Talk:Bertrand paradox (probability)

In #1: "therefore the probability a random chord is longer than a side of the inscribed triangle is one third." Should that be 2/3rds? -- —Preceding unsigned comment added by 76.195.14.166 (talk) 16:27, 28 September 2007 (UTC)[reply]

Hi, please consider the 2 entries in wikipedia (I am refering to the Bertrand's Paradox entry and the Bertrand Paradox entry). Are they refering to the same paradox? Or did Bertrand came up with 2 different paradoxes. I search for Bertrand's paradox on google and it appears that the paradox on probability is the more common find: that is what is the probability that a randomly picked chord of a circle will have a length greater than the side of an equilateral triangle inscribed in the circle

To whom it may concern -- I'm planning to add diagrams to this article. It's pretty text-heavy at the moment, diagrams will definitely help. Thanks for your patience. Wile E. Heresiarch 05:10, 20 Aug 2004 (UTC)

How can case 3 have anything to do with it? It's an area argument while chords are lines -- that's a dimensionality disagreement. Also, it suggests that a random chord only intersects the middle are every fourth time -- which is not true. Is there an explanation for this, or is this paradox resolvable with proper mathematical definitions? ✏ Sverdrup 19:09, 21 Oct 2004 (UTC)

blunders at Bertrand's paradox

The answer is here: http://www.sezampro.yu/~seik

any comment? http://user.sezampro.yu/~seik/Bertrands%20paradox.htm ---- english version

Incorrect statements in article?

The article says "Bertrand intended to show that the classical definition of probability is not applicable to a problem with an infinity of possible outcomes. According to the classical definition, the probability of a compound event is the ratio of the number of favorable cases to the total number of cases. Such a definition is inapplicable here, for there are an infinity of chords. Extrapolating the classical definition to ratios of lengths (solutions 1 and 2) or areas (solution 3) seems to yield inconsistent results." However, based on my reading of the earlier part of the article, this isn't true at all. The paradox says nothing about probabilities becoming invalid because of infinities. It says that the probability will vary based on what method is used for randomly picking a chord. I don't see any way that the use of infinity messes up the problem. Perhaps someone can clarify? If not, I'll remove or reword that passage. MrVoluntarist 00:39, 18 August 2005 (UTC)[reply]

Im not sure how much sense the statement you are referring to makes, but the problem is related to the use of infinity. Each method is capable of producing an infinite number of chords, but each one also produces them at a different rate for a given degree of precision. Anotherwords all infinities are not the same.

Well, it is related to infinity, but only in passing. It's not a fundamental part of the problem. the problem were posed with the circumference discretized, and you used the same three methods, but "rounded" to the neared permitted point, that would clearly not involve infinity, but would raise the paradox. It's just that people would assume that "the method" of picking a chord involves picking a random two points, but the paradox still shows in that there is no method specified and different ones produce different answers. MrVoluntarist 17:02, 28 August 2006 (UTC)[reply]

The paradox is indeed very much related to infinities. The classical definition of probability isn't concerned with randomness at all; it's all about counting possibilities. The number of favorable possibilities divided by the number of total possibilities gives the probability. By definition, period. No randomness involved here. However, that definition came under heavy attack by probabilists during the second half of the nineteenth century, and Joseph Bertrand were one of its critics. (John Venn and George Boole were others.) As a result of their criticism, and by the efforts of these men and many others, the modern view of probability grew. Today we know that for a probability problem to be well posed we have to specify the random mechanism involved. Even when the sample space is finite. But to prove the classical definition wrong, Bertrand had to use an example with a continuos parameter. iNic 23:47, 25 February 2007 (UTC)[reply]

Different point of view

Maybe you should see also http://www.bertrands-paradox.com ...

Ramifications

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]

Fool's question ?

If there's three different methods of randomly selecting a chord, wouldn't the answer be the probability when randomly selecting one of the three methods??? Sophistifunk 02:25, 15 February 2006 (UTC)[reply]

No, because why limit it to those three methods? There are potentially infinite many methods of selecting a chord, and that would leave undefined how to choose between those infinite methods. If you restrict it to these three methods and have an equal probability of selecting each method, then the solution is, you are correct, well-defined. MrVoluntarist 03:20, 15 February 2006 (UTC)[reply]

For example, here are a couple of other possible methods of picking a "random" chord (they may be equivalent, but it doesnt mean they arent different methods);

pick a random point inside the circle and a random direction, make a chord in the direction through the point
pick a random point inside the circle and a point on the circle, join them up and extend into a chord
pick a random diameter and a distance away from the diameter, draw a chord parallel to the diameter at that distance. (there are 2 chords in each case, but allow the distance to be negative and then pick an "up" let positive distance mean take the chord "above" the diameter etc)
pick a random diameter and a random point inside the circle, draw a chord through the point perpendicular to the diameter
pick a random diameter and a random point on the circle... etc

Teutanic 13:28, 27 March 2007 (UTC)[reply]

An Unconditional Solution

Each of the three solutions presented (1/3, 1/2, 1/4) are conditional probabilities. They are conditional on the method used to select the "random" chord. Given that the method of chord selection follows the method detailed in the first solution, then the conditional probability that the "random" chord is longer than the side of the inscribed equilateral triangle is 1/3, etc. Each of the three solutions makes assumptions about the experiment for which there is no direct evidence in the wording of the problem. As a result, the solutions are all accurate (in that they provide the correct conditional probability) but also all irrelevant, in that they fail to provide the simple, unconditional probability called for by the question. The only information that exists on which to base the calculation of the probability is:

1) There is a circle, in the abstract sense, as opposed to any specific circle.

2) There is a length associated with a side of an (abstract) equilateral triangle inscribed in the circle.

3) There is a "random" chord, which therefore has a "random" length.

There is no mention of how the "random" chord was selected, except that it was "at random".

The question restated is: What is the probability that the "random" length is longer than the length of the side of the inscribed equilateral triangle? And that's it. No other information is implied on which to base the calculation.

To obtain the unconditional probability, it is sufficient to note that:

1) Absent any specific guidance in the wording of the question, it is assumed that any possible "random" length will be as equally probable as another other possible "random" length.

2) The length of the "random" chord will therefore follow a Uniform distribution between 0 and twice the radius of the circle (when the chord's length is equal to the circle's diameter).

3) The length of a chord created by a side of an inscribed equilateral triangle is equal to ${\sqrt {3}}$ times the radius of the circle.

4) The probability that a chord whose length is uniformly distributed between ${0}$ and ${2r}$ will exceed ${{\sqrt {3}}r}$ is equal to ${1-{\frac {\sqrt {3}}{2}}}$ , which is approximately equal to 0.134.

This solution suggests that the "paradox" is not a result of uncertainty in the meaning of term "random". Rather, it ironically arises from certainty in what "random" means. Once the method of obtaining the "random" chords is defined, the probability computed becomes conditional on that method. The only assumption that is necessary to answer the question unconditionally is that of the equi-probability of the lengths of the "random" chords. Each of the three solutions previously presented make this assumption as well, albeit indirectly.

63.194.28.214 22:12, 18 September 2006 (UTC) G.M. Sesiter[reply]

Please read Wikipedia:No original research. If you found this in a journal or something, that would be one thing, but if you just made it up it's a bad idea to publish it on Wikipedia. Publish it on your own website, or somewhere like PlanetMath. —Keenan Pepper 00:43, 23 September 2006 (UTC)[reply]

Please don't remove sections of talk pages. If you want to take back something you've written, cross it out like ~~this~~. When talk pages get too long, they are archived. —Keenan Pepper 18:33, 23 September 2006 (UTC)[reply]

I believe that the entire point of Bertrand's paradox is that if one is interested in an unconditional probability, according to the chain rule they have to start with a known prior, and where do you get that? In the case of the circle thing, it's the probability of a particular chord being chosen to begin with. If you make that a probability given a certain method, that then requires you to know the probability of that method. Considering that there are an unbounded number of algorithms by which to chose a chord from a circle given a random number, how do we assign probability densities to this set of mappings? That's why it's a paradox. The article should be better though. - JustinWick 00:57, 27 September 2006 (UTC)[reply]

Knowing the various prior probabilities is not necessary. What makes the solution unconditional is simple--it doesn't condition on anything. There are no assumptions made or specified about how the "random" chord is selected. It is clear from the myriad solutions that the answer obtained is not independent of the way the "random" chord is selected. This, in and of itself, is not particularly surprising, and certainly not paradoxical. What may be paradoxical is why anyone would feel the need to bring something more to the problem that what is simply stated in its formulation. Why fill up the outcome space with endpoints or midpoints, or distances of midpoints from other points, or other such stuff? The crux of the problem is comparing lengths, and the outcome space should be full of chord lengths. True, the length of a chord is a function of the "random" endpoint of the first solution, or the "random" midpoint of the other solution, but why be indirect? The reverse is not true, at any rate, the "random" endpoint of the first solution is not a function of the chord's length, etc! Such an indirect approach, through functions, might be useful if the outcome space we wanted to consider was too abstract to deal with directly, but this is clearly not the case. The possible chord lengths will range between 0 and twice the radius with probability 1. We also know, and unambiguously so, that a transcribed equilateral triangle will have sides equal to ${\sqrt {3}}$ times the radius of the circle it is inscribed in. We only need to assume a Uniform distribution of values between those minimum and maximum values to come up with a direct and unconditional solution. Using simple calculus and the cumulative distribution function for a Uniform random variable, the solution is just 0.134.

63.194.28.214 18:23, 28 September 2006 (UTC) G.M. Sesiter[reply]

About previously mentioned "an unconditional solution"

On page http://user.sezampro.yu/~seik/Bertrands%20paradox.htm I have explained why it can't be considered as solution. In all problems that include probability we are interested about number of elements of some set not about their diversity. For example, if we have two red and seven blue elements what is probability to choose "at random" red element. It is not p=1/2 for sure (because of 2 colors)!!! (On page http://www.bertrands-paradox.com you can find simplified explanation about my point of view)

I think it's your example that is invalid. It's not a similar comparison at all. For each of the infinite possible lengths of chords between 0 and the circle's diameter (i.e., the diversity of elements) there are an infinite number of chords in the circle with that particular length (i.e., the amounts of the types of elements). This is completely different from your example, where you have 2 red and 7 blue elements. A more appropriate analogy would be to imagine and infinite number of red elements and an infinite number of blue elements. The probability of selecting red would be 1/2 in this case, since both colors are equiprobable. The limited example you provided suggests that you do not think that the Uniform distribution is valid for solving some problems that deal with probability. Likewise, the simple experiment of rolling a fair die should not be a valid premise for any questions concerning probability, since the number of kinds of elements in the outcome space is exactly equal to the amount of elements. Here is a useful thought experiment. Imagine that we have an infinitely sided fair die that has each of the possible lengths of chords on its sides. We roll this die to obtain a length. Once this is done, we pull out another infinitely sided fair die that contains the coordinates defining the endpoints of the line which makes up a particular chord that has a length equal to the result of the first die. So we have our first die, which contains all of the infinite possible lengths, and for each of those infinite possibilities, we have a secondary die that defines where the chord is located on the circle. In this way, we can hypothetically account for every element in the outcome space. Technically, the set of all the infinite secondary dice can uniquely identify all of the possible chords on their own. However, here's the important part. The probability question posed in Bertrand's paradox, without adding anything to it, is a question of chord lengths, i.e., the provenance of the first die. It does not in any way care about the secondary die roll, i.e., where any specific chord of any specific length is located within the circle. Any selected chord will either be longer than the target length, or it will not, it doesn't matter that you condition on its being one of the infinite number of specific chords of that length. What this means in terms of our thought experiment is that: p(1st die's result is longer than the side of the equilateral triangle|second die's result) = simply p(1st die's result being longer than the side of the equilateral) since the second die's result is independent of the first's. I've followed your links and have enjoyed your discussion of the topic. However, I believe that you make the same mistake that all of the other solutions (1/2, 1/3, 1/4, etc.) make. You are more worried about the second die's result in our thought experiment than the first's. If we could go back in time to put word's in Bertrand's mouth to change the text of the question he posed, then certainly we could make any solution plausible. We might make Bertrand say, for example, "Look, I have a circle with circumference 20. Here are two possible chords, the first is length 10 and the second is length 5. Now, what I want to know is...what is the probability that a chord selected at random from the two options available (5 or 10) will be longer than the side of this equilateral triangle that I have inscribed within my circle?" We could thus force the solution to be 0! Or anything else we want it to be, so long as we are free to change the problem. I don't claim to know what Bertrand meant when he posed the problem. But I do think there is a solution for the the actual text that he used, and absent any presumptive modifications or additions to this original text, the solution is 0.134.

63.194.28.214 23:22, 9 November 2006 (UTC) G. M. Sesiter[reply]

Be carefull! Consider following problem..:"A line has length 9cm where 2cm is red and 7cm is blue. You choose by chance a point from that line. What is the probability that your point will be red? " p=1/2 or p=2/9? You have now infinite number red and blue elements but we know that we have exactly 7/2 time more blue than red elements. So, we should allways compare their infinite numbers before we make statement based on their diversity (i.e. p=1/2)! Only if we compare their numbers we can say something about probability. (I think that the Uniform distribution is valid for solving some problems that deal with probability but it has to be used on apropriate way!)

Bozur Vujicic

Ah, this is a better example than your previous one, but still not entirely analogous. What makes it misleading is your preoccupation with presenting a choice between a clearly incorrect and a clearly correct answer. It is a straw man argument. There is no confusion on my end for determining the probabilities of mutually exclusive compound events made up of simple events in the outcome space. In essence, your line example does have an infinite number of simple events in the outcome space. However, this is a bounded infinity. You cannot, for example select a point that does not fall on the line (i.e., it has too small or too big a value relative to the end points. The compound events you defined, "red" and "blue" have unbounded infinities with respect to counts, or the number simple events in each compound event. I am entirely clear on this distinction and do not base my support of the 0.134 solution on the wrong "infinities". In fact, I also believe that Bertand's problem is similar to your line example, and will explain. Consider, we may always define the elements in the outcome space to reflect more variability than is necessary. For example, we are interested in rolling a fair die. In general, we are only concerned with the value of the roll, be it 1, 2, 3, 4, 5, or 6, which we define as the simple events in the outcome space. However, we may define the outcome space as the physical orientation of the roll in 3 dimensions. Now we have an infinite number of possible outcomes (simple events) in the outcome space. However, it is clear that this extra information is irrelevant, as once the value of the roll is determined, the other information is extraneous and independent. For example, if I want to figure out the probability that the roll will be greater than or equal to 5 (a compound event), it does not matter if the number on the top of the die is rotated 45 degrees from some arbitrary reference dimension, it just matters if it is greater than or equal to 5. For the purpose of determining this probability, we may collapse all of the infinite variations of a roll of, say "4", and call them, simply, "a roll of four". So a roll of 4, itself, becomes a compound event in the artificially expanded outcome space. Returning to your new example, you essentially have a line of 9 cm. This line contains an infinite number of elements (simple events) corresponding to points in one dimension. You choose to partition a portion, or portions of the line into a compound event, which you call "the red portion(s) of the line". The complementary portions of the line form a mutually exclusive compound event, "the blue portion(s) of the line". Since the combined length of the red portions is 2 cm, and correspondingly, the combined length of the blue portions is 7 cm, then it is obvious that a randomly selected point in the outcome space will have 2/9 probability of being in a red portion, and 7/9 probability of being in a blue portion. In actuality, your new example is a great example for why 0.134 is the correct solution to Bertrand's paradox. For Bertrand's paradox, the outcome space is made up of an infinite number of simple events, just as with the outcome space for the line in your example. Those simple elements are the possible lengths of chords in the circle. These lengths may also be represented as "points in a line", conceptually, to follow your example. The length of this conceptual line is the diameter of the circle, and it has endpoints at values of 0 and this diameter. We are interested in defining a compound event, codenamed "red", which is the segment on the line where the points are higher in value than ${\sqrt {3}}$ times the radius of the circle. The remaining part of the line, codenamed "blue" include all of the points not in the first event. It is pretty clear that the probability of randomly selecting one of the red points will be 0.134, even though there are an inifinte number of points (corresponding to chord lengths) in each of the red and blue sections of the line. Just as with my die roll example, where the number facing up was a function of the die's position in 3-dimensional space, the length of the chord is a function of it's position relative to the circle in 2-dimensional space. Also, just as in my example, where adding the infinite 3d position information was extraneous and irrelevant, and simply resulted in artificial simple events in the outcome space, information about which of the infinite chords of length X was selected results in artificial simple events in the outcome space for Bertand's question.

63.194.28.214 19:26, 13 November 2006 (UTC) G. M. Sesiter[reply]

Intention of my example was to show how we can't make decision based just on diversities of some set. So, I should expand my example one more time. Suppose that we consider curve y=x2 where x ordinates belongs to above mentioned line [0,9]. The task is to determine probabilities of chosing red line (L<4) at random! Lenght of lines are from 0 to 81, and critical value is 16. Obviously, correct solution is not 16/81 but 2/9 is correct answer! That is the point where you have lost correct interpretation of the problem. In probability theory first step is to determine basic set (i.e. how many elements (lines) we have in our problem) then how many elements belongs to searched subset, and result is subset/set. Color, lenght, weight etc. are not interesting primarily if we are not able do determine relation between number and their characterisrics like in my last expanded example.

Vujicic Bozur

Ah, another example. But again, the example is arbitrary and irrelevant. I'm not disputing that one needs to base the probability of an event on the number of elements in the event and not on the number of types of elements. I never have disputed that, and in fact, it's an artificial argument on your part. You seem to be stuck on it, but it's not remotely informative to this discussion, since it has no relevance to the solution to Bertrand's paradox that I favor. This is where your reasoning is flawed. My assertion has always been that we need to find an appropriate density function for the probability distribution of the chord lengths, since Bertrand's problem is essentially one of comparing a randomly chosen chord to a target length value. This is why the outcome space needs to be made up of chord lengths. If Bertrand had specified a particular side of an inscribed equilateral triangle, and asked "What is the probability that a randomly selected chord will bisect this side (line)?", then we need to keep the position of the chosen chord, or it's coordinates, in mind. However, it is clear that the distribution of valid chord lengths is Uniform between 0 and the diameter of the circle. It's a fairly easy mental exercise: Consider the top half of the circle. There are an infinite number of parallel chords between o and D in length. Each chord appears once. The same is true of the bottom half, by symmetry. Now, any chord not parallel in the top half is simply a translation of one of the parallel chords. In this case, again, there will be an infinite number of chords parallel to it, between 0 and D, and again, each length will appear just once in this new parallel set. So if we focus on the possible lengths that chords may take, it is clear that the counts, or numbers of valid chords for each possible length are infinite and unbounded. Also, for any instance of a chord of length X, there is one instance each of chords with length between 0 and D not including X. See, to come up with an analogy, I'm stating that all mammals are warm blooded, and you're pointing out an alligator and trying to argue that since the alligator is cold blooded, I shouldn't be making assertions about mammals. Consider the curve y=x2? Honestly, what does that have to do with anything? It may or may not be an interesting example of something, but it's a non sequitur.

63.194.28.214 19:07, 21 November 2006 (UTC) G. M. Sesiter[reply]

Here is that arbitrary and irrelevant example about "mammals". http://www.crnogorskoprimorje.com/fotke/administrator/2006-11-22_064449_example.JPG What is the solution? Is it still without relation to Bertrand's paradox? The lenght are between 0 and 81 and critical lenght is 4!

"It's a fairly easy mental exercise: Consider the top half of the circle. There are an infinite number of parallel chords between o and D in length. Each chord appears once" ......."However, it is clear that the distribution of valid chord lengths is Uniform between 0 and the diameter of the circle. "??????????????????????

What a mistake! regards Bozur

Back from holiday. I see that you're still at it. Well, you're right, I can definitely see some mistakes. I'll point them out.

Here is that arbitrary and irrelevant example about "mammals". 1. I thought I stated that it was an analogy, not an example--there is a difference. Oops. Mistake #1. 2. Your diagram. Actually, I didn't need a diagram at all. It was a fairly simple question you posed. However, it is contrived and irrelevant. 4+2 = 6. Does that mean 4x2 = 6? Obviously they are different problems. Your example was contrived so that in the restricted range from 0 to 9, the length of the lines under the curve (Y-axis values) are functions of the position on the the endpoint of the line of the x-axis. Additionally, due to the artificiality of the example, the position on the x-axis is also a function of the length of the line! This is not true with chords of a circle, a point which I believe I made some time ago. A given length does not imply any particular one of the infinite number of possible chords corresponding to that length. So your example, while correct strictly for the putpose of your contrived, failed refutation, is irrelevant as it's an unrelated point. Oops. Mistake #2. 3. However, it is clear that the distribution of valid chord lengths is Uniform between 0 and the diameter of the circle. "?????????????????????? Since you are artistically inclined, why don't you show me a diagram of a semi-circle that shows it containing 2 different chords of the same length, where chords are parallel to the base (diameter)? This is the diagram that would disprove my accounting of the various chords contained in the circle. Oops! You missed your chance with that last bit of artwork, but I'll wait for it.

Best wishes, 63.194.28.214 20:45, 29 November 2006 (UTC) G. M. Sesiter[reply]

I didn't understand #1 explanation. Obviously you don't have linear relation (function) in the circle e.g. y=7x (y - lenghts of chord, x -number of chords) so you can't prove that set of lenghts are linear transformed from set of number of chords because of nonlinear transformation (and we need number of elements to count probability). As in my example, it is not the same to count probability among y and x axis. That is crucial, and that was the purpose of my example. I didn't understand even #2 explanation. I simply didn't realise what are you trying to say. Problem with nonlinear transformation is changing the density. I think that you don't want accept that fact.

Božur

I may have been unclear, so I'll try to explain better. First, I apologize for not initially seeing the usefulness of your example. I think I can use it to make an analogy that I'll hopefully be able to transition into a more direct discussion of Bertrand's paradox. I'll refer back to your diagram, since you've taken the time to create it and it looks nice.

To begin with, I don't dispute the solution you arrive at, namely 2/9 as the probability. In fact, I don't dispute any of the other mainstream solutions to Bertrand's problem, 1/2, 1/3 (the solution you favor), even 1/4. Each is correct, but only by making certain assumptions. Looking at your diagram, we have various lines under a curve defined by y =x2, between the values of 0 and 9 on the x-axis. Obviously y is changing at a faster rate than x. A line exists in 2 dimensions, and can be (if you want to) defined by its endpoints. In your diagram, one endpoint is always on the x axis. The other endpoint is elsewhere on the plane. You chose to select a "random" line (in 2 dimensions) from the standpoint of the x axis (a single dimension). The x coordinate will always be the same for the line's endpoints, while the y coordinate will be 0 for one endpoint, and equal to x2 for the other. OK. If this is true, then the solution you have presented is correct. But only if this is true. Let's look at what happens when you vary your "random" picks at the rate that the x-axis is increasing. You end up imposing a specific distribution on the lengths of the lines. It's a question of scale of measurement. Variations on the x-axis are less fine (precise) than variations on the y-axis. Let's say the most fine measurement I can make on the x-axis is at the 1's digit. So I can define points at x=1, x=2, and so on, up to x = 9. In going from x=1 to x=2, I jump from y = 1 to y = 4. A lot more ground has been covered on the y-axis. I've skipped over potentially many more possible values for y that I could have used, relative to the increase in x. Now, let's say I can define lines at the 10ths digit on the x-axis. Once again, every meaningful unit change in x skips over possible y values measured at the 10th's level of precision. Etc. So, are you wrong to base the probability in your diagram on variations in the x-axis? Of course not! If that's what you choose to do, then your solution is great. However, let's say that I don't like the x-axis. I want to base my "random" selections on the y-axis. Now let's look again at your diagram. My selections will be a "random" y coordinate between 0 and 81. This value will fall somewhere on a portion of the curve y = x2. This point will have a corresponding x coordinate. My second point will have the same x coordinate, but the ordinate will always be 0. These two endpoints will also define a line in your diagram, also between the curve and the x-axis. This method will impose a different distribution on the line lengths. Now, following my preference, what is the probability that a "randomly" selected line will have length less than 4 cm? Now it is 4/81. Is this a better solution than 2/9? Not really, but it is the correct solution following the way I choose to model the "random" selection. I could argue that since the ordinate for the bottom endpoint is always 0 (constant) for lines under the curve, and both endpoints' x coordinates are also constant, that the variability in line lengths is entirely dependent on variability in the location of the ordinate for the top endpoint, which may make it somehow "better" to base my random pick on values in the y-axis. Obviously, there is a relationship between y and x, and both are needed to define endpoints for the lines, but since the lines are vertical, the lengths are increasing at the same rate as the y-axis values. But I clearly don't have to use the y-axis. Using the x-axis, or any other way, for that matter, will necessarily constrain the precision at which possible lengths could be selected, yielding a different, but still correct result, but from a different perspective. An analogous situation exists with Bertrand's paradox. Bertrand never left instructions for how to constrain the length of the "random" chord. Another way to put it is that his question does not include instructions on how to impose a distibution onto the chord lengths. If only he had been more specific, then the question would lose its ambiguity. I understand that this means different methods result in different solutions. My position is that since no method of constraint has been specified, then we should impose none of our own making. However, technically, we need to assume some sort of distribution to arrive at a solution. I defend defaulting to equiprobability (Uniform) in the absence of any compelling reason for favoring a different distribution for the chord lengths. This way, our "random" selection method is based on selection values that increase at the same rate as the length values themselves. In this sense, it is "unconditional" on the method of chord selection. The solution this yields, 0.134 is a lower-bound solution. Any other reasonable distribution placed on the chord lengths will necessarily have a higher probability as the solution, like 1/4, 1/3, or 1/2. This suggests, to me, that these particular methods are unecessarily simplifying the sample space. For example, I roll a fair die, The probability of rolling a 3 is 1/6. If I ignore or skip over all even rolls, then the conditional probability of rolling a 3 given that I roll an odd number is 1/3, etc.

P. S. While I was looking over you diagram and trying to follow the point you were making I realized something funny. Are you aware that you've provided an argument for the solution of 1/2? I found this strange, since I know you favor 1/3 as the correct solution. I'm not quite as artistic, so you'll have to bear with me. In your diagram, replace the y=x2 curve with the top left quadrant portion of the circle. Assuming a unit radius, the x-axis will range from 0 to 1, as will the y-axis. Between the curve of this quarter-circle and the x-axis will be half-chords. A circle with unit radius will have a length of ${\sqrt {3}}$ for the side of an inscribed equilateral triangle. Half of this length (to get a half-chord) is ${{\sqrt {3}}/2}$ . So we want to know what is the probability that a randomly selected "half-chord" under this curve will be greater than ${{\sqrt {3}}/2}$ . A vertical line between the curve and the x-axis will have a length of ${{\sqrt {3}}/2}$ when x = 1/2. So, following the logic of your diagram, the probability that a randomly selected half-chord will be greater than ${{\sqrt {3}}/2}$ is equal to 1 - 1/2 = 1/2! Since you don't favor this answer, I'm sure you can see why the logic of your diagram example is not necessarily the best or only way to approach the problem. I hope this has explained my position a bit better. I actually have enjoyed your critique for the most part and I'm sure you'll let me know if you still think I'm a crackpot.

63.194.28.214 23:49, 30 November 2006 (UTC) G. M. Sesiter[reply]

I think that you didn't hear at all about "jacobian" that has to be used in nonlinear transformation. So, if lines are verticaly ordered e.g. you can't get p=4/81 if you like y-axis or any other axis. It simply does not matter what you like. YOU HAVE TO COUNT CHORDS ! You can't change the origin set of lines if you want to count over y-axis. Just read your sentence: "This suggests, to me, that these particular methods are unecessarily simplifying the sample space. For example, I roll a fair die, The probability of rolling a 3 is 1/6. If I ignore or skip over all even rolls, then the conditional probability of rolling a 3 given that I roll an odd number is 1/3, etc." CORRECT! But it is not "unecesessarily simplifying" it is "not allowed transformation" which yield to wrong result. You can't change the original set as you wish. And where is the problem then? A you agree that for a fair die probability of chosing 3 could be 1/6, 1/2 (it is 3 or it is not), 1/3 or whatever ??????? Is it allowed to change the basic set as we want ?????? My example 2/9 has only one solution!

Yes I state that the answer is p=1/3. I have explained why other examples can't be used because they contain nonlinear trasformation (http://www.bertrands-paradox.com). In those two examples (p=1/2 , p=1/4) we started with some presumptions that are neglected in conclusion. In p=1/2 we naglect the fact that surface of the circle can't be covered uniformly by lines (semi radiuses), and in p=1/4 we neglect presence of midpoint of the circle that defines infinite number of chords.

Intention of my example here (p=2/9), wasn't analogy to paradox, just to clarify "jacobian", so any connection to paradox is groundless.

Best wishes Božur p.s. http://www.crnogorskoprimorje.com/fotke/administrator/2006-12-05_065216_example2.jpg

Intentions aside, I believe you did provide an analogy to Bertrand's paradox. Your graph of a segment of the curve y=x2 and my example of the segment of the circle x2+y2 =r2 are analogous. Therefore, either you should not support 1/3 as the solution to the paradox, or you should not support 2/9 as the solution to your diagram. You can't have it both ways. At any rate, I am familiar with Jacobian matrices but do not see the relationship to Bertrand's paradox. The paradox has always arisen out of uncertainty over the distribution of chord lengths. This uncertainty is not the result of not being able to find a correct unique solution. The point is that there is no unique distribution. By modeling selctions on just the lengths of the possible chords, themselves, ala the Uniform distribution, I do already have a linear relationship between the possible values selected and the lengths, I don't need a linear approximation. In your diagram, since we are concerned with the lengths of vertical lines, these lengths are entirely dependent on y coordinates. There is no need to change variables to come up with a length distribution. Your diagram is not a digram of a density function. If it where, you would have the length of the line (determined by y coordinates in your diagram) on the abscissa and densities on the ordinate. In fact, that's the one thing that is not explicit in the diagram--the distribution of line lengths. It is obvious that there is not a unique method to select lines in the area defined in the diagram. As a result, I'm afraid it does matter what I like, and I may base the distribution of lines on the y axis if I choose to do so, despite your prohibition. The new (second) diagram does not add anything new to the argument. However, if you mean strictly greater than 2, then the probability would indeed be 0 regardless of the method of line selection, since there is no length greater than 2 in the outcome space. Still, nothing new. If I select lines on the basis of y coordinates, then I will have one possible distribution with which I can determine the probability of getting a line between any two lengths between 0 and 2. If I base it on the x axis, the main difference is that I can find the probability of obtaining a line of exactly length 2, since this length is constant over all values of x from 2 to 9. Different distribution, different results, what else is new? Honestly, I think the only way that I can be convinced to change my preferred solution would be to use the source material as evidence. If you can show me where Bertrand was explicit in specifying a particular method for selecting chords (and therefore a particular corresponding distribution of chord lengths), then I will be convinced. This is the only new example I'd need to see. If not, then I'm afraid that I will be a lost cause to you. Either way, I must add that it has definitely been interesting, and no hard feelings.

63.194.28.214 23:17, 5 December 2006 (UTC) G. M. Sesiter[reply]

1. I didn't provide analogy to Bertrand's paradox. If we consider lines under x2+y2=r2 it is not Bertrand's paradox, just one approach to resolve it, but wrong because we can't cover surface of the circle with radiuses uniformly what is the idea for that way. (with sets of parallel chords)

2. I put corection in the last picture before your answer (not longer then 2, but equal 2 )

3. If I understand you well my last example is also paradox because it has as many results as we wish, because if we take care about lenghts (not about number of elements) we will get different result. Ok, I didn't precise density function for elements on x axis, but that schould be an unconditional solution to consider that as uniform on x axis, like the number of chords that we can find on the circle.

4 In my examples if we try to find result over y axis with Jacobian we will get correct answer.

5. Consider fair die with numbers 1,6,7,8,9,10. What is probability to get number greater then 5. If we overlook number of elements and just consider their values we will get 1/2 !!!

6 ........"By modeling selctions on just the lengths of the possible chords, themselves, ala the Uniform distribution, I do already have a linear relationship between the possible values selected and the lengths, I don't need a linear approximation."...... Unconditional solution is that on concircumference are endpoints of chords uniformly distributed rather then their lenghts, isn't it? Because we schould make uncoditional approach on the number of elements not on their characteristics.

regards Božur

If your argument is that we cannot cover the surface of the quarter-circle (my example) with lines under the curve unifromly then you should make the same claim for your diagram with the section of y=x2. Your solution of the diagram is therefore one possible approach, but not the correct one, as you put it. If this is an irreleveant point that you are trying to make, then due to symmetry, the solution to Bertrand's paradox should also be 1/2, since the probability of selecting a half-chord equal to the square root of 3 divided by 2 will be 1/2 following your same methodology of solving the problem you posed in your diagram. The endpoints of a given chord on the surface of the circle, where the chord intersects any other perpendicular chord drawn in the circle, the chord's length, are all characteristics of the chord, so I fail to see the point you are trying to make in #6. Furthermore, I have never disputed that once a particular method of selecting lines is decidied upon, then the distribution of chord lengths will be determined. So if your argument relies on selecting endpoints of chords then I have no argument with you as to your different solution. My argument has always been that it is not clear that that is what you should be doing. Bertrand never specified to do so, and obviously there are other approaches to follow that result in different distributions. Also, I did not follow your example in #5 about the fair die. Even ignoring the number of elements (if that were possible to do) I'm not sure how you arrive at a solution of 1/2. With a fair die, assuming no numbers are repeated on multiple sides, the number of simple events is equal to the number of different values. So I think that the interpretaion you made for your example, with p=1/2 does not hold water. What you might have said was that the die had 1 on one side and 6 on the other 5 sides. And then you could have tried to use this as an example of why the solution is not 1/2. But that would have been pointless, since I have never supported basing probabilities on anything other than the number of elements in the outcome space. That was your invention, of a misguided interpretation of what I was doing. My point has always been that when the number of elements in the outcome space is equal to number of values, then there is no distinction between the two. Finally, since you were not able to provide a definitive rationale for your solution by referencing the actual wording of Bertran's paradox, I assume that you cannot do so. Or is it still forthcoming? This the only type of evidence that will be conclusive. Other than that, I probably do not need any more examples that are "groundless" with respect to Bertrand's paradox, since after all, that is what we were ostensibly discussing. Unfortunately, I will not have time in the future to respond to anything else you post here, unless it is clear evidence that Bertrand was specific in describing a distribution for chord lengths that he had in mind. If you are able to do so, then I salute you! At any rate, I appreciate the time you have taken to argue your point of view, and have been impressed by your convinction. Not convinced, but certainly impressed. Best wishes.

63.194.28.214 20:12, 6 December 2006 (UTC) G. M. Sesiter[reply]

The surface of the circle could be uniformly covered only with sectors not semi radiuses!

http://user.sezampro.yu/~seik/Paradox%20Bertrand_22_23.GIF

http://user.sezampro.yu/~seik/Paradox%20Bertrand_24.GIF

That is why we can't obtain 1/2 as result. If we rotate paralel set of lines (that uniformly cover surface of the circle) we wouldn't uniformly pass over surface of that circle because farther part will go faster. So, one set of lines doesn't represent all possible chords. That is crucial.

I would agree with you if the task of Bertrand's paradox is: Find the probability that the length of a random chord will be greater than ${\sqrt {3}}$ and shorter then 2.

But we have inserted information that chords are on the circumference: Find the probability that the length of a random chord will be greater than the side of an equilateral triangle inscribed in that circle.

You sad:"My point has always been that when the number of elements in the outcome space is equal to number of values, then there is no distinction between the two." I agree, but I can't see where we can use it in Bertrands paradox ( if we apply that as number of elements=number of chords; their values=their lenghts ). If you use that as uncoditional solution, I think that we have more reason to use it considering points on the circumference as uniformly distributed along that circumference (that is always used in similar situations). So, I think that you schould provide proof for your quoted sentence i.e. that number of possible chords has linear transformation to number of possible lenghts or if it is unlinear, there schould be jacobian too. Wihout that it is not valid solution. Best wishes

Božur

OK, I will post one more reply. You state that we have inserted information that chords are on the circumference. This information is not inserted, it is unavoidable, by definition of a chord. By this same reasoning, we have "inserted" information that the chord will have a midpoint somewhere in the the circle, the chord will intersect a perpendicular radius at some point in the circel, and the chord will have some length. These are all characteristics that follow once you specify that we have a "chord" and they are each the basis of a different solution. You cannot focus on just one characteristic and assume that the others are not do not also necessarily exist for any given chord. Secondly, I can easily provide a proof that my preferred solution is linear to the number of possible lenghts. It's not complicated at all. My solution "randomly" picks from the possible lengths themsleves, so the correlation, which indicates the degree of linearity is equal to 1. If you plot my possible picks on the x-axis and the corresponding chord length on the y-axis you get a line with unit slope. Simple. Finally, my example with the solution of 1/2 that you cited does not in any way involve covering the surface of the circle with semi-radii. In fact, it doesn't invlove the entire circle at all. I merely select a quarter of the circle between, say, (-radius,0) and (0, radius). This section defines a curve. Just like in your 2/9 example. Again, following the rationale of your 2/9 solution, the probability that a line under my curve will be greater than the square root of 3 divided by 2 is equal to 1/2. I'm extending this as the solution for the entire circle, by symmetry. Now, what is crucial for you to understand is that I do not personally support this solution. Obviously, I support 0.134. However, my point is that if you hold to your 2/9 example as the only correct solution then you should also hold 1/2 as the solution to Bertrand's paradox, since it follows the same logic. Since you don't do so, and have stated that you think your 2/9 example is "groundless" as a basis for approaching Bertrand's paradox, then I ask you: Why did you present it at all? We are ostensibly discussing solutions to Bertrand's paradox here, so why devote numerous diagrams and paragraphs to something that is, in your opinion, irrelevant and off topic? Since a chord has numerous characteristics associated with it, and any one of those characteristics may be used as a basis for establishing a probability distribution for "randomly" selecting chords of a circle, what we need to definitively answer Bertrand's paradox is explicit guidance on which charcteristics should serve as our basis for approaching the problem. If you favor selecting enpoints, and imposing the resultant distribution on the problem, so be it. But that is an assumption, or preference on your part, not a conclusion based on anything explicit in Bertrand's paradox. Other people will have different preferences, as with anything else in life. Your prefer to have uniformity on the surface of the circle, but this choice will not have uniformity for the actual chord lengths, as the solution of 1/3 shows. Someone else may choice to have uniformity in the chord midpoints, and will therefore impose a different distribution on the chord lengths, which again will not be uniformly distributed. You can pick your chord characteristic and make it uniform, and then every other characteristic will not be uniform. So what? No big surprise there. The characteristic I favor to make uniform in my scheme of "random" selection is the actual length of the chord. All other approaches are indirect selections of lengths, through arbitrary (in the sense that they are based solely on personal preference) functions. For example, by selecting endpoints, you are slecting endpoints, and you get lengths secondarily to that selection. Since there is no way of making the length distribution uniform other than by directly selecting lengths, and selections based on other chracteristics of the chords yield different solutions, then from the standpoint of the chord lengths they are conditional solutions. If you chose to indirectly select lengths this way (using this function), you get this result. If you indirectly select lengths using this other characteristic (and this other function) you get this different result. Etc. If we do not indirectly select chord lengths, then the chord lengths themseleves are uniform, yielding the unconditional solution. Finally, you should know that by the time you are sitting down to determine the Jacobian you have already brought your assumptions in to change the problem since you will be incorporating your personal choice for selecting chords into the process. Of course once you have a particular system for selecting chords it is difficult to argue against the solution for that system. However, the problem is that there are multiple ways to do so, and there is no compelling reason to have to use any one particular method to the exlcusion of the others. I'm leaving in a couple of days on a personal matter and will likely have very limited internet access where I am going. I'd love to keep up this correspondence, although I think this section of the comments has gotten quite long already. It's getting to the point where I need to scroll up for what feels like a few minutes just to click the link to add my comments!At any rate, I doubt I'll be able to follow up with adding any replies, at least for the next few months. Oh well, take care and best wishes.

63.194.28.214 20:19, 7 December 2006 (UTC) G. M. Sesiter[reply]

I see that we schould clarify p=1/2 first.

Your sentence :"By this same reasoning, we have "inserted" information that the chord will have a midpoint somewhere in the the circle, the chord will intersect a perpendicular radius at some point in the circel, and the chord will have some length. "

1. Have you remark that each chord couldn't have separate midpoint on the surface of the circle? Infinite number of chords that are diameters have common midpoint .... so, doing on that way we change origin set of elements and that is not allowed !

2. Could you explain why we choose perpendicular radius on that chord? Wy not some line like http://www.bertrands-paradox.com/slike/1a.gif or http://www.bertrands-paradox.com/slike/1b.gif or http://www.bertrands-paradox.com/slike/1c.gif . Why radius? Why is radius important in considering problem?

Please focus your answer on #1 and #2 because that two questions are basis for understanding blunders in Bertrand's paradox. Because if you can't defend those states you wouldn't be able to defend multiple solution for paradox and your way which is simplest one among them.

2/9 is without relation to Bertrand's paradox because their tasks are completly different, and we can't change experiment and than show that it has similarity to Bertrand's paradox. I put example 2/9 only to show that lenght variation couldn't be taken in calculation if we don't have linear transformation between number of elements and number of lenghts.

Best wishes Božur

Well, I'm back. Sorry Božur, but where I come from, you need to first open the closed door before you can walk into the room. There is absolutely nothing in the statement of the problem at the center of Bertrand's paradox that suggests your solution. Why should we care about either of the #1 and #2 points above? Did Bertrand specify that the chord was to be chosen such that the density of points along the circle was to be Uniform. Or that the distribution of midpoints of chords was to be Uniform along a radius? He didn't specify anything about how chords would be drawn--hence the paradox. I believe our main difference is that, while I noted an alternative solution to the ones presented on the main page, I did not claim it was the only solution--a claim that would need to find justification in the wording of the problem. I only claimed that the solution was unconditional on the method of selecting the chord, since all possible chord lengths are treated as uniformly distributed. I also respectfully disagree with your rejection of your own 2/9 example, since the logic underlying it is consistent with another reasonable solution to the problem, assuming that chords are selected following the procedure otlined on the main page. The only way you can satisfactorily justify your position is to find support for it in the wording of the problem. Since I have made this challenge to you many times during our discussion, and since you have not yet responded to it, I don't believe there is anything more to be gained from this circular argument.

Take care, 63.194.28.214 00:53, 24 February 2007 (UTC) G. M. Sesiter[reply]

The contents seem to be too far down the page. It's pointless having a contents panel unless it is very near the start. Maybe there should just be one para in front of the contents and the detailed description of the problem moved after the contents.

Hi again :-))

you sad: "Why should we care about either of the #1 and #2 points above? Did Bertrand specify that the chord was to be chosen such that the density of points along the circle was to be Uniform. Or that the distribution of midpoints of chords was to be Uniform along a radius? He didn't specify anything about how chords would be drawn--hence the paradox."

If we suppose that midpoints are uniformly distributed along a radius we can't neglect that in the following step when we try to explain why that should be a radius. What connection has a radius with a circle in problem? I can't see that connection. If you see, I will be glad to hear that. If we suppose that midpoints are uniformly distributed on circle we can't neglect that in the following step when we say ... center of the circle represent infinite number of chords. Those two methods are anacceptable, we can't change our statments during problem resolving!!!!

1 condition ... start with uniformly surface and then change that density with something on your wish
2 condition ... start with uniformly distribution among one line without any explanation what does represent that line (Is it representiv particle of a circle???)
your condition ... forget the problem, use only limits, no matter how should looks the origin set of chords.

What challenge? I will repeat ... my example hasn't connection with paradox!!! Maybe it seems to you that it has but it hasn't. It only shows that if we have to find origin set we can't do whatever. Yes, if we are free to change basic problem we can say ... look it is like a part of the circle in #2, but it is not serious it is funny approach.

regards

Hello to you too :). This is fairly straightforward. What you claim to be "unacceptible" is entirely a matter of your opinion. Since Bertrand did not specify a method of selecting chords--it's not unacceptible to Bertrand! To answer your question about why the radius is important I note that it does not have to be, but it may be, depending on whether or not it is used as a basis for selecting chords. The position of a chord may be determined by its distance from the center of the circle, a distance which will vary between 0 and r. Therefore the length of a chord will be greater than the side of the inscribed triange if its distance from the center is less than r/2. Here's the importnat part: assuming that a chord is selected such that it's distance from the center is uniformly distributed between 0 and r, the solution to the paradox is 1/2. To make it more concrete, assume we have a random experiment set up to select random chords according to the following method. We have a circular disc with radius r that is thrown onto a table ruled with parallel lines exactly 2r apart. As a result, only one of those lines whould form a chord by crossing the disc. This method of selecting chords is certainly legitimate with respect to how Bertrand worded the problem, so if you insist that 1/3 is the only correct solution I'll need to see a proof that this experiment will yield a solution of 1/3! Obviously, as stated it will result in a solution of 1/2. The only way 1/3 can be the single true solution is if there is only one single true method for selecting chords, but you won't find that assumption supported anywhere but in your imagination. Since you cannot seem to find any support in Bertrand's wording of the problem, I will not require you to do so directly. Just prove that the above experiment will result in a chord longer than the inscribed equilateral triangle with probability of 1/3 instead of 1/2--that will suffice. I'm sure it will be an interesting exercise.

63.194.28.214 20:09, 6 March 2007 (UTC) G. M. Sesiter[reply]

Hello! :-))

Your modeling of the Bertrand's task with distance from circle center have one mistake. Zero distance will have infinite number of chords and all other chords will have only one point(distance) that represents them. So, your experiment have wrong assumption and it will lead to the wrong result, that is clear. (Imagine fair die, and probability of chosing 6 ..... ok, you say, all other numbers I will replace with one element and number six is the second element and my experiment proves that correct result is 1/2 .... NO, IT IS NOT CORRECT EXPERIMENT, you can't change number of elements on your wish! ... let aside the fact that in Bertrand's paradox we do not have discrete number of elements, the idea is in representative set that we can use in modeling)

Would you go along with my state about p=1/2 now?

Sorry to state the obvious, but any distance, D, from the center of the circle will have an infinite number of chords associated with it. This should be fairly apparent, since as you note, 0 distance has an infinite number of chords (since there are infinitely many radii). Each radius can be bisected perpendicularly with a chord exactly D units away from the center. However, this is a relatively pointless argument to have to make. The experiment I described is certainly a legitimate method for selecting a "random" chord since it does not violate any conditions stated by Bertrand in the wording of the problem. It is not your place to make up imaginary conditions in order to conclude that the experiment is "not correct". The only place it's not correct is in your imagination. If you want to convince anyone besides yourself, I would think that the bare minimum you would need to do would be to cite the specific part in the wording of Bertrand's paradox that forbids the consideration of experiments such as the one I have described, and only permits experiments of the type that will yield your arbitrarily preferred solution. You could also start by solving the experiment I previously described. It specifies a method for "randomly" selecting a chord, which is all that Bertrand requires as a condition for addressing the problem. If you insist that 1/3 is the only correct solution to the paradox, then it should be the solution to my experiment example, since the experiment is absolutely OK as far as the conditions specified by Bertrand go. If, on the other hand, your position is that different methods for selecting chords may yield different solutions, and despite having no specific support from the statement of the problem as posed by Bertrand you personally prefer the one that yields 1/3, then I heartily congratulate you on helping me to get this discussion to stretch so far down the page based on such an obvious conclusion. If you really stop to think about it, you'll see that I'm correct. Bertrand himself produced both the 1/3 and the 1/2 solutions. Don't you think that if he meant for the problem to have a unique solution he would have reworded it to make it more specific at that point? Whether you realize it or not, your personal interpretation of the problem brings assumptions to it that are purely arbitrary. If Bertrand had chosen to weigh the problem down under the baggage of specific conditions intended to restrict the methods of picking chords, it would no longer be a paradox. Now if you want to reword Bertrand's paradox so that the solution is clearly 1/3, then by all means, go ahead. Just don't call it Bertrand's paradox after that. You could call it Božur's problem.

Since Bertrand posed the paradox, it has been looked at countless times by many, many people. It's been considered from just about every angle. The solution I favor is not OR by me. I simply noticed it was lacking on the main page as one of the presented solutions. You can find another presentation of the solution here (and probably many other places as well):

http://mathworld.wolfram.com/BertrandsProblem.html

I personally prefer this solution (0.134), although I stop far short of claiming it is the only solution, or the "true" solution, whatever that means, since it is obvious that Bertrand intended there to be multiple possible solutions. I prefer it for the reasons I have already stated about 3-4 meters up this page. As the link I provided you points out, has the additional benefit of being insensitive to rotation, scaling, and translation of the circle, as does the solution (1/2) favored by the Bayesian scientist E. T. Jaynes. As Jaynes noted, these properties arise only if the solution is uniform over the radius, which is true for both the 1/2 and 0.134 solutions. 63.194.28.214 21:33, 8 March 2007 (UTC) G.M. Sesiter[reply]

I am sorry if you are little bit nervous about this topics, it wasn't my itention to be maybe unpleasent.
I know that many people had been looking on Bertrand's problems, and I know that Bertrand thought there are multiple possible solution.
"any distance, D, from the center of the circle will have an infinite number of chords associated with it" ... I know that too
"Each radius can be bisected perpendicularly with a chord exactly D units away from the center." --- ok
"since there are infinitely many radii" --- hm hm, yes there are infinitely many radii but infinitely many radii can't cover that surface uniformy. It means that one radius doesn't represent representative set of possible chords. Between each two radii we have a sector, and sector should be than representative set of possible chords in that case, but than we have problem with the circle center.
In other words, if we pick up points from the circle surface (that represent chords) we can colect three times more points/chords between tho radii that have distance longer than R/2. That is the reason why we can't use radiuses if we say that each point have one relative chord (let aside circle center).
That blunder is more obvious in example wich yield in p=1/4, where we compare surfaces with that preasumption where each point represent only one chord and vice versa, despite of fact that we all know that circle center will hide infinite number of chords!!!! I still belive (better to say "I know") that is not valid modeling/transformation!
My position is that 1/3 is correct answer because other examples involve mistakes in calculation/modeling.
I have studied all views about Bertrand's paradox, and interesting contemplation by E.T.Janes too, few years ago.
I see clearly what you are trying to say, but I'm not sure do you realy have read what I want to explain.

Nonsense! I haven't taken anything you've written to be unpleasant. On the contrary, I can honestly say that I generally enjoy reading your replies. I do understand the point you're trying to make. But with an inifinte level of precision, it becomes very abstract. True, between any given two radii there will be some angle that defines a sector. True, the number of points defined at a given level of precision between two raddi will be greater the farther away you move from the center of the circle. However, for any point at any level of precision identified on the surface of the circle you may darw a unique radius using that point and the center of the circle. As the level of precision approaches infinity, the angle betweentwo adjacent radii approaches zero, and the difference between the relative number of points between two adjacent raddi at various distances from the center aproaches zero. Also, as the level of precision in measurement approaches infinity, we may draw the circle smaller, and radius approaches 0 the distance between relative number of points between two adjacent raddi at various distances from the center aproaches zero. It's a bit more abstract in this case, however. Consider the first solution on the main page, specifically the diagram accompanying it. If we look at a fixed point on the surface of the circle, namely the point where a corner of the equilateral triangle falls on the circle, and then base the probability using the same rationale described for solution 1, then we have the same problem! Because one end of the chord is always fixed, you realize that the second point(s) for the chord, all located on the opposite side of the circle, are not increasing at the same rate--ie.e there are sectors between them and the surface-fixed point. It doesn't matter if you fix a point within the circle, like the center, or you fix a point on the surface, the same problem arises. The crux of this problem is that we are asked to determine the probability of a 1-dimensional element taken from a a 2-dimensional space. Consider, if we have a 2-dimensional figure, we can divide it using a line expressed in 2 dimensions (although it is 1-dimensional in itself). We can then define probabilities based on the areas created by the line. Likewise, if we start with a 1 dimensional line, we can define a point on the line to divide it into sections and then base probabilities on this division. In both cases the lower-dimensional element serves to create divisions in the higher dimensional figure, and then probabilities are based on measures of the higher dimensional figure (i.e., areas for 2 dimensions, and distances for 1 dimension, etc.). The probabilities are derived from the higher dimensional measures! But with Bertrands paradox, we are asked not to do this, but to find a probability corresponding to selecting a lower-dimensional element from a higher dimensional figure, a subtle difference.

I don't have any problem with your preferred solution. I think you've defended why you prefer it most admirably, and I'm actually fairly fond of it overall. It's a very Archimedean approach to the problem. For example, assuming unit radius for the circle an inscribed equilateral triangle has sides of length ${\sqrt {3}}$ . Following the rationale of picking chords described in the first solution on the page, you use 2 sides to define boundaries for chords less than the side of the triangle, so 2/3 of the chords will be of lower length. If we instead inscribe a square, and again use 2 sides to describe bounds in a similar manner, we have 2/4 of the chords being less than the side of the square ( ${\sqrt {2}}$ ). The difference in probabilities is 1/6, so 1/6 of the distribution of chord lengths falls between ${\sqrt {3}}$ and ${\sqrt {2}}$ . If we then define an inscribed pentagon, there will be 2/5 less than it's length. With a hexagon, 2/6 of the chords will be lower, etc. If the number of sides approaches infinity and we sum up all the differences in probabilities between the solutions for each adjacent polygon, then, you guessed it, the sum of the probabilities below a length of ${\sqrt {3}}$ approaches 2/3, which supports the solution of 1/3 for the probability that a chord will be longer than the side of an inscribed equilateral triangle. Very old school. See, even I can come up with ways to justify your preferred solution--I have nothing particularly against it. But to see my point...you need to focus on something that you wrote, perhaps without giving it much thought at the time. You wrote, and I quote with emphasis added by me:

In other words, if we pick up points from the circle surface (that represent chords) we can colect three times more points/chords between tho radii that have distance longer than R/2.

I agree, if we do one thing, like pick chords from the surface of the circle following an experiment that mirrors the one described for the first solution listed on the main page...we get a solution of 1/3. But if we do another thing, we get a different solution. The crucial point, which I keep needing to repeat, is that Bertrand did not specify that any method for selecting chords was to be preferred to any other method. The only thing that needs to be taken into account is that chords are to be selected "at random". So I return to the example experiment that I described with the disc and parallel lines ruled on the table. It satisfies Bertrand's sole requirement, that the chord be selected at random, so it cannot be simply dismissed as invalid. For any given level of measurement precision, it will result in a solution of 1/2, that's just the way it is. In fact, for any solution to Bertrand's problem (even the 1/4 solution that you detest the most) a valid experiment could probably be created that would result in supporting the solution. There is no unique probability distribution across all experiments. And that's the paradox. Now that doesn't stop anyone from arguing in support of a personal preferred experimental framework for tackling the problem. But such people necessarily bring preferences and justifications that are by definition arbitrary. But that doesn't mean it can't be fun to argue about such stuff. Or that such argument can't be well reasoned or elegant. Jaynes' take on the issue is fairly compelling, but not convincing (an impossibility). Your own arguments have generally been well reasoned. As for mine, I'm not so sure, but I've never claimed that the solution I mentioned was the only valid one. I just presented it becasue it wasn't listed on the main page. Somehow or other I ended up embroiled with you in this debate. I guess no one else was avaliable? :) At any rate, I hesitate to contribute anything else to this discussion, since, honestly, I don't think I have anything else worth contributing. I also worry that if some misguided soul tries to print out a record of this discussion there will end up being a world-shortage on printer ink and the loss of several trees! Best wishes and take care.

63.194.28.214 01:41, 15 March 2007 (UTC) G.M. Sesiter[reply]

Hi! :-))
(about your previous comment: The angle between two radii will approach to zero but related rates (number of points that we are interested in) will still the same. The same case is if we make radius shorter.)

Ok, I will summarize my point of view now as short as I can.
p=1/4: if we replace one chord with one point on the circle surface we will neglect infinite number of chords because of the circle center --- it is obviously not correct; no matter about distribution or whatever (in Bertrand's paradox postulate we do not have obligation to resolve this on that way, so we reject this method)
p=1/2: If we use one radius (as the representativ set of possible chords) it means that all radii comprise all points (i.e possible chords) but separatly! It is not correct, because of previously mentioned angle/sector rates, and we have to reject this method immediately.
p=aprox.(0.134): Considering only limits of chords we neglect need of counting elements, we involve our new set of elements that, generally speaking has no connection to problem.

I think it is too obvious. Your explanations were long but they didn't affect those crux problems(/mistakes) in resolving. Best wishes.
Božur

Hi Božur,

I'm afraid I didn't explain myself correctly, or you misunderstood the point I was trying to make, so I'll try one last time. If you look at the diagram for the first solution on the page (1/3), you'll see that there is an angle between all chords enclosed in the space between the two sides of the equilateral triangle. This space contains all chords longer than 1 side of the triangle. The points on chords near the corner that falls on the circle have a different rate than the points on the chords that are farther away from that point, i.e., those points that are near the opposite end of the circle. So, at a fixed level of precision, those chords will share many of the same points. This is true for any solution that relies on a fixed point on the surface of the circle, or within the cirlce, or whatever. Due to the nature of a circle, the rates change the farther you move from the fixed point. If you use the diameter (1/2 solution), and set up a system of parallel lines, then you can cover the entire interior of the circle with chords, and each point within the circle will fall on just one chord. At any rate, this is not a crucial argument.

The important point is that the problem, as stated, does not have a unique probability distribution. I have complained in the past that you have ignored my request to (1)find textual proof in the wording of Bertrand's paradox that explicitely calls for a unique method for selecting chords. You have not done so despite a rather lenghty discussion. I also now feel the need to complain that you are ignoring my disc example. (2)If you must insist on 1/3 being the only solution then you need to rationalize that argument with the experiment of tossing a disc onto a table that I described some time ago. Otherwise, your argument falls flat. Simply repeating it over and over again is not proof. I'm glad you find it too obvious. If you wish me to find it obvious as well, then you need to bring something new to the argument, becasue I am wholly unconvinved by your attempts so far. You could try to address either (1) or (2) above. If you cannot deal with these two points that are in obviously conflict with your reasoning, then I'm afraid that we're at a final impasse. While this has been an interesting discussion to this point, I cannot justify pointlessly going around and around in a repetitive exercise. If you see fit to adequately addess those points, then I will respond to congratulate you. Otherwise, I'm afraid this will be my final contribution to this discussion. Good luck, and take care.

63.194.28.214 17:27, 19 March 2007 (UTC) G.M. Sesiter[reply]

Hi Sesiter,

I am not forcing you to answer. If you wish to write you are welcome. Yes in 1/3 there is an angle between radii and middle points are not uniformly distributed than, and etc etc ..., but way you worry about that? We can have any distribution (Student, t, Gauss, uniform etc) but if we use some nonlinear transformation in our moddeling we MUST include in calculation distribution that is changed. So, which distribution you propose to use then, among those changed rates?

(1) There in no textual proof in the wording of Bertrand's paradox that explicitely calls for a unique method for selecting chords !!!!!!!!!!

But if you change during your moddeling origin set, than you MUST change its distribution in your calculation. So if you state in 1/4 that points are uniformly distributed on the surface, what kind of distibution we had in origin set (before yor neglecting/erasing infinite number of chords). Do you claim that in origin set each chord hadn't equal probability to be chosen, or maybe they had??? Why you are free to do that, to reject infinite number of chords???

(2) Is that your example with lines that are 2r apart? If so, I cant's see where is the difference. You can't get all possible chords by throwing so you have to use math to count them all on some way. You can measure distance from the center (illegal way) , their position on the surface (also illegal), etc, etc .... Because you are allways interested about all posible chords, you have to count them all, and choose some correct method for calculation.

"If you use the diameter (1/2 solution), and set up a system of parallel lines, then you can cover the entire interior of the circle with chords, and each point within the circle will fall on just one chord." It is not true, only rectangle could be covered on that way, where each point would have one appropriate line (in this case with rectangle we are able to have only two different set of parallel lines). If we want to calculate using uniform distibution among a radius, do we state simultaneously that it is not valid distribution, because chords with further distance will have less probability to be chosen????

best wishes
Božur

Hi Božur,

I think you've at least attempted to answer the points, so I don't think a response from me will be pointless at this time. My point about the angle between chords leading to not all points falling on unique chords was actually your complaint, from a while back in the discussion. To save you the trouble of hunting for it, I quote you:

That is why we can't obtain 1/2 as result. If we rotate paralel set of lines (that uniformly cover surface of the circle) we wouldn't uniformly pass over surface of that circle because farther part will go faster. So, one set of lines doesn't represent all possible chords.

So now you are basically refuting your own problem with the 1/2 solution! I'm glad to hear it.

(1) I find your response to this point is unsatisfactory. You are assuming that Bertrand meant to formulate the problem a specific certain way. I cannot imagine what this assumption is based upon, as "a chord is drawn at random from a circle" does not explicitly support the approach you are taking. In fact, it is fairly obvious that Bertrand did not intend there to be a unique solution, and thus was very careful in how th e problem was worded.

(2) Again, your answer is not satisfatory to me. You claim the experiment to be "illegal". Putting aside the unusual choice of description, how can you possible hope to prove this? "Illegal" with respect to what, exactly? With respect to how Bertrand framed the question? Certainly not! "Illegal" with respect to your bias? Perhaps, but who cares besides you? The experiment exactly satisfies the only requirements set by Bertrand, to wit: There is a circle and there is a chord drawn from the circle "at random". Bertrand wished to know the probability that the chord will be longer than the length of a side of an inscribed equilateral triangle. And if we have an experiment set up to offer a solution to this specific question, then we can solve for any other valid length as well. Your argument is misapplied: If we want to calculate using uniform distibution among a radius, do we state simultaneously that it is not valid distribution, because chords with further distance will have less probability to be chosen???? It falls flat precisely because the experiment will result in a Uniform distribution of lengths across the radius. It is difficult to argue with obvious reality. So your complaint is invalid---it cannot exist as a function of the experiment!

From my perspective, where you're making a mistake is in the consideration of the problem. You are attempting to find an area of the circle, which when divided by the total area will give you your solution. But we need to consider chords, which are infinite. A circle is not a probability distribution! Finding the area within the circle is not the same as finding the area under a density function. You can tell them apart easily enough, the density function has the possible values for chord lengths on the x-axis, and densities on the y-axis. There is no unique density function for drawing chords "at random" from a circle. This is because the probability distribution is generated and defined by the method by which the chords are selected "at random", i.e., by the experiment set up to select chords. Bertrand only requires that any experiment selects chords at random. The disc experiment will yield an outcome space that is complete (exhaustive) and well defined. In that outcome space, 1/2 of the chords will be less than the side of an equilateral triangle. Your preferred solution defines a different outcome space. It is different because the method for selecting the chords is different. It contains more chords less than the side of an equilateral triangle, therefore the solution 1/3 is lower than 1/2. My preferred solution of 0.134 contains even more chords less than the side of the triangle. And such an outcome space can also be easily set up by a carefully considered experiment. The same is true for the 1/4 solution, an experiment can be set up such that the outcome space is fully defined and exhaustive with respect to the experiment. That is the only criterion necessary for determining a solution!

I appreciate the vigorous defense your are making for your position, but my interpretation of Bertrand's paradox is not compatible with it. I think you have good points, and a careful rationale when talking about your preferred solution. But I cannot support that it is the only valid solution based solely on your arguments, becasue I think you are arguing from the wrong interpretation of the problem. We seem to be positioned at two different perspectives with little common ground. Oh well, I suppose neither one of us can convince the other. But no hard feelings! I have enjoyed this discussion with you and can honestly say that I've benefited from your thoughtful analysis. Take care.

63.194.28.214 17:16, 21 March 2007 (UTC) G.M. Sesiter[reply]

I am not refuting my state, I wonder way you say so. I sad just that one radius doesn't have representative set of possible chord's. Your conclusion is really strange. ...... Ops (1),(2).... again and again ... your conclusions are arbitrarly and superficial. I don't think that Bertrand meant one way for resolving.....oh there are so many questions and our discussion is now too wide. I will try again to focus our attention on just one question. If we make progress we can go further.

BERTRAND's TASK: Consider an equilateral triangle inscribed in a circle. Suppose a chord of the circle is chosen at random. What is the probability that the chord is longer than a side of the triangle?

SOLUTION SO CALLED 1/4: A chord is fully determined by its midpoint (WRONG, center of the circle represent INFINITE number of chords, and and all other points only ONE). Chords whose length exceeds the side of an equilateral triangle have their midpoints inside a smaller circle with radius equal to 1/2 that of the given one. Hence, its area is 1/4 of the big circle which also defines the proportion of favorable outcomes - 1/4.

DO YOU AGREE WITH THIS SOLUTION OR NOT. (TO NEGLECT INFINITE NUMBER OF BASIC ELEMENTS WITH THIS MODDELING OF THE PROBLEM)

JUST.. YES OR NO.
IF YES, I DO NOT SEE RATIONAL REASON TO CONTINUE THIS DISCUSSION, IF YOUR ANSWER IS NO, WE CAN TALK ABOUT 1/2 AND 0.134 SOLUTIONS.

Of course, no hard feelings. Just to make it faster and more productive.

all the best Bozur

Hi Božur,

I'm afraid I don't see it the way you do. I see the probability distribution arising from the method of selecting the chord. As for the solution of 1/4, yes, I do believe it's possible to set up a valid experiment that has 1/4 as the correct solution. Just as descibed on the page. The probability derives from the location of the midpoint, and on whether the midpoint is in the smaller circle, or not. If you condition on the location of the randomly selected midpoint (inside the smaller circle or outside), then p(longer|inside smaller circle) = 1 and p(longer|outside smaller circle) = 0. Using the law of total probability, the area of the smaller circle = 1/4 the total area of the circle, the area outside of the smaller circle is 3/4 the total area of the circle and the solution is 1/4. And why are you overly worried about a single point(the center)anyway? Surely you realize that the position of the midpoint is a continuous variable? And if we have a continuous probability distribution the presence or absence of a single point makes no difference. I mean p(x < a) = p(x ≤ a), and p(x > a) = p(x ≥ a), etc. But you do raise another valid point. This discussion has long been cycling back and forth with no ground gained on either end. Probably the best thing for you to do would be to try to publish your view on the paradox in an appropriate peer-reviewed medium. You're obviously a bright person, with some good math skills, so this seems to me to be the most promising next step that you could persue. If you can get enough acceptance for your position from the mathematical community, then perhaps the Bertrand's paradox page could be modified to reflect the single "true" solution to the problem that you claim exists. Once you publish a solid proof, without obvious flaws, then your solution should have no trouble gaining mainstream support. At that point, this entire argument will be moot, since it's just not rational for you to be overly concerned with what I personally think. Good luck!

63.194.28.214 18:25, 3 April 2007 (UTC) G.M. Sesiter[reply]

Yes, I know that position of the midpoint is a continuous variable, and probability of choosing that point is zero. I can accept that 1/4 is good experiment but for some different problem; it is without connection to bertrand's task, bacause if we go backwards we can't recreate all possible chords from points of the circle surface using that state that each point is midpoint of only one chord. We can't recreate infinite number of diametars. If circle center had replaced finite number of chords, we would be able to use that experiment.
Thanks for your wishes, but till now neither my professors, nor math community in my country had understanding for this.
bye
Božur

Objection to the last two sentences of the article

It seems to me that the article is fine in almost all respects, to the extent that it reports judiciously on the literature ("judiciously" in that there's a lot of literature on the problem definitely not worth reporting on).

The one misgiving I have is at the end, "As Jaynes pointed out in his paper, the main question is whether one should think of Bertrand's paradox as a physical experiment. Nobody said it should be one, and if one does not think of it as physical, then the common-sense solution described here does not apply."

This is not the main question at all. For it to be so, Jaynes must commit the logical fallacy, if P implies Q then Q implies P. Having shown that a certain physical experiment suffices to imply a certain distribution of chords, he then wants that conclusion to depend necessarily on physical considerations.

A better statement than "method 2 is the only solution that fulfills the transformation invariants that practical physical experiments would have" would be "of the three methods, method 2 is the only one based on a distribution that is independent of the radius and position of the circle." This is a simple and natural geometrical condition making no mention of physical experiments yet incorporating the core of Jaynes' "common-sense" solution.

(This might seem to be contradicted by the highly nonuniform distribution of midpoints for method 2. However the concentration of midpoints at the center of the circle is not a property of the distribution itself but rather of the distribution combined with the circle, since the circle determines where the midpoints go. The "Chords chosen at random" figures aren't intrinsically biased by the circle in this way, and the lack of influence of the circle on method 2 is quite visible. In contrast method 1 pushes the chords away from the center of the circle to a slightly noticeable degree, giving the distribution that most looks like a ball of yarn, while with method 3 that tendency is very visible, as it needs to be in order to offset the large bias introduced by the circle if it is to achieve the uniform distribution of midpoints shown at the upper right.)

Physics can be shown to be even less relevant by coming up with other physical experiments justifying either of the distributions associated with methods 1 and 3, by appropriately incorporating the circle into the experiment. Not only is method 2 not the mathematically correct solution, since there is no requirement that the distribution be independent of the circle, it is not the physically correct solution either, for the same reason. Physics is a total red herring here. That Jaynes' article appeared in Foundations of Physics may have more to do with the (very) eclectic tastes of that journal's editors than with its relevance to physics.

So while Jaynes may be correctly quoted here, I don't think it's appropriate for the article to accept his visibly fallacious reasoning at face value. To do so understates the defensibility of method 2 by ignoring a much simpler and mathematically natural justification for it, that it's the only distribution that doesn't follow the circle around. Vaughan Pratt 01:24, 8 April 2007 (UTC)

If you say that it seemd to you that article is fine, does it meen that you also agree with neglecting infinite number of chords in method denoted with "3"?
Because center of the circle is common midpoint for many chords so it hides infinite number of chords that are diametars.
Is it allowed to make a model/experiment on that way, and provoke erasing infinite number of elements?

Božur

I agree with Vaughan--method 2 is not the physical solution. In a physical problem we can ignore the exact center of the circle (because it can never be physically determined), and therefore ignore the problems associated with the infinite number of chords that eminate from it. Ignoring, then, these diameter chords, any chord can be uniquely described by its midpoint within the circle. Of the three methods presented here, the only one that has a distribution of midpoints that is translationally and rotationally invariant is method 3. Therefore, this method could just as easily be described as the correct physical solution. Snydley (talk) 02:08, 6 December 2007 (UTC)[reply]

Image for selection method 3 is not visible

This image:
http:/upwiki/wikipedia/commons/thumb/8/8b/Bertrand3-figure.svg/161px-Bertrand3-figure.svg.png
is not visible in my browser (Opera 8.54) either on the article page or on the image page:
http://en.wikipedia.org/wiki/Image:Bertrand3-figure.svg

Curiously, the link on the image page works:
http:/upwiki/wikipedia/commons/8/8b/Bertrand3-figure.svg

I am not familiar enough with how images work in Wikipedia to see what needs to be tweaked to make this image visible in the article.

Please help. Ac44ck 16:27, 14 July 2007 (UTC)[reply]