Talk:Bertrand paradox (probability)

Ramifications of the Paradox

Can someone in-the-know please talk about the ramifications for probability theory of the paradox? This is more important than having a lot of nice-looking images illustrating the distributions. I am not familiar enough with deep probability theory to write this. Thanks - JustinWick 18:37, 9 February 2006 (UTC)

Good idea! What happened was that this and other paradoxes of the classical interpretation of probability forced a more stringent approach to probability, frequency probability, which is the current scientific view. I'll try to write a note about that shortly. iNic 13:51, 11 December 2006 (UTC)

You might look at the Stanford Encyclopedia of philosophy entry "interpretations of probability" for starters. Many proponents of the "subjective" interpretation of probability as rational degree of belief (or "credence") see the Bertrand Paradox as one motive for having a theory of imprecise credences. The paradox arises because applications of the principle of indifference to different parameterizations of the problem yield inconsistent results. The wiki is very, very helpful in showing the problem as originally posed and very, very nicely illustrates Jaynes' reply. However, the transformation invariance response given by Jaynes doesn't seem to generalize, at least according to Howson and Urbach in their "Scientific Reasoning: The Bayesian Approach" (2005) (see pages in the 270s/280s). Also, the wiki is not very helpful in showing the generality of the problem of relativity to parameterization for applications of the principle of indifference. Maybe a discussion of van Fraasen's "cube factory" could be added to show how the question about parameterization generalizes. So the short answer is that if you interpret probability as rational degree of belief then the Bertrand paradox might motivate you to give up trying to use the principle of indifference to fix precise prior probabilities and instead motivate you to adopt a theory of imprecise credences. Hope that helps some.

Van Fraasen's Cube Factory: The following example (adapted from van Fraassen 1989) nicely illustrates how Bertrand-style paradoxes work. A factory produces cubes with side-length between 0 and 1 foot; what is the probability that a randomly chosen cube has side-length between 0 and 1/2 a foot? The tempting answer is 1/2, as we imagine a process of production that is uniformly distributed over side-length. But the question could have been given an equivalent restatement: A factory produces cubes with face-area between 0 and 1 square-feet; what is the probability that a randomly chosen cube has face-area between 0 and 1/4 square-feet? Now the tempting answer is 1/4, as we imagine a process of production that is uniformly distributed over face-area. This is already disastrous, as we cannot allow the same event to have two different probabilities (especially if this interpretation is to be admissible!). But there is worse to come, for the problem could have been restated equivalently again: A factory produces cubes with volume between 0 and 1 cubic feet; what is the probability that a randomly chosen cube has volume between 0 and 1/8 cubic-feet? Now the tempting answer is 1/8, as we imagine a process of production that is uniformly distributed over volume. And so on for all of the infinitely many equivalent reformulations of the problem (in terms of the fourth, fifth, … power of the length, and indeed in terms of every non-zero real-valued exponent of the length). What, then, is the probability of the event in question?

Van Fraassen's cube factory is actually solvable: the answer is 1/2 (as posed above). To reach this answer, we merely rely on the metaphysical fact that particles and measurements are discretized. That is, whether we consider the material used to construct the cubes, or the means by which we measure the quantities in question, the available values in our 'random' range are limited. If, for example, we stipulate that we can only measure side lengths at 1/4 foot intervals, our available cubes based on side length are ¼, ½, ¾, and 1. Of these, exactly half are less than or equal to ½ in side length. If we maintain that limitation on measurement, but consider the question of face area, we find that the available cubes based on face area are 1/16, ¼, 9/16, and 1. Of these, exactly half are less than or equal to ¼ in face area. This symmetry persists across volume as well. The idea here is that we recognize that particles and measurements are quantized; in spite of the apparent difference in range, the actual values available in that range have the same count, and the values which correspond to the 'same cube' are exactly the same in number. Applying this to Bertrand's chord problem is trickier, however, because that problem is ill-posed. I am yet convinced that it has a unique solution using this reasoning, but to date I have not taken the effort to formally prove it. Chrisfl.wiki (talk) 18:34, 14 August 2013 (UTC)

We could include this, but there is a big difference to Bertrand's Paradox: with Bertrand's Paradox, we have to apply transformation groups in two-dimensional space, while the cube factory problem is just about finding an uninformative prior probability density on a scale parameter, which is less difficult. Do you think this page would profit from more mathematics? Like a section "mathematical details"?Hanspi (talk) 18:32, 1 September 2009 (UTC)

In order to have a random selection you must establish a process in which each element in the population has equal probability to be chosen. In the case of Van Fraasen's Cube Factory described above if you focus on the random selection of length, then you have to consider a unifom distribution of lengths in the population which will relate to a non-uniform distribution of area. If you focus on the area you must consider a uniform distribution of areas which relate to a non-uniform distribution of lengths. In Bertrand’s paradox the focus is the length since the length of the chord is to be compared with the length of the side of the triangle. Therefore the lengths uniformly distributed in the population is the unique solution. This will result in a probability of the chord being greater than the side of the triangle of (2R-1.732R)/(2R) = 13.4%, where R is the circle’s radius and 1.732R is the side of the triangle. Prodrigues1953 (talk) 12:23, 23 November 2010 (UTC)

...can we not...?

...use the distribution, "select two points on the circle randomly with uniform distribution" and use the chord that those two points to make to solve the problem? It seems to give the least bias to the problem. ZtObOr 22:02, 20 November 2008 (UTC)

This is equivalent to method 1.

Hanspi (talk) 08:45, 29 June 2009 (UTC)

POV

This article looks extremely "POV" to me. Michael Hardy (talk) 16:31, 15 March 2010 (UTC)

In what sense is it POV? My only real issue with it is the title including "paradox" when it is not a paradox. That seems like calling the Monty Hall problem a paradox. To be fair, this is a bit more "paridoxical" in that the Monty Hall problem, while counterintuitive, is well-posed whereas here the issue is that there seem to be multiple interpretations of the problem. However, upon close inspection, it appears that one must conclude that either the problem is ill-posed or else conclude that there is one best interpretation basically using Ockham's razor. Either way, there is no real paradox. —Ben FrantzDale (talk) 13:23, 22 March 2010 (UTC)

I've never heard of a formally-defined and accepted definition of "paradox" in mathematics or science. One could consider Hughes's or Quine's types in the Paradox#Logical_paradox, or maybe look over List_of_paradoxes#Mathematics, but that's not the point. The point is that it's referred to as a paradox and to say it's not when there's no formal definition is POV. And honestly, most paradoxes in science literature are "resolvable" in some sense, or the real-world oddity has a formalizable explanation, unless they are ill-posed or contradictory to begin with. After all, coming up with solutions is the fun of talking about paradoxes! SamuelRiv (talk) 21:40, 8 August 2011 (UTC)

Description of methods

As described, methods 1 and 2 are not random as the chord is derived from the triangle. I intend to change the description eg so that method 1 says 'pick two points on the circumference'.

This distracted me for quite a while whilst I was reading the article. Jimbowley (talk) 10:05, 6 July 2010 (UTC)

Jaynes' Solution

Jaynes' Solution seems to only make sense if they're saying that there's a random line, and you're looking at the chord where the line passes through the circle. This is a way of constructing a chord with a line and the circle, and you could just as well get method three by doing the same thing with random points. Shouldn't it be that line segments are taken at random, and it only works with the line segments that happen to end at the borders of the circle? If you do it that way, the correct answer would depend on how you're distributing the lengths of the chords. — DanielLC 03:41, 5 September 2010 (UTC)

Jaynes's error is that he is generating random lines in a plane instead of random chords of a circle, thereby ignoring part of the problem statement. Ant 222 (talk) 14:24, 4 February 2018 (UTC)

False claim about midpoints identifying chords

Presently there is a claim that a chord is identified by its midpoint, "A chord is uniquely identified by its midpoint", but this isn't true in the case of the centre of the circle which identifies an infinite number of chords (also diameters), so it strikes me that this claim is incorrect. —Preceding unsigned comment added by 60.240.67.126 (talk) 23:05, 29 October 2010 (UTC)

I thought of this just now as well. But it turns out, the center is the only point where this is not true. The reason is, there is exactly one line that crosses through the center of the circle and the midpoint of a particular chord. If the midpoint of the chord and the center of the circle are the same point, then it is no longer possible to specify uniquely.--75.80.43.80 (talk) 08:07, 30 March 2011 (UTC)

Since the points are being chosen at random in a subset of the real plane, you can just safely exclude (r,θ)=(0,[0,2π)) as a point since its probability of being chosen is infinitesimal (arbitrarily close to 0) and we aren't doing anything countoury or topologicalish. SamuelRiv (talk) 21:45, 8 August 2011 (UTC)

The midpoint doesn't uniquely identify any chord in a circle. Choose a random point within the circle, perhaps U(-r/2,r/2) for the X and Y coordinates and discarding any points falling outside the circle. Once you have this point, an infinite number of chords could be drawn through this point, some of which would be longer than the side of the triangle and some of which would be shorter. Maybe choosing a U(0,2π) angle reproduces Method 3, but that isn't stated in the article. Frank MacCrory (talk) 18:38, 30 July 2021 (UTC)

Unique Solution

The correct answer is 13.4%. The problem asks you to compare the LENGTH of a chord with the LENGTH of the side of the triangle. Therefore the parameter to be chosen randomly is the LENGTH, not angle, nor radius and definetely not point within an area. As an example, the only way to choose randomly a length from 100 chords is to generate 100 chords where the difference in length between two consecutive chords is the same. This way each chord will have the same chance to be chosen. In the problem in question, being R the circle radius, the side of the triangle will be 1.732R. Given a randomly specified length, the probability of the chord this length being greater than the side of the triangle will be (2R-1.732R)/(2R).100%=13.4%. If the 100 chords were generated by dividing the 90° angle in 100 equaly spaced angles, each chord corresponding to each angle will be greater than the corresponding chord generated by random length. Therefore the probability of the chord being greater than the side of the triangle will be higher. 33% greater than 1.732R in the case. If the 100 chords were generated by dividing the radius in 100 points and taking the lengths of the chords passing through each point, each length will be greater than the corresponding length generated by random angle. 50% greater than 1.732R in the case. In the random angle as well as random radius methods, the chord lengths are not equaly distributed in the range since the difference in length between two consecutive chords is not constant. Actually the difference decreases with increasing angle for instance. When you choose the second method you divide the 90° angle in 100 equaly spaced angles and take the chord corresponding to the angle. that will give you the 33% probability. The chords lengths are not equaly spaced. When you choose the third method you divide the radius in 100 equaly spaced points and take the chord passing through it. That will give you the 50% probability. The chords lengths are not equaly spaced. The unique solution is to divide the lengths in 100 equaly spaced lengths. This is equivalent to Jaynes solution by throwing straws onto the circle. Prodrigues1953 (talk) 10:42, 23 November 2010 (UTC)

What they are asking you to compare is irrelevant. The point is that "chosen at random" is not well defined. Your argument is heuristic, not mathematical. The mathematical solution to the paradox is to be more specific about what you are asking, since "chosen at random" can't be defined unambiguously for all possible situations. The article even discusses different physical models that give the different results. In those cases, the parameter compared is also length, but length is not the uniformly distributed quantity.--75.80.43.80 (talk) 14:23, 7 May 2011 (UTC)

Flies and molasses

In order to arrive at the solution of "method 3", one could cover the circle with molasses and mark the first point that a fly lands on as the midpoint of the chord.

Uh... Very colourful, but couldn't we just say one could throw a dart at a round dartboard, and use that as the midpoint? --Doradus (talk) 18:31, 1 March 2012 (UTC)

we don't compare lenght, we just have to split our set of chords by the limit lenght and the problem is defining that set. The set proposed here by aproach nr 2 or 3 are not regular, because they involve nonlinear transformation. If we assume that mid point represent one chord why we have to make projection of that point on flat surface of the circle?? That is main question here! If we are able to make nonlinear transformation why we can't repeat that again and again. Problem is well stated and it is excellent example how we can't insert our assumption. Unique solution is 1/3 -- see: www.bertrands-paradox.com — Preceding unsigned comment added by 217.65.193.35 (talk) 10:14, 11 June 2012 (UTC)

Connection to Aleph number

Is there a well-known citeable connection? Shyamal (talk) 12:31, 17 September 2012 (UTC)

Possibly easier way to explain Jayne's result?

It struck me that the maximal entropy solution is to define a "randomly chosen chord" by selecting uniformly from the set of all possible lines in $\mathbb {R} ^{2}$, and accepting only those that happen to be chords of the circle. However it is not difficult to show that this is the same as method 2.

Choose your line at random from all possible lines. Without loss of generality, choose a coordinate system centred on the circle, with the y axis parallel to the candidate line. (There are two ways to do this, rotated 180° from each other. For definiteness we choose the one that gives a positive x value if the line does not pass through 0. If it does pass through 0 then the two orientations are identical.) Thus, the candidate line is defined by its x-coordinate. By the principle of indifference, all non-negative x values are equally likely. Now this line is a chord iff $x\in [0,r)$; and since all we have done is discard the lines that fall outside this range, all values within that range are equally likely.

But now what we have is a random radius of the circle, and a chord defined by its intercept of that radius, uniformly distributed along that length. This is identical to method 2. -- 203.20.101.203 (talk) 23:04, 14 November 2012 (UTC)

The concept of "selecting uniformly from the set of all possible lines" seems ill-defined to me. 66.188.89.180 (talk) 19:14, 15 March 2013 (UTC)

Is there a lower or upper bound?

The problem demonstrates ways in which the solution might be 1/4 or 1/2. Are there methods which generate answers outside this range? If so, are they as simple as the methods described, or more complex? ± Lenoxus (" *** ") 23:10, 13 May 2013 (UTC)

There is no lower or upper bound (well ... ~0% and ~100%). Proof: Use the random radial point method, but instead of picking a straight line, pick a curve of any shape

Lets say your curve is a dense spiral in the middle, and from there a short line to the end, so that 99% of the line is close to the center. After picking any angle, the chance of picking a point in the center area (and thus creating a chord long enough) is 99%. Qube0 (talk) 08:49, 16 September 2022 (UTC)

How precisely

...are the six unsourced distribution graphs and the unsourced closing paragraph of the "Bertrand's formulation of the problem" section not violations of WP:OR and WP:VERIFY? I understand, for the erudition of its members, that Wikiproject Mathematics is given great latitude. But is this not egregious, to ignore and present our own research on the matter (presuming we are not reproducing that of another, and so plagiarising)? Reply at your leisure, but do reply. 2601:246:C700:9B0:E5E5:B1AE:733F:DB51 (talk) 16:15, 24 December 2019 (UTC)

I would note in addition to the legalistic argument—that we are a confederation held together by a commonly agreed upon set of rules, and so only as strong, in the end, as we are willing to adhere to them—there is a very practical argument in favour of presentation from source rather than ones original research or formulation of an argument. It is, that original work is overly dependent on the author, and that at an encyclopedia managed on a volunteer basis, such a dependence is impractical. Every editor/reader query, "What was meant by... ?" is either managed by a trip to the source cited, or by a post-and-wait episode for the author (or their supporters) to reply to. The former is our way, the latter is not, and is an entirely impractical way for this encyclopedia to be maintained. 2601:246:C700:9B0:E5E5:B1AE:733F:DB51 (talk) 16:26, 24 December 2019 (UTC)

WP:SOFIXIT. --JBL (talk) 17:46, 24 December 2019 (UTC)

How is relevant evidence absent here?

The principle of indifference is only applicable in the absence of relevant evidence. The relevant evidence here is that as the point on the radius in method 2 approaches the circumference at a steady speed the shrinkage of the chord accelerates, favoring the longer chords. Applying an inapplicable principle hardly qualifies as a "paradox".

A few years ago I taught a freshman seminar at Stanford titled "Paradox: bug or feature?" where we encountered some actual paradoxes, some harder than others. This one wouldn't even have qualified as a paradox. Vaughan Pratt (talk) 21:55, 24 October 2020 (UTC)

I'm not a mathematician, but it seems like this article doesn't mention the most important point

The key to all this -- the single principle that this whole paradox is a demonstration of -- is simply that there's no such thing as a uniformly random selection from an infinite set. That's the *entire* thing here, and it literally isn't mentioned. That's weird.

There are infinite chords on a given circle, and it is not possible to select an item at random from an infinite number of items, with equal probability for every item. This is a regular and known mathematical fact: https://math.stackexchange.com/questions/14167/probability-of-picking-a-random-natural-number 69.113.166.178 (talk) 01:30, 6 July 2021 (UTC)

It's not Mathematical paradox, this is an example of incorrect conclusions in mathematical reasoning

Are there any mathematicians among the moderators of this article?

So Joseph Bertrand ask:

"If a chord in the circle is randomly chosen, what is the probability that the chord is longer than a side of the equilateral triangle?"

Definition of Chord is:

"A line segment connecting two points on a curve."

I.e. this means that "If a chord in the circle is randomly chosen..." is equivalent to "If two points on the circle is randomly chosen...", so Bertrand's question transform in:

"If two points on the circle is randomly chosen, what is the probability that the chord, which these two points define, is longer than a side of the equilateral triangle?"

Because only the first Bertrand's argument includes the correct statement for choosing two random points on on the circumference of the circle "Choose two random points on the circumference of the circle and draw the chord joining them." this is the only correct answer too - 1/3. The other two Bertrand's arguments are not answers to Bertrand's question at all.

Argument 3: This argument effectively claims that "any chord that has a point that falls within a concentric circle of radius 1/2 the radius of the larger circle is longer than a side of the inscribed triangle" and on the basis of this statement draws the wrong conclusion that they should be matched "the area of the smaller circle and the area of the larger circle" when you calculate the probability if a random chord is longer than a side of the inscribed triangle. Total logical absurdity!

While it is true that every point that falls within the small circle is a part of chords which are longer than a side of the inscribed triangle, FOR THE OUTSIDE POINTS THAT ARE BETWEEN THE GREAT AND SMALL CIRCLES, YOU CANNOT MAKE THE OPPOSITE CLAIM! Chords can pass through these points, which can be longer or shorter than a side of the inscribed triangle, so you can't use areas of the circles to calculate probability that the chord is longer than a side of the equilateral triangle, in the way Bertrand do it.

So after concluding that "The chord is longer than a side of the inscribed triangle if the chosen point falls within a concentric circle of radius 1/2 the radius of the larger circle." and before you start doing probability calculation, you change Bertrand's question in:

"If two points on the circle is randomly chosen, what is the probability that the chord, which these two points define, will intersect a concentric circle of radius 1/2 the radius of the larger circle?"

https://drive.google.com/file/d/1JeLwdE26VJZp1EMLUrFaWyCQTNsG0RAJ/view?usp=share_link

Now you can do the correct probability calculation by simply removing one side of the triangle so that the other two form a tangent to the inner circle, and notice that this argument 3 reaches the same conclusion about probability, like argument 1, answer 1/3, because in practice it differs in nothing from it.

Argument 2: Another miscalculation of probability. After "The chord is longer than a side of the triangle if the chosen point is nearer the center of the circle than the point where the side of the triangle intersects the radius" you change Bertrand's question in:

"If one point on the quarter circle(quadrant) is randomly chosen, what is the probability that the perpendicular from this point, to one side of the quadrant, will fall within the bottom half of the side of the quadrant?"

https://drive.google.com/file/d/1-1fc-w260_BSExNhacRK0Ch7NQLUxUnC/view?usp=share_link

And, of course, probability again is 1/3, because the arc of bottom half has angle of 30 degree, and above arc angle 60 degree.

As I said, Joseph Bertrand, is not very intelligent, the question is why other mathematicians did not realize this much earlier.Enchev Emil (talk) 05:41, 25 November 2022 (UTC)