Talk:Bertrand paradox (probability)

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I think this solution paragraph needs adjusting: "The problem's classical solution depends on the method by which a chord is chosen "at random"." It depends on the density measure given to the set of chords. That may or may not depend on your method of choosing a chord. The key thing is the density measure.

"The argument is that if the method of random selection is specified, the problem will have a well-defined solution (determined by the principle of indifference)" If you specify exactly how the chords are chosen then you don't need the principle of indifference to determine the answer. E.g. uniformly selected start and end points on circle.

"in the absence of further information there is no reason to prefer one over another." This is true. But to be pedantic, the problem statement does suggest answer #1:

The problem says "a chord of the circle is chosen at random". It doesn't say "a line was chosen at random and then a chord made from it" so it isn't answer #2. It doesn't say "a chord midpoint is chosen at random" so it isn't answer #3.

It says a chord is chosen at random. A chord is a line segment AB where A and B intersect a circle. The uniform distribution of A and B on the circle seems to be a very sensible default interpretation of the problem statement. This gives answer #1, a probability of 1/3. — Preceding unsigned comment added by 140.253.50.106 (talk) 04:57, 9 April 2024 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

The nature of the "paradox" was that there is often more than one way to parametrize a problem involving a continuous variable. So for a circle of radius 1, the chords must be in the interval (0,2], but what is the distribution? Bertrand proposed three different solutions to show why this is a conundrum. Presently the main objection about Bertrand's chord paradox is whether or not it refutes the Principle of Indifference because a particular solution can't be appropriately justified. (The Principal of Indifferent allows a probability to be assigned to each state where there are a finite number of states, e.g. a coin flip or a dice roll.) In reality there isn't a math problem here, only a philosophical one. The philosophical problem is like debating "How many angels can fit on the head of a pin?" 99.190.32.88 (talk) 20:03, 26 October 2023 (UTC)[reply]

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)[reply]

...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)[reply]

This is equivalent to method 1.

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

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

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)[reply]

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)[reply]

POV - That's the whole point. Bertrand chose a simple problem that can be analyzed by geometric probability, so a lot of convoluted math isn't needed. This allows the reader to readily understand that the problem can be analyzed from different POV (frames of reference). Supposedly the different POV are due to the fact that a continuous variable is involved. 99.190.32.88 (talk) 20:09, 26 October 2023 (UTC)[reply]

No, these are not "points of view".

The article simply shows how different assumptions about the probability distribution on chords lead to different answers. Nobody is promoting any of these as the "correct" distribution. Opinion has absolutely nothing to do with this.

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)[reply]

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)[reply]

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)[reply]

Many authors have pointed out problems with Jaynes "solution," but I don't recall any that actually point out where Jaynes went off the rails. His statement of the problem is "Bertrand's problem (Bertrand, 1889) was stated originally in terms of drawing a straight line

'at random' intersecting a circle." He then goes on to say "presumably, we do no violence to the problem ( i.e., it is still just as 'random') if we suppose that we are tossing straws onto the circle, without specifying how they are tossed."

Well he is wrong about that, so the problems with his solution begin with how he paraphrased Bertrand's statement. A reasonable translation of Bertrand's statement is: "A chord is drawn randomly in a circle. What is the probability that it is shorter [longer] than the side of a inscribed equilateral triangle?" Note here that Bertrand asks about a shorter chord, but his solutions all solve for a longer chord. Thus the convention has been to pose the problem asking for a longer chord.

First a clarification. "a straight line ... intersecting a circle" is called a secant line.

Now where Jaynes goes off the rails. A large part the conundrum in Bertrand's chord paradox is puzzling about choosing "random chords" versus "random secant lines" because a chord can be extended to form a secant line, and a secant line has a line segment which is a chord. By stating the problem in terms of secant lines, Jaynes has stated the problem in such a way that there is a unique solution. Jaynes concrete model of straw tossing also has a definite solution, as does any concrete model. However as Friedman pointed out Jaynes is using a fixed length straw. To get exactly 1/2 a straw of infinite length must be used.

To understand how the 1/3 and 1/2 answers are related think of using a measuring circle concentric with the chose circle. Let's assume that the radius of the chosen circle is R and the radius of the measuring circle is r. Now we are going to select two points randomly on the measuring circle and use them to form a chord. Over the interval [0, R/2] the probability that the chord is longer than the equilateral triangles side is 1. Over the interval (R/2, R] the probability drops from1 to 1/3. Over the interval (R, infinity] the value increase to 1/2. Note that for about 10R the probability is very near 1/2.

The gist here is that random secant lines created by choosing two points on circle R are not random lines of the plane. Think about circle 1 meter in diameter and a point, P, 1 astronomical unit away from the circle. At P the lines from circle R are essentially parallel. This is far from isotropic. 99.190.32.88 (talk) 20:54, 26 October 2023 (UTC)[reply]

Another point about the 1/2 solution. It assumes the chord set is derived from the set of random lines in the plane that intersect the circle. Think of the flip problem. Instead of drawing a random line across a circle, how about tossing a circle randomly on a line? That problem unequivocally has an answer of 1/2, assuming an infinitely long line so the end of the line isn't hit thus only creating 1 intersection not two. (For a human tossing on purpose a couple of meters would seem more than reasonable for tossing a circle 25 cm in diameter.) To make that model more efficient, consider a multitude of parallel lines a distance 2R apart where R is the radius of the chosen circle. Tossing a circle is now virtually 100% efficient.

To mull the point over, consider two plastic sheets. One with a circle and a much larger one with parallel lines etched. Is throwing the sheet with the circle any different than throwing the sheet with the lines? As a physics problem there are two frames of reference, but the end result after contact is the same.

herace 99.190.32.88 (talk) 05:38, 27 October 2023 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

For any point, there's an infinite number of chords which pass through this point, but only one chord for which this point is the midpoint, that's essential to this definition. (Excluding the center, as noted above) --Daniel (talk) 23:25, 16 July 2023 (UTC)[reply]

You are absolutely right! Bertrand sneakily avoids that consideration. We all learned that a circle has an infinite number of diameters in grade school. Why shouldn't we use the fact that a circle has infinite diameters as a restriction on any possible solution?

In fact any solutions can be converted to a solution about the density of chords in the circle by a change of variables. Since there are an infinite number of diameters, when considering the density of chords, the center must be an asymptote. Hence the density of the cords going towards the center must go to infinity. Thus Bertrand's area solution is bogus.

Now Martin Gardner proposed making random chords from raindrops falling on a circle. But here Gardner sneaks in a different problem. Bertrand didn't ask about a circular area but rather a circle. Raindrops don't know anything about circles per se. 99.190.32.88 (talk) 21:09, 26 October 2023 (UTC)[reply]

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)[reply]

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)[reply]

Well, information theory says that you are wrong. Bertrand asked for the probability, not the distribution of the chords. So the problem can be rightfully cast into non-parametric statistics. So instead of recording chord length an experimenter just records true(T) or false(F) to the question "Is the chord longer than the equilateral triangles side?"

This isn't far fetched. Think of marking off the side length of the equilateral triangle and just comparing that to the random chord. Works fine!

Now Information Theory say that the 1/2 solution to the problem is the most random answer. Consider trying to guess if the next chord is longer or shorter. If only 1/4 are actually longer then one should guess F, and would be right 3/4 of the time. However if 1/2 are actually longer then one can never do better than just random guessing.

herace 99.190.32.88 (talk) 21:51, 26 October 2023 (UTC)[reply]

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)[reply]

Flies don't land in circles per se, but rather circular areas. Circles have an infinite number of diameters, but circular areas do not. 99.190.32.88 (talk) 21:14, 26 October 2023 (UTC)[reply]

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)[reply]

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

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 ${\displaystyle \mathbb {R} ^{2}}$, 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 ${\displaystyle x\in [0,r)}$; 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)[reply]

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)[reply]

Jaynes really starts off assuming random lines of the plane in his problem statement. Note that for a circle of radius R, the midpoint of the chord must be within a concentric circle of radius R/2. Crofton showed that for one convex body inside another that the ratio of their perimeters was a measure of the lines intersecting the inside body divided by the number of lines intersecting the outside body. 99.190.32.88 (talk) 21:35, 26 October 2023 (UTC)[reply]

I meant to say that "the midpoint of the chord longer than the equilateral triangle's side must be within a concentric circle of radius R/2."

herace 99.190.32.88 (talk) 22:32, 26 October 2023 (UTC)[reply]

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)[reply]

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)[reply]

...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)[reply]

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)[reply]

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

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)[reply]

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)[reply]

That isn't in fact an issue! Even though the probability of a continuous random variable taking on a particular value is zero, one can instead define a probability density that allows one to compute the probability of a random variable being an element of an open set, and this probability density will be non-zero in general. Read a bit here: https://en.wikipedia.org/wiki/Probability_density_function StrawberryLetter22 (talk) 03:27, 28 February 2023 (UTC)[reply]

This paper, which you can read here, is mentioned prominently in what is currently the "Recent developments" section. If you turn your attention to the two paragraphs at the bottom of page 7 of the paper, you'll see that the authors claim that an arbitrary probability density function ${\displaystyle \rho (x)}$ can be the limit of a sequence of functions ${\displaystyle \rho _{n}(x)}$. These functions, as defined at the top of the page, are nothing more than characteristic functions times a constant, so they obviously can't approximate an arbitrary density function. This is hidden by the authors' choice to focus only on the probability mass assigned to a fixed interval ${\displaystyle [x_{1},x_{2}]}$. FRuDIxAFLG (talk) 06:07, 11 December 2022 (UTC)[reply]

Can someone please give me an opinion on this. Because even if you just look at these two pictures from the post, it will be clear to you that Argument 2 and Argument 3 are the same argument, but someone deliberately put in an extra choice to change the probability between two arguments:

https://qph.cf2.quoracdn.net/main-qimg-76652df2fdc28fddde42b21fa80c2ae3

https://qph.cf2.quoracdn.net/main-qimg-1d7020012424e71e226b34e9d650138a

Original Quora's post: https://www.quora.com/profile/Emil-Enchev-16/Did-you-know-that-the-Bertrand-paradox-is-the-best-indicator-that-mathematics-has-become-a-den-of-crooks-and-low-intelli

P.s. I am the author and I am curious if any mathematician will finally dare to say something against what is written in the post. Are all mathematicians crooks and cowardly conformists?

77.85.215.165 (talk) 14:36, 29 January 2023 (UTC)[reply]

H isn't the midpoint of the red and blue lines in the 5th picture. FRuDIxAFLG (talk) 05:03, 1 February 2023 (UTC)[reply]

The last sentence of method 1: 'The length of the arc is one third of the circumference of the circle, therefore the probability that a random chord is longer than a side of the inscribed triangle is 1/3.' Surely that's a typo and is intended to say 2/3, right? The chance of choosing a point (the second chord endpoint) that doesn't lie within an arc of one third the circumference of the circle is 2/3.

Happy to defer to someone on this since I'm not a mathematician, but unless I'm utterly misreading this should be corrected; I'll set a reminder to myself for 2 weeks from now to come back and change it if no one disagrees.

[UPDATE - correcting, 5/16/2023]

eggsyntax (talk) 01:56, 2 May 2023 (UTC)[reply]

Hold on, there are many googleable sources that show that this should indeed be 1/3, see https://www.cut-the-knot.org/bertrand.shtml or here http://web.mit.edu/tee/www/bertrand/onethirdmath.html or here https://www.uio.no/studier/emner/matnat/math/MAT4010/v17/notater/w-bertrand-paradox-%28probability%29.pdf or here https://medium.com/quantum-physics/bertrands-paradox-9a6789dcf02e. You get the point. It should definitely read one third.

I think you've misinterpreted slightly what the paragraph is saying? You're right in that choosing a point that doesn't lie within that special opposing arc is indeed 2/3, but that's the probability that the created chord is shorter than the side of an equilateral triangle, while the actual question is "What is the probability that the chord is longer than a side of the triangle?"

You can think of the first point as being fixed, with the tip of an equilateral triangle placed on the first point. Then, the second point is chosen around the circumference at random. Clearly, only the opposite arc generates chords that are longer than the side of the equilateral triangle, and this chord covers precisely one third of the perimeter, therefore the probability (according to this method of course) is indeed one third. AlteredDesk (talk) 00:52, 19 May 2023 (UTC)[reply]

Ahhhh, you're entirely right, I misread it. Thanks very much for catching it and setting it back so quickly! I was concerned that I was misreading, which is why I brought it up on the talk page well in advance, but despite my attempted caution I had it wrong. eggsyntax (talk) 01:43, 19 May 2023 (UTC)[reply]

In the references one of the names is misspelled. The name should be Tissier, not Tissler. 99.190.32.88 (talk) 20:54, 23 October 2023 (UTC)[reply]