random()*width-width/2] ] def chordCalc(points): def quadraticEquation(a,b,c): descriminant=(b**2 - 4*a*c) if descriminant <= 0: return None x1 = (-b+sqrt(descriminant))/(2*a) x2 = (-b-sqrt(descriminant))/(2*a) y1 = sqrt(1-x1**2) y2 = sqrt(1-x2**2) return [[x1, y1],[x2,y2]] m = (points[0][1] - points[1][1]) / (points[0][0] - points[1][0]) b = points[0][1] - m*points[0][0] qa = 1+m**2 qb = 2*b*m qc = b**2-1 return quadraticEquation(qa, qb, qc) def distance(points): return sqrt((points[0][1] - points[1][1])**2 + (points[0][0] - points[1][0])**2) chords=0 largeChords=0 sqrt3=sqrt(3) while True: if random() > 0.9999: print float(largeChords)/chords points = randomPoints(500) #adjust me down to 0.5 and see probability go to ~0.33 intersections=chordCalc(points) if intersections is None: continue chords += 1 if distance(intersections) > sqrt3: largeChords += 1

murbard2 11y ago

For extra credit compute the probability under Jeffrey's prior.

amalcon 11y ago

One could just as easily make an argument for a ratio of sqrt(3)/2. The chord is selected by randomly picking a length out of the range [0, 2], placing it arbitrarily on the circle, and then rotating it randomly between zero and one revolution[1]. Before we can select a random chord, we must first map the set of chords onto the set of real numbers (or, equivalently, tuples of real numbers). This is required in order to meaningfully apply a distribution to them. The problem is that there are multiple ways to do this. Which mapping we select has as much effect on the outcome as the probability distribution. Likewise, if we want to "count" all random chords, we need to do the same thing (map them onto the set of real numbers). We will get different "counts" depending on which mapping we select. [1]- Edit: This is not a very satisfying approach, but it does demonstrate the problem well. Being satisfying isn't a requirement for a mathematical construct.

andrew-lucker 11y ago

There should be infinitely many "solutions" to this if you are allowed to define the sampling method arbitrarily. Intuitively, one would think that random points are uniformly distributed, but that is not maintained in two of these solutions. If you randomly generate two points in [0,2pi) with a uniform distribution then you will quickly get the 1/3 probability empirically or provably by the limit. However, if you are allowed to derive the second point from the first (such as the cross section example) or create some other dependency, then you are given great liberty as to what the resulting "probability" will be. Sure you can even guarantee that every two points on the circle will be present "eventually" but infinity is very large and how you approach it will determine your eventual outcome. Here we could subsample preferred points first and actually derive any specific probability (0,1) and maybe even make it sound reasonable. This behaviour will even be maintained into the limit, but it is certainly not intuitive, at least to me.

xyzzyz 11y ago
The other two solutions are also results of generating points from uniform distribution: radius and point on radius result from uniformly drawing pair from [0, 2pi) x [0, 1) (radius, distance from center), while "random midpoint" draws from the uniform distribution on the open unit disc. There's no "dependency" at the drawing time. However, if one remembers the substitution formula for multiple variable integrals, this hints why the Jacobian multiplier is necessary in it: all three underlying spaces mentioned ([0,2pi)^2, [0,2pi)x[0,1], B^2\boundary) are diffeomorphic. They clearly don't have the same measure though -- the first one measures 4pi^2, second 2pi, and third pi. However, one can obtain one value from another by integrating Jacobian of appropriate diffeomorphism.
- andrew-lucker 11y ago
  On most discrete samples for the 1/4 or 1/2 case, the probability of choosing any two points in the subset may be uniform, but the location of the permitted points is definitely not. If you add the requirement that all discrete point samples should be equidistant, then you fix yourself on 1/3 and there is really no way to break that conclusion. Sorry, in this case "uniform" is ambiguous and I didn't mean the usual definition related to probability, but rather to point spacing.

ihm 11y ago

I got 1/3, but with a "different" distribution. (I took symmetrized pairs rather than pairs. But I realized after I wrote this comment that folding a square in half preserves relative area so symmetrizing makes no difference.) I started off with a flat torus (two ordered points on the circle) and then quotiented out by (x, y) ~ (y, x) (since the chord corresponding to both is the same) to get a triangle as in the diagram [0]. By geometric intuition/trigonometry/etc it's clear that a chord has length < sqrt(3) iff the difference between the angles x and y is less than 2pi/3. The region corresponding to pairs with distance distance less than 2pi/3 is shaded in [0]. One adds it up and sees that the unshaded region is 1/3 of the whole. [0]: http://i.imgur.com/ow8Dkgd.png http://i.imgur.com/ow8Dkgd.png

reilly3000 11y ago

The average of the provable probabilities is the resultant probability. About 36%.

agnes49 11y ago

The Bertrand paradox is a problem within the classical interpretation of probability theory. http://www.trainingintambaram.in/web-designing-training-in-chennai.html http://www.trainingintambaram.in/web-designing-training-in-c...

aaron695 11y ago

You and another incredibly smart human have to chose an answer (of the three) in secret. You differ you die. Which one do you chose? To me this is the correct answer. I don't think the other two would even be in consideration. I also think aliens would agree, although I'm not sure that would be fair since the original question (on this website) is being asked to me, a human. [edit] Interesting this is also not the exact Bertrand paradox since they perhaps have added units to it to get around Edwin Jaynes solution *See Wiki

lordnacho 11y ago

IMO the most natural interpretation of "random" with no qualifiers is the one that gives you 1/3. Pick a point, pick another point, a third of the time the second point in within the far arc of the triangle. I'm not sure the ninja method gives you what I'd call a representative sample of possible chords. Looks like he only cuts vertically, so he can never give you a skew-slice. Similar with the dragon.

chadzawistowski 11y ago
It doesn't matter if he only cuts vertically in the animation. The pizza could be on a spinning wheel, and as long as it's a perfect circle, that wouldn't matter to the ninja's cuts. User antimagic thought it was most natural to draw lines by picking random (x,y) coordinates on a plane, then overlaying a circle. This "natural" choice corresponds to selection method #2, so they ended up getting the 1/2 answer.

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flowerjeni 11y ago

It seems every methodology relies on upon the way arbitrary harmonies are chosen. Every methodology yields diverse probabilities of selecting a harmony with a certain property. http://www.trainingintambaram.in/ccna-training-in-chennai.html# http://www.trainingintambaram.in/ccna-training-in-chennai.ht...

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