|Hanspeter Schmid Switzerland 4/18/2006 9:28:56 AM|
|E. T. Jaynes gave a very good mathematicalm argument using transformation groups in his 1973 paper "The Well-Posed Problem". See
for a brief explanation and reference.
|Bozur Vujicic Serbia 4/25/2006 9:03:49 AM|
|You simply didn't catch the point. I have explained here why we can't use that second aproach in consideration. But if we do so, we could say the same thing .. it is invariant in some cases! But, be aware - that aproach for our problem is not appropriate!!! |
|Brian McMullen USA 10/15/2006 7:32:42 PM|
|The problem is not the definition of the word random, it is the definition of the word chord. You want your solution as ignorant as possible about the relationship between circles and triangles. Method 3 does not work because you are generating chords based on polar coordinates from an arbitrary orgin, weighting the area near the orgin far more heavily than the rest of the circle. It would be ideal to generate chords based on two random points within the circle, because the easiest way to represent chords evenly in respect to area is to generate them with respect to area, via random coordinates within the circle, you should get the same result wether you use polar coordinates weighted for distance from origin(center of circle)(the probability of a point would need to increase at a square of r) or unweighted xy coordinates(easiest to program id guess)
To confirm, You randomly generate two points within the circle, then you extrapolate the chord represented, if it intersects the small circle you enumerate a variable. the probability is collisions/iterations. I will probably post more when I am more rested.
my guess is something like 70-75% probability the chord will be longer than the leg of triangle.
even looking at example 1(the highest probability here), the area represented by 50% of the chords in your integration is far less then 1/2|
|Brian McMullen USA 10/15/2006 8:37:22 PM|
|It would be the same result if you used a fixed point on the circumfrence and a random point within the circle to generate the solution|
|Bob USA 10/16/2006 1:56:13 PM|
|Brian McMullen USA 10/16/2006 8:46:09 PM|
|I ran it with 100M iterations with a probability of 0.2169|
|Brian McMullen USA 10/16/2006 8:54:55 PM|
|With Better collision detection 0.2171(using a circle of radius 5000)|
|Bozur Vujicic Serbia 11/6/2006 4:44:46 AM|
I did not understand yor statment well. You sad:"you are generating chords based on polar coordinates from an arbitrary orgin, weighting the area near the orgin far more heavily than the rest of the circle". In respect to what I weighted some chords in calculation?
Your example involves presumption that longer chords have more probability to be chosen because they consist more points (among diametar we can choose more points to determine one chord than in other direction). Why? In that case each chord is not equaly probable and we are changing the original problem! It is not allowed. Do you agree?
Consider similar problem where the circumference is not closed, where small part "dl" of it is missing (like shape of the letter "C"). How can you connect area with the number of chords then? So, here is crucial not to make fake transformation. If you have to count chords you can do it via area only if you are able to prove that transformation as bijection (one to one transformation).
Thanks for your comment Brian!
|zcer malaysia 3/15/2007 3:55:08 PM|
|i have a possible resolution here at http://miniverse.blogspot.com/2006/12/solution-to-bertrands-paradox.html
i'm no mathematician, so it's not at all rigorous. But somehow, i think it's correct. i'd really like your opinion|
|Bozur Serbia 3/22/2007 9:20:17 AM|
|Hi I foud your work accidentally and scanned very fast that page. I have to look it again tenderly.|
|Bozur Serbia 6/14/2007 5:57:55 AM|
|zcer you have to agree that midpoint of the circle makes hidden infinite number of chords, so that way can't be acceptable.
|Jo Germany 9/10/2007 7:06:48 PM|
|I think your refutation of the first example is wrong. It is not easy for me to understand what you exactly mean, but to me it seems that your main objection is that (in your opinion) not every chord can be caught by choosing an arbitrary point on an arbitrary radius as midpoint of the chord. This is wrong since every chord has a midpoint and each point, especially every midpoint is lying on a radius. Furthermore it is wrong that one cannot get the surface of a circle by integrating r. If you integrate r over [0,R] and over [0,2*Pi], this Integral yields the correct result Pi*R^2.
Maybe I didn't get your point exactly. But I think Bertrands paradox should not be called a "blunder". If there was a simple mistake mathematicians would of course have discovered it in the meantime. However, I have to admit that I am not content with the explanation, all three solutions would equally be correct; I for myself sympathize with the solution 1/2 ...|
|lukomor Ukraine 9/24/2007 8:29:31 AM|
|The right answer is 1/2, but no one of three classic solutions is not right.
|Bozur Serbia 10/23/2007 6:19:35 AM|
|Jo, you sad: ''If you integrate r over [0,R] and over [0,2*Pi], this Integral yields the correct result Pi*R^2.'' --- yes, but in that case you didn't integrate radii but sectors, and that's big difference. By integrating radii you can' get circle shape (but rectangular)!! And on the surface of the sector where angle is extremely small you allways have three times points more on the farther part (r>R/2)!!
''If there was a simple mistake mathematicians would of course have discovered it in the meantime. '' --- I think it is not just simple mistake and I deeply belive that math is improving all the time as any other science.
I can't say that I sympathize any of approaches but some of them I can't accept (math rules doesn't allow that), so the remaining solution is 1/3.
|Bozur Serbia 10/23/2007 6:20:51 AM|
|Lokomor, could you explain your example which yields to 1/2 for Bertrands paradox?|
|Lukomor Ukraine 10/24/2007 10:54:17 AM|
|Bozur (Serbia 10/23/2007 6:20:51 AM)
Lokomor, could you explain your example which yields to 1/2 for Bertrands paradox?
- - - - - - - - - - - - - - - - - - - - -
Of course, I can explain it.
But the explanation could be extrimely long.
I can't do it in a few words.
|Bozur Serbia 10/31/2007 5:19:59 AM|
|but this field can accept as many words as you wish|
|Lukomor Ukraine 11/1/2007 11:55:32 AM|
|Well, let's start.
First of all I should remind, that "practice is the criterion of truth".
The only practical (physical) experiment was made by E.T.Jaynes.
You may accept or not accept his theoretical reasonings, but some quotations from his "Well-posed problem" (1973), in which "the unique solution has been verified experimentaly", can convince anyone.
1. "It will be helpful to think of this in a more concrete way; presumably, we do no violence to the problem (i.e., it is still just as "random") if we suppose that we are tossing straws onto the circle, without specifying how they are tossed.
We therefore formulate the problem as follows.
A long straw is tossed at random onto a circle; given that it falls so that it intersects the circle, what is the probability that the chord thus defined is longer than a side of the inscribed equilateral triangle?"(E.T.Jaynes)
2. "The Bertrand experiment has, in fact, been performed by the writer and Dr. Charles E. Tyler, tossing broom straws from a standing position onto a 5-in.-diameter circle drawn on the floor. Grouping the range of chord lengths into ten categories, 128 successful tosses confirmed...[p=1/2]... with an embarrassingly low value of chi-squared".(E.T.Jaynes)
|Lukomor ua 11/5/2007 6:43:30 AM|
Well, in "Part I" we have convinced, that physical experiment gives an unique solution of P=1/2.
So, we should find the point where "the wonderful mind" falls down to a blunder.
In my humble opinion, this point is, that the term "a chord" is complex.
"A chord" is the result of combination a circle and a straight line.
Henry Poincare in "Calcus des probabilites" (Paris, 1912) shows that there are two ways to get
"a random chord".The first way is a combination of a fixed circle and a random straight line.
This way leads to paradox . Another way is combination of a fixed straight line and "a random circle".
This way gives an unique solution.
If we fix a straight line on a plane, and will tossing a circle (for example a coin with radius R) "by
chance" to a plane,
the coin will cover a part of a straight line, if the distance from the line to the center of coin will less than +/-R,
And the covered part of line will be longer then a side of triangle, inscribed to a coin, if the distance from a straight line to the centre of coin will be the less, then then +/-(R/2), i.e. d=R.
Thus, we'll find out, that probability is: P=d/D=1/2.
In "Part III" I'll shaw that both two ways are equivalent.
|Bozur Serbia 12/14/2007 9:01:39 AM|
|About Part II ===== There in no problem how to perform experiment but how to count chords as result of that experiment. Why to measure distance from our center to that chord? The problem is the length of that chord and function between its distance and lenght is not linear transformation and that will obviously lead to mistake in counting!!! (y=x2, x(0,4); y(0,16) p(x<3)=3/4 among x, p(y<9)=9/16 among y, p(x) in not equal p(y) because of nonlinear transformation, we can't use p(x) for p(y) and vice versa !! )
Abaut Part I ===== If we tossing lines we will do it uniformly in all direction, and we know the result before counting.
Experiment for p=1/3 == Fix an end of one line on one point of the circumference than rotate that line by chance around that point and count how many times appropriate chord has lenght longer than critical, probability will be p=1/3. Do you agree? Is it valid experiment? So, your experiment can't be argument in our discussion. We should go back on motives for counting on p=1/2. Did you read my state about p=1/2, what do you think, is it reasonable?|
|Lukomor .ua 12/18/2007 9:28:58 AM|
|L:First of all I must be apologized for so long a silence.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Bozur: -Experiment for p=1/3 == Fix an end of one line on one point of the circumference than rotate that line by chance around that point and count how many times appropriate chord has lenght longer than critical, probability will be p=1/3. Do you agree? Is it valid experiment?
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
L: I cannot agree with this assertion.
You arbitrarily fixed the point at a distance of d=R from the center of circle.But this is not necessary condition.
For example, I can select the point at a distance of d=2*R from the center, then probability is equal: P=arcsin (R (2*d))/arcsin (R/d) of =arcsin (1/4) of /arcsin (1/2) of =0,48....
It is not possible to base the selection of point precisely on the circle.
|Bozur Serbia 12/20/2007 8:55:07 AM|
|Let's suppose your point is not on distance of d=2*R but d>>2R it leads to solution p=1/2. So, your example belongs to that approach. (I am not picking the point from the plane surface but from the cirumference that is somewhere in 3D space, so I didn't take one point that is accidentally on the circumference on that imagined surface, but for that solution only possible choosable points are on that circumference. How would you find probability in case the circumference is not plane?)
My idea is that for Bertrand's paradox problem only correct solution is p=1/3 because in that case we can't find infinite number of chords that are neglected!!!
In approach that leads to p=1/4 we neglected infinite number of chords whose lenght are d=2R. If we can do so, we can repeat that more and more and neglect all chords that has similar lenght and get new solution p->0.
In approach that leads to p=1/2 we neglected infinite number of chords that has distance smaller than critical. Because for the same grid, we can create more (shorter) chords that couldn't be covered with that grid. Or, in other words, in that approach one set of possible chords are normally on one radius (its midpoints), so all radii consist all possible chords. Surface of the cirle can't be covered with infinite radii! so we neglected shorter chords and increase real amount of probability. Or let's go back on your point with distance d>>R (for simplicity I took that point, not d=2R). If you change angle for the same small amount when the line from our point is near the center of the circle and when the line is almost tangent, it is obviously that the curve line that we pass over in second case is longer and on that part we could create much more chords than in the first (between your two chords - part of lines). Therefore our chords are not equally probable. Theory says, we can neglect only finite set in one infinite set. We could neglect infinite subset if his rate to appropriate infinite set is zero, but we haven't that here. |
|Lukomor Ua 12/22/2007 2:07:21 AM|
Bozur (Serbia 12/20/2007 8:55:07 AM)
Let's suppose your point is not on distance of d=2*R but d>>2R it leads to solution p=1/2.
= = = = = = = = =
Let's suppose the point is not on distance of d=R but d=R+delta_R, or d=R-delta_R. (delta_R<|
|Lukomor UA 12/22/2007 2:11:54 AM|
|Sorry, I repeat...
Note #1.========================= Bozur (Serbia 12/20/2007 8:55:07 AM) Let's suppose your point is not on distance of d=2*R but d>>2R it leads to solution p=1/2. = = = = = = = = = Lukomor: Let's suppose the point is not on distance of d=R but d=R+delta_R, or d=R-delta_R. (delta_R<|
|Lukomor UA 12/22/2007 2:13:53 AM|
|Lukomor UA 12/22/2007 2:16:57 AM|
Note #1.========================= Bozur (Serbia 12/20/2007 8:55:07 AM) Let's suppose your point is not on distance of d=2*R but d>>2R it leads to solution p=1/2. = = = = = = = = = Lukomor: Let's suppose the point is not on distance of d=R but d=R+delta_R, or d=R-delta_R (delta_R less then R).
In this case the probability will differ from P=1/3. And we must examine this case (we should not neglect it).
|Lukomor UA 12/22/2007 4:57:21 AM|
|Part III: ****************************************It is first necessary to give some determinations. ***Determination №1. A random straight line on the plane - a straight line which passes: 1). With the equal probability through any point of plane; and 2). With the equal probability in any of directions. ***Determination №2. Random circle on the plane - the circle, whose center with the equal probability is located at any point of plane.
Now let's conduct the classification of the problems, which appear with the examination of the mutual arrangement of straight line and circle. ****************************************Problems of the first level. ****************************************Problem 1*) What is probability that the random straight line will cross the fixed circle of a radius R? ****************************************Problem 1#) What is the probability that the random circle of a radius R will cross the fixed straight line?**************************************** Problems of the second level. ****************************************Problem 1.1*) What is the probability that the random straight line will cross the circle of a radius of r=R/2, when it did cross the fixed circle of a radius R? ****************************************Problem 1.1#) What is the probability that the random circle of a radius r=R/2 will cross the fixed straight line, when it's concentric circle of a radius R crossed this straight line. ****************************************
Problems of the third level
**************************************** Problem 1.1.1*) What is the probability that the random straight line will cross the circle of a radius r=R/2 under the conditions that: 1). This straight line crossed the fixed circle of a radius R; 2). this straight line was carried out through the point, which lies infinitely far from the center of circle. ****************************************Problem 1.1.1#) NO. ****************************************Problem 1.1.2*) What is the probability that the random straight line will cross the circle of a radius of r=R/2 under the conditions that: 1). This straight line crossed the fixed circle of a radius R; 2). This straight line was carried out compulsorily through the point, which lies at a distance d less than R from the center of circle; 3). This straight line is carried out compulsory to direction perpendicular from the selected point to center circles.
**************************************** Problem 1.1.2#) NO. *****************************************Problem 1.1.3*) What is the probability that the random straight line will cross the circle of a radius of r=R/2 under the conditions that: 1). This straight line crossed the fixed circle of a radius R; 2). This straight line was carried out compulsorily through the point, which lies at a distance d=R from the center of circle; **************************************** Problem 1.1.3#) NO|
|Lukomor Ukraine 12/24/2007 7:29:28 AM|
Let us examine problems set in Part III in more detail. Problems of the first level [1*) and 1#)]. They have a result P=0, since an infinitely small quantity of random straight lines will cross the circle of final diameter on the infinite plane. The problem of second level 1.1*) has a result P=1/2, although this is unevident. However, obviously that a result P=1/2 has the task of second level 1.1#). These tasks maximally correspond to that posing of the problem, which we find in Bertrand’s problem. The tasks of the third level have additional conditions; therefore the probability, which they determine will be conditional. These additional conditions, which are absent in the original problem of Bertrand, change the result of solution. Only in task 1.1.1*) the influence of additional conditions does not change the result.
|Bozur Serbia 12/25/2007 4:52:11 AM|
|Note #1 ===== If delta_R < R then p=1/3+delta_p and as delta_R increases p becomes p=1/2. I emphasize case when delta_R is large because mistake is then more obvious. If we are aware of that we are not interested in any other delta_R (we are not neglecting it). ===== (soon I will post answer for part III and IV)|
|Bozur Serbia 12/25/2007 9:05:25 AM|
|Your preassumption on the beginning ..."2). With the equal probability in any of directions." ...caused p=1/2. So, there is no difference if we are tossing lines on the circle or circles on the line, the result must be p=1/2 because of the first statement that lines will fall uniformly in regards to angle. Bertrand's paradox offer several approaches with different uniform distribution. One with uniform in regard to angle which will lead to p=1/2, second is uniform distribution in regard to surface of the circle which leads to p=1/4 and third in regard to distance (and angle from choosen point) which will result in p=1/3. Experiments of each of these approaches could be performed in more than one way and we can discuss about them like here, I don't see why is important only discussion about p=1/2 i.e. why we have to reject immediately other approaches. ----- Especially when it is easy to verify that p=1/2 has weaknesses, as I explained before.
...................................... Is it difficult to find answer for one similar task where one small part delta_l of the circumference is missing?.|
|Lukomor UA 12/26/2007 4:31:18 AM|
|B:If delta_R < R then p=1/3+delta_p and as delta_R increases p becomes p=1/2.
L:And what abaut d=R-delta_R.
Why should we neglect that chords?|
|Lukomor UA 12/26/2007 4:37:03 AM|
Experiments of each of these approaches could be performed in more than one way...
L:But there is the only experiment if we take a fixed straight line + random circles.
And the only result of that experiment is P=1/2, isn't it? |
|Bozur Serbia 12/28/2007 7:21:20 AM|
|If we choose one point inside the circle, we can't create all chords using that point. For example, if we choose delta_R=R/2 (then d=R-R/2=R/2) it is impossible to create chord whose lenght is d=R/10, R/20, etc. So, it is meaningless. Similary situation, but less obvious is for d=R-delta_R.|
|Bozur Serbia 12/28/2007 8:26:53 AM|
|Yes, if you take fixed line and random circles you will get p=1/2; but question is "why should we use that experiment for Bertrand's problem". Someone can say we have closed curve line (circumference)and we are picking points from that line that represents ends of chords. Probability for first point is 1 and we do not have to do that, and for second that fulfill Bertrand's task is p=1/3. Or we can play darts and then if choosen point is middle of the chord, probability is p=1/4. But your experiment and experiment with darts have nonlinear connection with Bertrand's task and they are causing mistakes. (for p=1/2: sets of parallel lines can't represent all chords uniformly, and for p=1/4 one of the problem is center of the circle)|
|Lukomor Ukraine 12/28/2007 11:55:26 AM|
|Bozur: "If we choose one point inside the circle, we can't create all chords using that point."
L:But we should choose: 1).ALL the points of the Plane one by one.
2).ALL the angles one by one over each of the point.
|LUkomor Ua 12/29/2007 6:54:23 AM|
|Happy New Year, Bozur!
|Bozur Serbia 12/31/2007 4:21:11 AM|
|Yes, we can split all points on the plane in three groups. I: inside the circle, II: outside and III: on the circumference. ========= I: If we choose one point inside the circle we can't create some chords (for the center there are only radii and other chord's that consist that center point can't be made), II: if the point is outside then chords are not equally probable. So, that approach are useless (or ultra difficult for counting), and there is no need to move our basic set to points on the plane surface? Yes, chords and circumference are in the same plane but involving that plane doesn't give any benefit. If we have line y=x2 with lenght 4cm and task is to determine probability of choosing point that is farther then 3cm from one end line. We can observe problem in XY plane and look only on x or y axis (without appropriate transformation)and get obviously wrong result (or straight distance from (0,0) for example). So, p=1/3 has to be only correct solution for Bertrand's paradox!|
|Bozur Serbia 12/31/2007 6:04:57 AM|
|Lukomor, wishing you happy New year and a merry Christmas!
|Esteban Treviño Mexico 4/26/2008 3:21:22 PM|
|Good to find someone else recongize the blunder... and see that the congnitive illusion surrounding 'the paradox' ... Note that the probability of 1/4 results from taking the 1/4 area vs 1 area when all we logically know is that the random chord resides either in the 1/4 area or the 3/4 area... thus the probability is (1/4)/(3/4) = 1/3. I could go on to show how the other examples I explored also involve some inference error that when corrected also leads to 1/3...
The implications are prodound precicely because this example is often used as a model for explanation when we can't determine something, (and get different aparent correct answers) specially in theory of probability. ... the irnoy is that there is one precise correct answer...
Email at firstname.lastname@example.org|
|Bozur Serbia 5/9/2008 9:43:28 AM|
|Esteban, I think that preassumptions in some approaches are wrong, and I doubt that they can be corrected in your way. Because geometrical probability states on p=(part of the area)/(total area). Question is why to use nonlinear transformation between number of all possible chords and points on the circle surface that represent their mid-points. There is no valid explanation for that, because center of the circle represents infinite number of chords as I explaind above.|
|Esteban Treviño Mexico 5/17/2008 12:02:10 AM|
|Indeed preassumptions in some approaches lead to the wrong answer... never the less they can be corrected...(I hold) Assuming that one will get the right answer by using the geometrical probability of p=(part of the area)/(total area) is questionable... Why not assume that the probability p=(an area of larger chords)/(an area of lesser chords)... the mid-point representation for all possible chords leads to P= (area of larger chords- a cicle)/(area of lesser chords- a donut) that is (1/4)/(3/4)...==> 1/3 ... The assumption here is that the random mid-point chord choosen falls either in one area or the other area with equal chance because thats the only logical sustainable-congruent position one can hold.
A 'valid' explanation for ignoring the center mid point given to me was that calculating the probability of that particular point is zero... and thus can be ignored (and with the same logic so is the probability for any other point)...
My approache assumes nothing about linear/nonlinear transformation it simply considers the larger chords compared to the smaller ones ... and in both cases the linear and the nonlinear representation I get the same answer... leading me to hold that its the answer...|
|Bozur Serbia 5/19/2008 5:16:09 AM|
|Theory of probability is based on following formula, p=(set of elements which probability we want to determine)/(set of all elements).|
So, if we say that each chord is determined with its mid-point (and they are spread over all surface of the circle) and chords that have lenght longer then critical have mid-points only inside circle with two times smaller diametar, probability becomes p=(R2/4)/(R2)=1/4.
Yes ... probability just for mid-point is zero, but that point represent infinite!!! number of chords, so that model where we do nut substitute one chord with one point (or finite number of chords) can't be discussed as option for resolving our problem (and any other problem in probability without using appropriate jacobians)!
What is probability of getting numbers 4,5,6 using fair cube (dice)? With your model of probability (p=(one set of elements)/(second set of elements)) becomes p=1!. p=3/3=1 and that doesn't have any sense.
Ok, let's rotate our circle for 90 degrees around y axis. Now, after one additional nonlinear tranformation we can see one straight line and probability is p=(2xR/2)/(2R)=1/2 !? I doubt that you will say p=1/1=1. How would you correct it now?
|Fermats Brother UK 1/24/2009 3:56:53 PM|
|The correct answer is p = 0.5.
All other values can be proved to be invalid !|
|Bozur Serbia 2/5/2009 4:18:36 AM|
|Could you explain why?|
|Esteban Treviño Mexico 9/2/2009 12:21:41 PM|
|Been a while! The probability of getting one set of 3 vs one from the other set of 3 is equal thus p = 0.5. If I am conceiving your other example correctly, based on the 1/2 answer, I still see that when the cord equals the diameter there are multiple possible cords and thus the model used distorts the answer. This can be corrected by instead of using the bisector and the length of the cord using the perimeter and the length the cord... and ding this give the 1/3 probability.
|Lukomor Ukraine 12/21/2009 6:56:26 AM|
See this article:
|Bozur Serbia 12/21/2009 7:09:23 AM|
|Please, refrase the conclusion, I am not sure what is your point.
|KL Singaproe 5/16/2011 1:17:44 AM|
|It all depends on how the chords are randomly drawn. |
|Bozur Serbia 6/2/2011 7:51:55 AM|
|No it doesn't depend on that. In all examples chord is observed as line that connect two points on the curve. But afterwards (in two problematic solutions) they involved nonlinear transformation trying to present all chords with its middle point. If it can be done with Jacobian those two examples would be acceptible.|
|KwajKid USA 10/19/2011 11:46:10 AM|
|The correct answer is 1/2. I could not follow your dialogues, but consider this argument:
Note for case II, we assume a uniform probability density to the center
of the chord over the interior area of the circle. In case III, we
assume a uniform probability density to the angles of intersections of
the chord on the circumference. In case I, we assume a uniform
probability density to the to the linear distances between the centers
of chord and circle. It is clear that in II and III, we do not have
translational invariance. But in I, we do have translational
invariance. You have to consider the distribution of the chords in all three cases. If translating the circle a little bit changes the distribution of the randomly selected chords, then we are dealing with a case that does not assume and require translational invariance. Consider raining down straws without skill from a long way away over a very large area, and consider our circle to be a very small area in the midst of this rainstorm of straws. These chords are going to be uniformly distributed. That is, the distances between the center of the chords and the circle will be uniformly distributed in such a rainstorm. In such a storm there will not be a uniform distribution of the centers of chord over the interior area of the circle (because a chord center uniquely determines the chord, and a uniform distribution of chord centers yields a distribution of chords that is not uniform - and therefore not translationally invariant). Likewise, in such a storm we also will not observe a uniform density to the angles of intersections of the chord on the circumference. We will observe a uniform distribution to the linear distances beween chord centers and the circle.|
|roro eire 3/13/2012 2:06:10 PM|
|This page and conversation is a bit strange. It all depends on the word "random". A random variable (or chord in this instance) is not properly understood until you know the underlying distribution (i.e. random uniform, random normal, etc). The three approaches actually use chords that are drawn from three different distributions. These arise from the way that the chord is chosen. The approach that yields the result of 1/2 uniformly covers the area of the circle. This would be many peoples intuition of "random". The other approaches create chords that, while valid, do not cover the area of the circle uniformly when many chords are drawn. KwajKid is pretty much correct.|
|Bozur Serbia 4/6/2012 3:02:18 AM|
"Note for case II, we assume a uniform probability density to the center of the chord over the interior area of the circle" ...why we do that?
" In case I, we assume a uniform probability density to the to the linear distances between the centers of chord and circle."...again, why???|
|Bozur Serbia 4/6/2012 3:23:16 AM|
|answer for roro:
I agree with your state that it all depends od the word "random", but if question is not defined closely we should stick to the question and that is the observation of chord which is straight line that connect two points on the curve. Anything in addition shouldn't have nonlinear transformation to the basic set. Otherwise we have paradox in every single problem. For example what is the solution of the following problem: "what is the probability to choose number 3 among 1,2,3,4,5". First, is it 3 or not yields to p=1/2. Second, it is odd number or not, third p=1/5, and so on... because we didn't well explained how we choose numbers?! So, there is no problems that can be defined so thoroughly to avoid mentioned free approaches.|
|Fardistantobserver UK 5/25/2012 7:57:58 AM|
|Bozur, I favour your conclusion but (maybe I should say "and") I think you underplay one of your arguments in case 2. The fact that an infinite number of chords can be drawn through the centre destroys the bijection in cases 1 and 2. In case 1 infinitely many qualifying radii may be selected if the mid-point of the "random chord" coincides with the centre of the circle.|
|Bozur Serbia 6/4/2012 8:08:05 AM|
|Yes, center point makes explanations in examples 1 and 2 unsustainable. |
|dingdong EARTH 7/1/2012 9:28:18 AM|
|Could I get compressed conclusion of errors of Method 1 and 2 ? I can't get the point exactly, and can't convince whether I understood correctly what you said.|
|dingdong EARTH 7/1/2012 9:28:59 AM|
|also with the reason why Method 3 is correct.|
|Bozur Serbia 7/13/2012 6:12:06 AM|
|Answer for dingdong! --- In methods 1 and 2 we made some presumptions that involve hidden nonlinear transformation and method 3 doesn't have any.
In method 1 radii can't uniformly cover surface of the circle as it is supposed; in method 3 center of the circle defines infinite number of chords, not only one as it is supposed, so the choosen method doesn't fulfil the task as well. From method 2 we can conclude that real probability is higher then 1/4 and from method 1 that it is lower then 1/2.|
|Yannis Mariolis Greece 9/8/2012 11:50:19 PM|
|Dear Bozur, sorry for not reading all the posts. However, I read your imaginary dialogue and I believe you 've got it right, 1/3 is the correct answer and your reasoning seems also correct. However, you should get some help from a native english speaker, because it is really difficult to understand you. Perhaps this is the reason you cannot get your work published. Apart from the language you can also make your points more straightforward.
This is how I tried to resolve this paradox: The important assumption that is not explicitly stated is that when we say that we randomly peak a chord, we mean that we UNIFORMLY peak a chord of a given circle among ALL chords of that circle. This is the case only for the 1/3 setup. In the 1/4 setup although we consider all chords, peaking is not uniform since all the diameters share the same middle point, whereas in the 1/2 setup although peaking is uniform we do not consider all chords but only a subset of chords that are parallel to one another. This is the main idea. However, the above points need some clarification, since the symmetries of the circle can create some confusion on what we can consider as the set of ALL chords. Thus, I have formulated a rigorous proof that accounts for these symmetries as well...|
|Bozur Serbia 9/21/2012 4:23:20 AM|
|Dear Yannis, thank you for your post! I will make some improvements regarding your proposals.|
|neopolitan Netopia 11/12/2012 6:15:09 PM|
|Hi, I'm partly with you in so much as there really isn't a paradox, but I think that you have arrived at the wrong answer. It's as the standard texts, Fermats Brother, Kwajkid, roro, Lukomur and (maybe) Jo have said - p=0.5 I have worked through it a slightly different way, with a somewhat different intent, and will soon be posting the argument here - neophilosophical.blogspot.com (look for the tag "paradox").|
|neopolitan Netopia 11/14/2012 6:27:41 AM|
|It's now in place, ready for you to take a look at :)|
|Bozur Serbia 1/1/2013 2:00:25 PM|
At first sorry for my late answer! I have read all your three posts at mentioned site and as I see, your method is some kind of generalisation of classic approach which leads to result p=1/2. But your post is interesting, because at one moment you expose several ways how to get two points on the circle! By definition, chord is line that connect two points on the curved line, and the question is why we have to create specific method for choosing two points. If we do so, do we make some influence on bijection? Or, in other words, do we change basic set of all chords? There is no explanation of that so I can't see specific value of your method.
One more thing! We will have random distribution if we know what is our basic set of chords. If not, we are blind!
If I missed something, please put it down.|
|neopolitan Netopia 1/25/2013 3:38:53 AM|
|Did you look at the Farewell to the Betrand Paradox which showed the flow chart? That should make my contention explicit.|
|Bozur Serbia 2/25/2013 9:05:28 AM|
|Yes, I looked at your flow chart. It is interesting point of view but I think that problem appear in blocks that consist part "draw line perpendicular to..."! Why, how that .. do we have our primar set of chords defined till that moment or not? If we know what is our set of chords do we make some transformations with that block that are not allowed?
So without that I think that making any conlusion about Bertrands paradox is not valuable.
If we can change basic set oh chords rough example could be throwing a dice. Probability for getting 6 is p=1/2 -- it will be 6 or not! |
|Bozur Serbia 10/13/2014 10:16:19 AM|
|Kent Jolley USA 3/16/2015 8:01:41 PM|
|why is not the proper way to select a random chord to choose two points randomly within the circle and have that define the line|
|Bozur Serbia 3/19/2015 9:29:25 AM|
Why not outside the circle? Why not inside the circle but from a surface that is not flat? I suppose your surface inside circle is flat, is it correct? If so, why you think the proper way is from flat surface?
By definition chord is line that connects two points on curve line. So you have to prove that your transformation on any other set of chords is linear. If you are free to choose any set probability theory does not exist any more! Lets make a vivid example. What is probability that fair dice will show 6? 1/2 because it will be 6 or it wont; it is obviously absurd.
|test test 9/26/2018 12:29:40 PM|
|enter a comment|