Bertrand's paradox is not a paradox!

**Problem:** Find the probability that the length of a random chord will be greater than the side of an equilateral triangle inscribed in the circle.

A chord is fully determined by its midpoint. The chords whose length exceeds the side of an equilateral triangle have their midpoints closer to the center than half the radius, the probability becomes 1/2.

A chord is fully determined by its midpoint. The chords whose length exceeds the side of an equilateral triangle have their midpoints inside a smaller circle with the radius equal to 1/2 of the given one. Hence, its area is 1/4 of the big circle which also defines the proportion of favorable outcomes - 1/4.

A chord is fully determined by its endpoint. If we use one vertex of the triangle as the first point, the chords whose length exceeds the side of an equilateral triangle have their endpoints on the arc between two opposite vertices, the probability becomes 1/3.

Is it a paradox or just a blunder? Does it have a unique solution? Is it a well-posed problem? Is it allowed to translate chords onto the surface and why do we do that? How do we choose the surface shape in that case?

**This page offers unique and original solution!**

The problem was proposed by French mathematician Joseph Bertrand in 1889. The offical statement is that it is a paradox and at the same time a good example that words "chosen randomly" sometimes are not enough. In other words, we must know how to find probability. The explanation for statement is that there are a lot of different results of the problem which depend on the method of calculation. Therefore, this paradox has became the model for explanation when we can't determine something, specially in theory of probability. Explanation on this page is realized like a dialogue between one fictive visitor and me (V-visitor, B-that is me, Božur), and we are discussing about this paradox from first example to the third. You'll see, there is no need for complex math calculations.

Yet, when you are confused about those examples, we can go to read dialogue about the first. Chords that are longer then side of equilateral triangle are red. Explanation is realized like a dialogue between one fictive visitor and me.

Yet, when you are confused about those examples, we can go to read dialogue about the first. Chords that are longer then side of equilateral triangle are red. Explanation is realized like a dialogue between one fictive visitor and me.

A chord is fully determined by its midpoint. The chords whose length exceeds the side of an equilateral triangle have their midpoints closer to the center than half the radius, the probability becomes 1/2.

V: Ok, I saw one simulation in java that shows the same result like this!

V: I don't understand.

V: And, if we use wrong calculation we will get wrong result.

V: Then, why we are using them.

V: Yes!

V: And what is wrong here?

V: One point of the surface of the circle determine middle point of one chord. And if the point stays closer to the center than half the radius then is appropriate chord longer then side of inscribed equilateral triangle.

V: Because, it includes set of chords from shortest one to longest.

V: Yes! If it represents set of chords that we have here, that is enough.

V: But it gives different result.

V: It can't covers surface of the circle uniformly. With radius we can calculate on that way because the circle consist of infinity radii.

V: Yes.

V: With integration.

V: If we integrate radii we will not get formula for surface of the circle!

V: I see! We integrate little sectors.

V: No, it is not! But if we leave the angle very small we will get the radius.

V: Then it is not a sector any more! 100x0 equals 0, and we can't make sector.

V: No, we can't! Then is all wrong, and it is the same as we use that curves above for calculation.

V: So, we must calculate with sectors, to make in summary a circle?

A chord is fully determined by its midpoint. The 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 favourable outcomes - 1/4.

V: Like previous example we must analyse correctness of this statement.

V: I don't know what to ask, it looks ok. Here is the surface covered uniformly.

V: Hm, why the surface here must be flat? Why we don't copy midpoint to uneven surface, and then look for their rates?

V: Maybe, because it is minimum value of surface?

V: It seems in that way when we seek where falls midpoint on some irregular surface we will make some of the regions more probably then other. We can give for inner region better chance if is there larger surface like on picture at right side.

V: Because we can't change their rates!

V: No!

V: All of the chords are determined by only one point of the surface of circle.

V: It has also their chord!

V: Oh, yes! There is unlimited number of chords that is determined by middle point!

V: Yes, but that is only one point, we can neglect it! Probability of choosing one point is zero in this case when we have infinity points!

V: Sorry?

V: And?

V: Definitely not! If we do that we can also use in our calculations those irregular shapes above.

V: So, here we have wrong model because we do not know what is relation between our one chord and one point here.

V: I haven't more question for this example.

A chord is fully determined by its endpoint. If we use one vertex of the triangle as the first point, the chords whose length exceeds the side of an equilateral triangle have their endpoints on the arc between two opposite vertices, the probability becomes 1/3.

V: Here we calculate probability by using one point, and then we are looking where could be the second. Is that correct if we are looking for just one point?

V: Ok.

V: Hm, it is difficult.

V: In previous samples we had hidden irregular transformation.

V: Maybe. But here it looks so simple.

V: Well, in first example the middle point of chord is used to transform our set of chords to one radius. In second it is used to transform our set to the surface of the circle. And here we do not use anything.

V: Yes, but not for some transformation, we just counting chords here!

V: No, we haven't!

V: You are a hair-splitter.

V: Here we replace one chord with one point.

V: In the first example we also have bijection to one radius but there is one more transformation.

V: When we say that all midpoints are on all radii, but all midpoints can only be on all little sectors! If they are not, we are neglecting some of them in our calculations.

V: We have discussed yet about that. There is problem the center of the circle, it involves transformation that is not bijection. We are neglecting there infinity number of chords.

V: It seems then statement for this example is correct. And if someone find in the future some more different examples?

V: On many web pages about Bertrand's paradox I have found statements that the problem in Bertrand's paradox is to define on which way should we try to find probability. And because it is not clearly defined we have paradox here.

V: Well, that is the answer of Bertrand's paradox. It is a blunder!

That is all.

**So called Bertrand's paradox is not a paradox because there is only one correct solution for the problem and it is 1/3!**

If you don't agree, you can add your comment to the discussion bellow. Here is the deal: If you don't enter a comment that will be accepted like agreement :). Of course, positive comments are also welcome.

You can contact me directly via** bozur.vujicic(et)gmail.com**, but only if you don't want to discuss so called Bertrand's paradox. For discussion, please use form below.

If you don't agree, you can add your comment to the discussion bellow. Here is the deal: If you don't enter a comment that will be accepted like agreement :). Of course, positive comments are also welcome.

You can contact me directly via

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BlendergxjRussia2/27/2021 3:43:48 AMFoamlukRussia2/28/2021 10:08:32 AMAvalanchercrRussia3/5/2021 2:32:06 AMVintageqygRussia3/10/2021 11:54:32 AMPortablehisRussia3/10/2021 10:05:06 PMJames LambertItaly3/11/2021 2:12:11 AMMilwaukeemcuRussia3/12/2021 12:23:32 AMPouringgzhRussia3/13/2021 6:30:51 AMFlexibleaycRussia3/15/2021 8:42:27 PMSeriessbeRussia3/16/2021 11:17:40 AMScannerkouRussia3/17/2021 1:01:12 PMCutterohaRussia3/19/2021 3:50:00 PMYamahabzcRussia3/20/2021 5:46:48 AMSightcskRussia3/20/2021 11:36:11 AMSeriesvwvRussia3/21/2021 5:27:45 AMSuperchipsyjfRussia3/22/2021 2:43:03 AMSuperchipsqayRussia3/22/2021 3:37:35 AMAscentrlqRussia3/22/2021 8:36:41 PMIncipiozrvRussia3/24/2021 1:46:58 PMPlasticunxRussia3/27/2021 2:19:19 AMArtisanrmrRussia3/27/2021 4:33:05 AMBluetoothdkbRussia3/28/2021 12:30:08 PMSpeakergreRussia3/28/2021 5:10:29 PMEdelbrockcecRussia3/29/2021 1:01:32 PMDormanemeRussia4/1/2021 2:25:49 AMRaymondMofUSA4/3/2021 9:10:16 AMKeypadagtlRussia4/6/2021 9:02:56 AMSuperchipsjbzRussia4/9/2021 5:39:59 PMSunburstyhqRussia4/10/2021 4:13:22 AMHanspeter SchmidSwitzerland4/18/2006 9:28:56 AMBozur VujicicSerbia4/25/2006 9:03:49 AMBrian McMullenUSA10/15/2006 7:32:42 PMBrian McMullenUSA10/15/2006 8:37:22 PMBobUSA10/16/2006 1:56:13 PMBrian McMullenUSA10/16/2006 8:46:09 PMBrian McMullenUSA10/16/2006 8:54:55 PMBozur VujicicSerbia11/6/2006 4:44:46 AMI 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!

zcermalaysia3/15/2007 3:55:08 PMBozurSerbia3/22/2007 9:20:17 AMBozurSerbia6/14/2007 5:57:55 AMJoGermany9/10/2007 7:06:48 PMlukomorUkraine9/24/2007 8:29:31 AMBozurSerbia10/23/2007 6:19:35 AMBozurSerbia10/23/2007 6:20:51 AMLukomorUkraine10/24/2007 10:54:17 AMBozurSerbia10/31/2007 5:19:59 AMLukomorUkraine11/1/2007 11:55:32 AMLukomorua11/5/2007 6:43:30 AMBozurSerbia12/14/2007 9:01:39 AMLukomor.ua12/18/2007 9:28:58 AMBozurSerbia12/20/2007 8:55:07 AMLukomorUa12/22/2007 2:07:21 AMLukomorUA12/22/2007 2:11:54 AMLukomorUA12/22/2007 2:13:53 AMLukomorUA12/22/2007 2:16:57 AMLukomorUA12/22/2007 4:57:21 AMLukomorUkraine12/24/2007 7:29:28 AMBozurSerbia12/25/2007 4:52:11 AMBozurSerbia12/25/2007 9:05:25 AMLukomorUA12/26/2007 4:31:18 AMLukomorUA12/26/2007 4:37:03 AMBozurSerbia12/28/2007 7:21:20 AMBozurSerbia12/28/2007 8:26:53 AMLukomorUkraine12/28/2007 11:55:26 AMLUkomorUa12/29/2007 6:54:23 AMBozurSerbia12/31/2007 4:21:11 AMBozurSerbia12/31/2007 6:04:57 AMEsteban TreviñoMexico4/26/2008 3:21:22 PMBozurSerbia5/9/2008 9:43:28 AMEsteban TreviñoMexico5/17/2008 12:02:10 AMBozurSerbia5/19/2008 5:16:09 AMSo, 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 BrotherUK1/24/2009 3:56:53 PMBozurSerbia2/5/2009 4:18:36 AMEsteban TreviñoMexico9/2/2009 12:21:41 PMLukomorUkraine12/21/2009 6:56:26 AMBozurSerbia12/21/2009 7:09:23 AMKLSingaproe5/16/2011 1:17:44 AMBozurSerbia6/2/2011 7:51:55 AMKwajKidUSA10/19/2011 11:46:10 AMroroeire3/13/2012 2:06:10 PMBozurSerbia4/6/2012 3:02:18 AMBozurSerbia4/6/2012 3:23:16 AMFardistantobserverUK5/25/2012 7:57:58 AMBozurSerbia6/4/2012 8:08:05 AMdingdongEARTH7/1/2012 9:28:18 AMdingdongEARTH7/1/2012 9:28:59 AMBozurSerbia7/13/2012 6:12:06 AMYannis MariolisGreece9/8/2012 11:50:19 PMBozurSerbia9/21/2012 4:23:20 AMneopolitanNetopia11/12/2012 6:15:09 PMneopolitanNetopia11/14/2012 6:27:41 AMBozurSerbia1/1/2013 2:00:25 PMneopolitanNetopia1/25/2013 3:38:53 AMBozurSerbia2/25/2013 9:05:28 AMBozurSerbia10/13/2014 10:16:19 AMKent JolleyUSA3/16/2015 8:01:41 PMBozurSerbia3/19/2015 9:29:25 AMDaveUK1/27/2019 7:17:28 AMBozurSerbia2/3/2019 6:18:28 PMFingerboardzkjRussia2/23/2021 10:14:36 PMFingerboardzzrRussia2/24/2021 1:02:44 PMJuicerrskRussia2/25/2021 5:30:22 PMWirelessrapRussia2/26/2021 5:49:20 PM