Bertrand's Paradox (with 3blue1brown) - Numberphile

[@numberphile]

Extra footage from this interview: kzbin.info/www/bejne/pnvcfIBjbK6ad7c 3blue1brown video on the shadow a cube: kzbin.info/www/bejne/oqWvhpSao6isn5I

[@LightVortexMatrixStudy]

🤓 Hello young Elect. The answer is All Light Does Not Move At The Same Speed. Therefore it is indeed a Paradox as it is against the law light. Solve that paradox young one. ❤

[@PluetoeInc.]

@@LightVortexMatrixStudy can you please elaborate what you are trying to express gentle sir

[@LightVortexMatrixStudy]

@@PluetoeInc. Seek and Find... That is our way.

[@AppleoTexza]

Hello, sir....I think the 2nd and the 3rd way to solve the problem was wrong.... If we take all figures made of infinitesimal points, we can say that the movement of the glowing point [in the 2nd and 3rd solution] in any direction by any number of points would cause the endpoints of the chord to move by a different number of points so the movement of that glowing point would not be equivalent to the different cases. As in we will be missing a few cases or probably over-counting. By the definition of chord {- 'A chord of a circle is a straight line segment whose endpoints both lie on a circular arc.' ~ Wikipedia (hopefully it is correct)} it is the join of 2 points on the circumference. Thus, if we are asked to uniformly choosing a chord it is meant to be with respect to the points on the circumference. In the 2nd solution we are distributing the points over the area of the circle and in the 3rd solution we uniformly distributing radially.

[@Матвертикаль]

Hi there✌️Would you mind if I translate this video to my native language (Russian)? May I upload this translated video to my KZbin channel?

[@hamgelato8143]

It must be nice to collab with Grant since he did his own animation

[@Xingchen_Yan]

Next vid, Grant teaches Numberphile how to use Manim

[@vigilantcosmicpenguin8721]

@@Xingchen_Yan No, I'm sure he's keeping his tricks a secret for himself. Like a magician.

[@jayd2279]

@@vigilantcosmicpenguin8721 It's open source tho

[@francescocostanzo8225]

@@jayd2279 news to me now I'm gonna have so much power

[@Leyrann]

@@vigilantcosmicpenguin8721 Not sure if joking, but this summer he literally put it out there and held a contest who could make the best math video and then featured the winners.

[@33NANO33]

4:52 for anyone who was also confused why the area was not pi/4: He meant "The inner circle has an area of 1/4 [of the outer circle area]".😅

[@rosiefay7283]

For anyone who was confused when he said one 4th -- he meant, as you say, 1/4 [of the outer circle's area].

[@Brawler_1337]

Thank you. I was wondering how Grant missed the pi in the area, but considering he’s talking about the area in proportion to the outer circle, the pis cancel out.

[@backwashjoe7864]

For anyone who was confused when he said "the inner circle has an area which is one-fourth"; he meant, as they say, "1/4 [of the area of the outer circle]".

[@manhalrahman5785]

For anyone who is still confused still, if the area of the circle is A, then one fourth means A/4

[@manhalrahman5785]

For anyone still confused, go learn some English first

[@collardgreen]

Every other person: This is because of my bad drawing. Grant: Thats a "function" of my Bad Drawing. 7:22

[@alexshih3747]

I think this paradox is a perfect example of how slippery a lot of concepts in probability theory can be. Even Erdos got the Monty Hall problem wrong.

[@MultiPleaser]

And Gabriel's Horn is also a perfect example. In that paradox they use 2D math to calculate the surface area of the horn, but 3D marh to calculate the volume. Here in Bertrand's paradox, he is doing the same thing in methods 1 and 2. Only method 3 is correct, where he chooses random points in 3 dimensions inside the circle.

[@rickdesper]

​@@MultiPleaser All three are distributions on the set of chords. We're talking about an infinite set which is why the concept of a "uniform" distribution is tricky. A uniform distribution on a finite set is simple: every point has the same probability, and when you add up the probabilities they sum to one. This doesn't work with infinite sets because point probabilities are of necessity zero. With infinite sets, we don't define probabilities on individual sets, but on subsets of the entire space. And the measure used on the space is essential with determining which distribution is uniform. The basic idea is that, for sets X_1, X_2 subset of the space X, if the measure m(X_1) = m(X_2), then we must see the probability p(X_1) = p(X_2). All three distributions are uniform relative to their respective measures. To say which is the right answer, we need to understand what the underlying space of chords is. Are we selecting the endpoints with a uniform measure or the midpoint? You'll get different answers based on what the measure of the space is.

[@PC_Simo]

@@MultiPleaser Random points in *_*2 Dimensions._* You can’t have 3 dimensions in a circle.

[@user-he1cx9my2x]

Yes but only because it was not explained in a mathematically sound way - it. was no fault of his own logic and problem solving skill, as once it was given to him in a correct framework without ambiguity he got it correct

[@tomkoziol141]

​@@user-he1cx9my2xWhat is the ambiguity? 🤔

[@kesim]

A triple cheers for bringing this important example up and making it known to the greater public! And so wonderfully rendered and explained, too!

[@ophello]

*triple

[@michaellinner7772]

I doubt that the general public cares or could even grasp the information given here. It is however, interesting for those of us who can.

[@edonveil9887]

What do you mean?

[@The_Scattered_Man]

@@michaellinner7772 Huh? (

[@erikisberg3886]

I totally get it! Blue sharpies were invented to solve difficult math problems on certain kinds of brown paper.

[@KillerMZE]

This ended in such a cliffhanger, never felt so compelled to see the extra footage

[@bernhardkrickl3567]

Have you watched the extra video? Do they present a solution to this paradox?

[@ericy1817]

@@bernhardkrickl3567 I'm pretty sure there's no "solution" because the problem itself is inherently vague, and the phrase "choose a random chord" is ambiguous. The solution is just to specify a single distribution or a method to choose a random chord and that would easily resolve the paradox.

[@KillerMZE]

@@bernhardkrickl3567 They show someone who said that the solution should be invariant to translation and scaling of the problem space, i.e. imagine the cords are already in the space, and the circle is moved or scaled. This rules out some of the solutions

[@wolfcatcombo5859]

Me tooooo lolol xD

[@HollowInn]

The problem lays in the defined rules. Stricter rules about how the lines can be drawn.

[@nomekop777]

This is a beautiful video. The style of numberphile mixed with Grant and his animations makes for a wonderful combination

[@drakoz254]

It's fascinating: each method is a different way of seeing the topology of the circle. The first is the classic S1, the second is the disc filling in S1, and the last is the disk as [0,1]×S1. (In particular, quotienting out the S1 part). My intuition tells me that the first and second should be different, but the second and third should be the same. Interesting!

[@chriss1331]

The second and third are different because in the second one the points are uniformly distributed in rectangular coordinates, while for the third one the points are uniform in polar coordinates. This leads to a bias towards the center of the circle in the third one, but not in the second.

[@joseville]

I’m not familiar with the terminology, but feel like I understand what you’re saying. What is S1? I made a similar comment: It seems like method 2 and method 3 are two different ways to choose the chord midpoint, so method 2 and method 3 reduce choosing a random chord to choosing a random point within the circle. The funny thing is they come up with different answers because method 2 choses a random point by choosing uniformly over the area of the circle, while method 3 chooses a random point by choosing a theta and a r each indepentely from [0, 2pi] and [0, 1] respectively, then combining the theta and r to produce a point within the circle.

[@drakoz254]

@@joseville just thinking with more abstraction and less numbers :) S1 is the circle, the "1-sphere" (1 because it's "1-dimensional" in some sense). Essentially, we can think of the unit disc either as itself, all together, or as pairs (R, D) where R is a point of S1 (you can think of this as an angle, as you have) and where D is a point of [0,1]. Obviously we have to fudge things a bit at the center, but the idea is clear I think.

[@jaredparkes5003]

Any point outside the circle will have two tangent lines to the circle; the intersection points of these lines to the circle will uniquely define a chord of the circle, thus any point outside the circle uniquely defines a chord. There is a finite region around the circle (bounded by a circle of radius 2r centred at a origin) where the chord defined by any point within will be shorter than s; any point outside this region defines a chord that is longer than s. The probability that a randomly selected point outside the circle will fall within the region where the chord would be shorter than s is 0; therefore the probability that a randomly selected chord will be longer than s is 1. Select a random point on the circle, then construct a ray line from the centre of the circle which passes through the point on the circle which is 90 degrees clockwise from the selected point. A randomly selected point on the ray line, along with the selected point on the circle will uniquely define a line which intersects the circle at two points, which defines a chord. If the selected point on the line is within r/sqrt(3) of the origin, then the chord will longer than s. The probability that the selected point on the line is within r/sqrt(3) is 0; therefore the probability that a randomly selected chord will be longer than s is 0. Any chord will be between 0 and 2r in length. s=r*sqrt(3). s/2r=sqrt(3)/2. Therefore the probability that a randomly selected chord will be shorter than s is sqrt(3)/2.

[@gabor6259]

mind = blown

[@juanitome1327]

I loved the first reasoning. It takes even further the idea that randomness has to be bounded by some condition other than “choose random”. Also it was pretty intuitive to me; you can observe that any point within some kind of outer circle will have length greater than s, since the angle between the 2 tangents will be “big enough” and so you kind of see a solution. BUT, since that set of points is finite, and the outer one is infinite, the probability of choosing ANY random point in R2 outside the circle and it happening to fall within a region, is virtually 0. And so a few questions arise from here. In infinite sets, does it even make sense to talk about probability (in a uniform distribution)? Wouldn’t it be always 0 or 1? Why can’t we make a tangent line from a point inside the circle? Is there any type of space where this could be possible?

[@juanitome1327]

But now I’ve realised that the region of the outer circle technically is an infinite set (of points) too so, how is it different to the outer space? If one is a subset of the other, how are these two infinites opposite seemingly?

[@frogkabobs]

@@juanitome1327 the difference is that one set has finite area and one has infinite area. In general, measure theory comes in handy to answer this type of question.

[@softy8088]

The question is: Is the method of determining a chord "at random", in fact, BIASED towards selecting chords with certain properties? And the question that follows: What even *is* an UNBIASED chord? Is there a way for chords to "naturally" fall within a circle?

[@nileshpandey8032]

Back to back videos by Grant. He just released one video on his channel 3b1b. What an amazing day!

[@ixajot]

And he talks about that at 9:55 when he mentions a problem with choosing a random orientation 🙂

[@jumpierwolf]

I didn't even realize before reading your comment that this was numberphile. I was so confused why there was brown paper.

[@jrcarlyon680]

This collab is a better Christmas gift than anything I have ever got from my family

[@romanski5811]

That's sad. I truly wish you that your circumstances in life will improve. Nobody deserves such a bad family. I wish you all the best!

[@josephang9927]

🥶

[@NightwindArcher]

Damn... that's rough. I hope life gets better for you soon

[@TheNameOfJesus]

Most people don't even know about probability density functions, so this sort of topic will really challenge them. As someone once said, "The generation of random data is too important to be left to chance."

[@jatinsoni1979]

someone="Donald Knuth"

[@gabitheancient7664]

mathematicians have allways the best quotes

[@TheNameOfJesus]

​@@gabitheancient7664 Physicists like Heisenberg would feel "uncertain" about your assertion.

[@Neme112]

@@jatinsoni1979 Are you sure? When I google the quote, everywhere it says that it's by Robert R. Coveyou.

[@bennmurhaaya8518]

This reminds me a little of one of my math exams questions back at uni, where it was about hitting a dart in the inner 2/3 of the target. I simplified it to 1D (2/3 of a line vs 1/3) not taking into account the fact that I cannot just get rid of polar coordinates like that. Here the different distribution of cords in the circle remind me of that. That some simplification that lead to a wrong distribution took place...

[@rmsgrey]

You get different distributions, but that doesn't mean that one of them is right and the others wrong. In some cases (such as this one), it reflects there not being a well-defined, unique, natural way of picking something (uniformly) randomly.

[@EebstertheGreat]

If you want to sample points randomly from within the unit circle, the fastest way to do it on a computer is to sample from within the unit square and then reject any that don't fall within the circle. If you want a way that doesn't reject any points, then you can generate a variable uniformly on [0,1] and take the square root to get your r. Then θ of course is just uniform on [0,2π). That gives you a uniform distribution on the disk. This is "correct" in the sense that the probability of getting a result from any region in the disk is proportional to that region's area. In other words, the distribution is uniform over the Lebesgue measure on the unit disk.

[@michaellinner7772]

It would surely distort the sampling pool if Peter Wright, Michael van Gerwen or many of the other PDC players took part in a real world demonstration.

[@bennmurhaaya8518]

@@EebstertheGreat Wait. I don't think that generating a variable and taking a square root would yield the same distribution as as a plain 0 through 1 equal random distribution. Years ago, I was running some code where I needed some float numers and the inbuilt random function was giving me to "big" of a numbers in range 0 - 1, no real 0,0012121112113111 to be seen... so for a quick solution I settled on squaring the floats but that skewed the distribution. I had much more "small" floats and it was super hard to get anywhere close to like 0,8. So I feel that this "mapping" is exactly the part where a 1/3 probability turns into 1/4 or 1/2.

[@bennmurhaaya8518]

plus you don't need to take a square root of the [0,1] rand value. That point is already within a unit circle. So a rand[0,1] and rand[0,2pi] should cover the circle.

[@Aw3som3-117]

I didn't know where it was going exactly, but immediately when he said that you pick a "random chord" I instantly thought "Define random. What's the distribution?" And lo and behold, that turned out to be where the paradox comes from. - If you pick a random chord by taking two random points uniformly distributed on the edge of the circle you get student #1's answer: 1/3 - If you pick a random chord by taking a random point uniformly distributed within the circle and make that the midpoint of the chord you get student #2's answer: 1/4 - If you pick a random chord by generating a random radial line and then choosing a random point uniformly distributed on that line to be the center of the chord you get student #3's answer: 1/2

[@WolfgangGalilei]

but how would this problem be fixed? what’s right?

[@Tesserex]

Same, I was at about 4:40 in the video and I realized both the paradox and the solution.

[@Tesserex]

@@WolfgangGalilei there is no right answer. You have to just rigorously define your chord selection.

[@stefanlclark]

In both the 2nd and 3rd options it's not very random to have that point as the centre of the cord.

[@japanada11]

@@WolfgangGalilei asking for a random chord is sort of like asking for "the world's biggest building." Do you mean the tallest? The longest? The one with the most volume? The one with the most floor space? Each question will give you a different answer. Words like "biggest" and "random" are simply not precise concepts unless you specify what exact measurements you're taking.

[@Dhuality]

This collab instantly improved my day

[@rgfs71]

Great example of why it’s important to state your assumptions! The problem isn’t in dealing with infinite spaces, it is in how you state what you know about one aspect of the space vs another. In each example the probability of the first point being chosen is assumed to be uniform in a particular space. The first student, for example, assumes points are equally likely to lie on the perimeter, but in the second example points are equally likely to lie within the circle, not the perimeter. These don’t describe the same space and therefore lead to different results.

[@ymyinfinity]

So much this! Not much of a paradox when you're commingling the sampling of two completely different distributions and getting different answers because of it.

[@lukeearthcrawler896]

Correct. There is no paradox here. The way the problem was defined at the beginning of the video makes the first answer the correct one. He specifically defined the random cord as the length of a segment linking two points generated randomly on the perimeter of a circle.

[@AstroTibs]

@@lukeearthcrawler896 It is a veridical paradox.

[@CliffSedge-nu5fv]

I feel like there is a _degrees of freedom_ issue here.

[@AySz88]

Reminds me of the unsaid random (and non-random) distributions in the Let's Make A Deal / "Monty Hall Problem", the one with three doors and two goats and a car. It's (usually) just implied that Monty knows what's behind each door, and always chooses to open a door with a goat behind it, rather than choosing randomly and sometimes opening one with a car "by chance". Monty could even have been following a procedure that skews the probabilities arbitrarily (and, according to one of my college professors, likely did so on the actual TV show because the show's results didn't match the statistics). For an extreme example, "evil Monty" knows you're going to switch if he reveals a goat. So, he reveals a goat only if you'd happened to pick the car already, and reveals the car otherwise so that you don't even get a chance to switch to it.

[@nigeldepledge3790]

It seems to me that the biggest clue here was that the second method looked sparser in the centre than the first. Clearly, the three different methods of "choosing a random chord" bias the outcome towards the three different results. The first and third methods are about choosing one or two random points along a line, whereas the second method is about choosing random points within an area. With the first method, there is twice as much circumference that will give you a short chord as there is circumference that will give you a long chord. With the third method, the randomly-chosen radius is irrelevant : what matters is the point on that radius. This method is giving equal probability to those points that will give either a long chord or a short chord. The second method, randomly selecting a point within the area of the circle, has three times as much area that will give a short chord as there is area that will give a long chord. (Points that will give a long chord must occur within circle that has half the radius of the initial circle, and this smaller circle will possess an area that is a quarter of that of the main circle.) And, making a welcome change, I worked this out for myself before the end of the video!

[@DBB314]

So which method is the most correct?

[@nigeldepledge3790]

@DBB314 - If I had to choose one, I'd say the third method; but I'm not sure that any one particular method is the "most correct". I think the third one has the least built-in bias.

[@robhall3607]

for the 3rd method consider a sector of the circle - the probability that the chosen radius falls in the sector is a constant that depends on the included angle - and then theres 50% chance of taking a point within radius/2, and 50% of taking a point outside of that. however the outer part has far greater area. hence the distribution is concentrated towards the middle of the circle in a way which is hard to justify as uniform

[@diobrando8979]

Couldn't expect less form a collab from my two favorite math channels. Awesome stuff!

[@kubaissen]

The answer is simple. First, we simplify to a lower dimension. Now we randomly choose a value from the length of the circle (2 * pi * r), and then divide it by the leg length of the triangle sqr(3). So the answer is (2 * pi * 1) / sqr(3)

[@okuno54]

Awesome! I remember reading this example in a random book in college; it stuck with me, but this is the first I've seen it again. Now I actually know the name, too!

[@ross302ci]

Yes but how "random" was this book you were reading in college? ;)

[@arkadiygertsman1287]

When he stated the first of the three problems, I paused the video and wrote up a lengthy solution using pdf's and all that. Then I unpaused the video and he answered it in one sentence. Very humbling!

[@TIAustralia]

Place two points randomly on the circumference of the circle. This seems like an innately intuitive and logical method to draw a chord. Naturally I wrote a program to do exactly that and arrived (of course) at the same probability as Grant. I checked that the points were evenly distributed along the circumference when considering it as a straight line, but it's not a straight line, the ends are joined, so this method may be pre-disposed to creating shorter chords. I also analysed the distribution of the midpoints (method 1), looking specifically at the abscissa and ordinate separately, they are not uniformly distributed. So it is not surprising that the different methods produce different probabilities. All this just goes to show that this is a delightful problem and worthy of inclusion in any high school / university mathematics, indeed, coding curriculum. Equally delightful is the collaboration between Numberphile and 3blue1brown, two KZbin channels that continually produce accessible and engaging mathematics content. Just think how much more engaging high school mathematics might be if these teams (and others) got together to write the curriculum!

[@henrym5034]

Choose a random point (uniformly) outside the circle. The two tangents from that point to the circle intersects the circle at 2 points, which defines a chord. Chord length is greater than s distance of point to circle is greater than some distance d (I think it’s 2 but doesn’t really matter) The probability is given by area outside circle of radius d / total area with the limit tending to 1

[@rudiklein]

These video's are showing me how much I don't know or understand. The more I watch the more I realise I know so little of the world. However, I just love they enthusiasm of the math guys when they show us this magic.

[@davidgillies620]

Sphere and disk point picking are tricky. The paradox is resolved by noting that different methods of picking points have different differential area elements, so the distribution with which the space is sampled is different. As a side note, Grant Sanderson's manim package is amazing, and so easy to use to do really impressive mathematical animations. If you know Python, it's quick to learn how to drive it.

[@thej3799]

I should learn python.Jpg

[@1SLMusic]

Taking the average of these results gets you 13/36.

[@Redditard]

but, I got 26/72

[@philipwilson46]

@@Redditard 26/72 simplifies to 13/36.

[@stromboli183]

Great video, explains and visualizes the problem very well. Often, uniformly distributed random things are tricky, and ill-defined more often than you’d expect. Three other examples (unrelated to this coord paradox): 1. Take a random natural number. 2. Take a random rational number between 0 and 1. 3. Take a random real number.

[@seedmole]

Yeah, like given that we know any rational number must either terminate or repeat, what is the probability that it does either?

[@andrewkarsten5268]

@@seedmole No, he's saying you cannot have a well defined notion of "pick a uniformly random rational number between 0 and 1." The reason you cannot do this is actually the same reason you cannot have "pick a natural number." This reason is that when you define probability on countably infinite spaces, you don't get a probability density function but rather a probability mass function. What this means (skipping a lot of details that would be filled in a first year probability theory course) is that any way you try to come up with a probability mass function that is uniform would give an infinite sum that either blows up to infinity or stays at 0. When you go to uncountable sets (all real numbers between 0 and 1 for example), you gain some things back since you deal with integrals instead of sums. The third example he gave explains why you need your sets to be bounded when wanting a uniform distribution. You can actually demonstrate it has within it the same inherent issue as the first two (as well as other issues) by saying after you pick your real number just take the integer part. If your picking your real numbers uniformly, then you are picking those integer parts uniformly, which gives a way to pick natural numbers uniformly, and again you cannot do that.

[@PunmasterSTP]

Bertrand’s Paradox? More like “This collab between 3b1b and Numberphile rocks!” The icing on the cake for me was Grant’s enthusiasm during the explanation. Awesome job guys!

[@boydstephensmithjr]

My "favorite" thing about sampling infinite spaces it that you can't uniformly sample from the Naturals (or the Integers).

[@cube2fox]

Here is a paradox: There appears to be no random way to draw a natural number, but nonetheless it seems clearly true that if you pick a random natural number, the probability that it is even is 1/2. But that seems to presuppose that drawing a random natural number is possible!

[@boydstephensmithjr]

@@cube2fox I think that's us taking an intuitive limit. For any [0, n) subset of the Naturals, you *can* uniformly pick, and if you do P(even) is close to 1/2. For some n it is exactly 1/2, and for all the other n, as n -> Inf the probably monotonically moves toward 1/2.

[@manswind3417]

​@@cube2fox I think yours is a classic case of infinite/infinite being finite, which is essentially the idea of limits. Yes, P(choosing an even number) = 1/2, and drawing is a random number is also impossible, but so is drawing a random even number. So when you eventually divide the two "impossible numbers", you get a finite number = 1/2, as both the numerator and denominator are infintely large.

[@cube2fox]

@@manswind3417 Perhaps that is related to what Boyd Smith said, but my reasoning was somewhat similar. The size of a natural number interval from 0 to n approaches infinity as n approaches infinity, and the same thing holds for the interval from 0 to m even numbers: As m approaches infinity, the number of even natural numbers approaches infinity. But the difference is that the number of natural numbers grows twice as fast as the number of even natural numbers, making the ratio approach 1/2. That's why I like infinite limits better than transfinite cardinal numbers (aleph null in this case), because for the latter the number of natural and even natural numbers would be the same, and there would be no way to arrive at the 1/2 ratio.

[@boydstephensmithjr]

@@manswind3417 You have to be careful with this though. It's possible to choose a different series of subsets where the limit is the naturals, but each of the P(even) from any subset is NOT 1/2 and does not approach 1/2. So while it's "obvious" and "intuitive", it might not actually be true / provable.

[@GuanoLad]

That was brilliantly explained. I think I can see where the contradictions originate, though, helped a lot by those spiderweb diagrams he made.

[@OslerWannabe]

Grant is an amazing guy. He's been doing 3blue1brown for quite a few years now, and yet he's not even 17 yet. Now THERE's a paradox.

[@columbus8myhw]

That can't be right

[@RalphDratman]

When you speak of an average you must specify that it is an average over a particular set or distribution! Anyone who has done a lot of programming with pseudorandom numbers will have understood that in some form.

[@ZakX11]

Thats exactly what he mentions at the end of the video...

[@RalphDratman]

@@ZakX11 If Grant used the term "distribution" I must have missed that. "Set" would only be different from "distribution" if used to refer to a finite set. The concept of a distribution is not intuitively obvious until one has thought about probability and randomness for some time.

[@rickdesper]

This video excellently illustrates the problem with the phrase "select at random", at least how that definition starts to fall apart when looking at infinite sets. The crux of the issue here is that "select a chord at random" is not clearly defined. So each of the three examples uses a different distribution on the set of chords of a circle, and thus shows that the measure of the set in question has different measures according to the various distributions.

[@eacy7deacy]

I find it interessting how they represent chords on the circle using three different spaces, then choose random points on it. The first is two points on the boundary, which is a torus. The second is a random point on thhe disc. The third is one line from the center, which can be viewed as a point on the boundary and a point on the radius. The final one can be thought of as a cylinder. Edit: I have noticed some flaws in this thinking. The first map gives a bijection, but the second and third don't. The center of the circle has all of the diameter through it, and by choosing opposing radiuses you get the diameter twice. This could be interpreted as saying the first one is the best, but I n probability I don't think this should matter since the overlaps have measure zero in their ambient space. Edit 2: the first does not give a bijection, we work with unordered pairs.

[@frogkabobs]

Since the map in each case is surjective, we must get a homeomorphism after identifying fibers. Hence, after this process we should get the same space (up to homeomorphism). We can find what this space is by looking at the paramaterization in the last case. Each chord is paramaterized by a radial segment perpendicular to the chord and where on that segment the midpoint intersects, which we can write as (s,r) with s ∈ S¹ and r ∈ [0,1]. We lose injectivity only at r = 0, where the chord (s,0) is the same as the chord (-s,0). Thus, the chord space is given by a cylinder with antipodal points of one of the bases identified, or the mapping cylinder of the antipodal point map. This is actually homeomorphic to a Möbius strip (WLOG of width 1), where we map [s,1] to the boundary of the Möbius strip, and map [s,r] to the point a distance (1-r)/2 to where [s,1] maps, perpendicular to the boundary.

[@mosab643]

Shouldn't the third one be a solid helix?

[@7tonin]

Another way to interpret the question. 💡 Choose a point on the circle, see the tangent to the circle at that point, then choose an angle between this tangent and the chord. Its values are uniformly between 0° and 180°. Remember the inscribed triangle, and its 60° angle. P(l>s) = P(60° < angle < 120°) = 1/3

[@littlenarwhal3914]

Ive got another approach which also yields 1/2: you think of a chord as a line intersecting the circle, and you can suppose the line is flat as otherwise you could rotate your point of view to make it so (implicit assumption on some kind of uniformity on 2d rotations). Then the chord is determined by the height of the line relative to the circle. Take the base of the equilateral triangle and shift it up to make an inscribed rectangle: the height of that rectangle is the length of the interval in which a point of the line such that the chord is bigger than s. That height is 1 so comparing with the total height (ie diameter of circle) we get 1/2.

[@WarDaft]

A fourth student, clearly hungover, stumbles into class. "Given any point onna shircle, theresh 'zactly two chordsh with any given length from the innerval (0,1). Thus the answer is pershishely 1 minush root 3 over 2." The student then falls over and passes back out.

[@PluetoeInc.]

I can easily say after reading the title this will be one of the most popular videos of Numberphile. Low-key so proud I clicked so fast

[@PC_Simo]

8:00 It should come as being a bit over a half; because the closer the radial line is to the vertices of the triangle, the greater the portion of it is that falls inside the triangle; and if it points directly to a vertex, then it’s fully inside the triangle; so, the portion of it that goes inside the triangle, is in between 1/2*r-1*r, inclusive. The distribution should, therefore, come close to p(l>s) ≈ 3/4. 🧐

[@Ganerrr]

"It's implicit when you're told to choose a random number between 0 and 1 you would use some kind of uniform distribution" ...wait there's multiple kinds?

[@wymarsane7305]

Anybody who's ever, against better judgement, rolled for stats in DnD, would be familiar with gaussian distributions.

[@manuelka15]

@@wymarsane7305 But gausian distribution is not uniform. The video implies that all 3 distributions are uniform on their own ways.

[@Speed001]

@@wymarsane7305 Actually yeah, why didn't they ever mention D&D in probability and statistics. It's one of the many things in my life that naturally led to probabilities. It's probably too simple because it only covers 2 cases.

[@gdvirusrf1772]

@@wymarsane7305 "Against better judgement"? The expectation for 4d6 drop lowest is higher than the average score using point buy.

[@kashu7691]

yes, a probability space (i.e the collection of everything you need to do calculations and get results) includes as a basic component the probability measure/distribution which must be specified before any probabilities can be calculated. for continuous outputs (sample spaces), the probability distributions are given by a density function which is integrated over. a uniform distribution is simply that which has the same value over all possible outcomes, just like when grant sketched a straight line pdf over [0,1], in comparison with a gaussian

[@8Clips]

Fascinating video. It took me a moment to wrap my head around the idea that none of the methods are wrong here. They're all valid, but over different probability spaces. Thank you Grant and Brady!

[@migarsormrapophis2755]

"They're all valid" But only the third one gives you the true answer

[@epajarjestys9981]

@@migarsormrapophis2755 You have not understood the video. The problem is not well-defined. It is not stated from which distribution a secant should be randomly chosen. There is no correct solution without further specifying that.

[@migarsormrapophis2755]

@@epajarjestys9981 You're just being obtuse. It is trivial to say that we should be choosing randomly from the set of all vertical chords ranging from -1 to -1/2 (all of which are less than the value), -1/2 to 1/2 (all of which are greater than the value) and from 1/2 to 1, all of which are less than the value. You may notice, half of those are shorter and half of those are longer. I'm sure I don't need to explain why anything true about this set of vertical lines is true about the set of all lines that can be drawn in the circle. So, no, the only thing being 'misunderstood' here is 3B1B misunderstanding what is meant by the word 'randomly,' (which, if you watch the second video, he does actually admit to somewhat). The only reason to think this problem is incompletely defined is because you haven't understood what's being asked. If you asked me, "what's 2 + 2" and I replied, "7!", it wouldn't be fair for me to then say, "but you didn't properly define what you meant by 2! For all I knew, you meant 3.5!" The word 'randomly' is much the same. You can take any word to mean anything, but that doesn't change the fact that certain words are understood to have certain meanings.

[@epajarjestys9981]

@@migarsormrapophis2755 Oh, boy. "Randomly" doesn't mean much until you have specified a distribution. _You may notice, half of those are shorter and half of those are longer._ It is not defined what "half of those" means here, talking about an uncountably infinite set without specifying a distribution. You think 3B1B hasn't studied math and knows what he's talking about? You should stop the meth.

[@migarsormrapophis2755]

@@epajarjestys9981 You've just committed a formal logical fallacy called an appeal to authority, but nevertheless, I think 3B1B specifically agrees with me in the next video. Did you _watch_ the next video?

[@kavetovaify]

Will draw every possible horizontal line that crosses that cycle. Only those lines that are in the upper forth part of the cycle and down forth part of the cycle will have length less than S. All lines in the middle (which will be 1/2 of all drawn lines) will have length greater than S. Then do that same process for every possible angle (not just horizontal lines). For each angle will get the same answer - 1/2. So the last method probably has the correct answer - 1/2

[@noammanakermorag9538]

No, none of the answers are correct! The question just isn't well-defined. There is no single way to select a "random chord".

[@tinnguyen5055]

@Noam Manaker Morag I'd say "chose a random chord" implies that each chord has the same propability of being chosen. Only the last result achieved that. The first two methods are biased against chords that go through the center region, making them less likely to be sampled.

[@noammanakermorag9538]

@@tinnguyen5055 I understand the confusion, but you must remember that there are an infinite number chords! The probability of drawing any specific chord is always 0. Generally speaking, the term "randomly choose" is only well defined for infinite sets under very specific circumstances. For example, randomly choosing a point on a circles circumference is well defined. You are correct that only in the third case are the chords distributed uniformly in *space*, but there is no reason to expect that random chords *should* be distributed uniformly in space! We were asked to randomly pick a chord on the unit circle, not to randomly pick a line in space that intersects the circle. These are fundamentally different things.

[@IulianSora]

How would this solution work for a line that has the length of the diameter ?

[@WarDaft]

You've chosen your chords so that they have an even distribution by one metric... but they won't by others. That's the whole point. Every metric is a valid way of choosing a random unique chord, every chord chosen by one metric can be chosen by every other.

[@ethanfranzen8684]

These answers are all correct; they just define the probability distributions differently. Method 1: The arc length is a uniform distribution from 0 to 2π. The chord length is sqrt(2-2cos θ) for a uniformly random θ. θ is the length of the arc. The offset f would be cos (θ/2). Method 2: The probability distribution of the midpoint of the chord is spread uniformly to all area in the circle. The chord length is 2sqrt(1-A) for a random A. A is the proportion of the main circle's area that is closer to the center than the chord's midpoint. The offset is f=sqrt(A) and the arclength is θ=2arccos sqrt(A). Method 3: The offset/intercept of the chord/line is uniformly random between what it could be to fit in the circle. The chord length is 2sqrt(1-f^2) for a random f. f is the offset, the distance from the centre to the midpoint of the chord. The arclength is θ=2arccos f. You could just as easily define the chord length, itself, to be randomly distributed between 0 and 2. In this method, the offset is f=sqrt(1-l^2/4) and the arc length is 2arcsin (l^2/4). (where l is the chord length)

[@maxdemian6312]

The three approaches have different results because they start from different assumptions on what events should be equally likely

[@Ardub23]

Just four days ago I learned about Bertrand's Paradox because it related (a bit tangentially) to Bostrom's simulation argument, which was a topic in my logic course. Kind of amazing to realize that this video was in the works while I was learning about that.

[@hafizajafarova1566]

Thank you for your job! It’s always interesting to see something new at this channel 😃

[@chimetimepaprika]

Without loss of generality are the four sweetest works in the English language.

[@umaresrar717]

In solution 2 and 3 you are taking in account of the chords that are only perpendicular to the line joining the point and the center of circle . But there are infinetly many chords possible that passes through that point. For example take the chord that passes through the same point and the center of circle (diameter) which is definitely greater than the triangles side.

[@invenblocker]

And all of those cords have their own point at which they do intersect a radius perpendicularly. The significant bit is that each such point has exactly one cord that intersects the radius perpendicularly, making it qualify as a unique identifier for cords.

[@umaresrar717]

@@invenblocker now I got it Thanks

[@invenblocker]

@@umaresrar717 Glad I could be of help.

[@reedclippings8991]

My intuition tells me the first definition of random is by far the best. It is the only one where I can say with confidence that the infinity within each of the 3 sections should be exactly the same size. The others make assumptions that either the midpoints or the number of chords should be spread out perfectly evenly with no gradation.

[@TK-dy7gp]

The second approach seems wrong to me, as if the mid-point is the center of the circle, then there are more than 1 chord defined by it (it defines infinity chords). But you can also argue that on a plane the probability of selecting a center is zero. Then my brain stop functioning...

[@yuvalne]

Reminds me of the Two Children Paradox, where the answer could be either 1/2 and 1/3 depending on how you define the question.

[@squeakybunny2776]

Could you describe the paradox :)

[@buttonasas]

tl;dr: Two children. One is a boy. What is the probability that both are boys? Here, wording is very important because "at least one is a boy" is different from "that one is a boy" or "the older is a boy", or etc.

[@cmyk8964]

Assuming male and female children are equally likely, there are 4 equally likely outcomes: MM, MF, FM, and FF, where M is a male child and F is a female child. “At least one is a boy.” Out of 4 combinations, 3 fulfill the precondition: MM, MF, FM. The probability is 1/3. “A specific one (e.g. The older one) is a boy.” This time, 2 fulfill the precondition: MM, MF. The probability is 1/2. “Exactly one is a boy.” There is no chance the other one is also a boy if exactly one is a boy. The probability is 0.

[@Anonymous-df8it]

@@cmyk8964 Where you said that the probability is 1/3, I think you meant 3/4

[@cmyk8964]

@@Anonymous-df8it No. The probability that one of them is a boy is 3/4. The probability that both of them are boys, _given_ that one of them is a boy is 1/3.

[@EconAtheist]

**cries** **takes another bong hit** **chillaxingly ready to watch extra footage**

[@nicholasleclerc1583]

Yay, 3Blue1Brown is now a regular Numberphile guest !!!

[@CliffSedge-nu5fv]

What is the frequency of periodicity that makes it regular?

[@arvinderbali]

Grant is revolutionizing maths understanding among peers.

[@SafetyBoater]

Chords that go through the center of the circle, diameters, can not be uniquely defined by their mid-points. Also, the distribution of points along the radius wouldn't be uniform. There are more points in a ring that is further from the circle.

[@gaston1473]

Yes, but it's implied that given any length (especially the diameter), the probability of a chord being precisely this length is 0. So excluding diameters doesn't change the result.

[@SafetyBoater]

@@gaston1473 There are infinitely many chords that pass through the center. I think excluding them turns into some sort of Banach-Tarski paradox, maybe. I wonder what this problem does if we look at the ratio of the length of the selected chord to its perpidicular bisector.

[@AimeePlaysMSM]

Your second point is a well-known issue with generating random points within a circle, and is one of the submissions to 3B1B's math videos 'challenge' (nubDotDev's submission), and largely focuses on the pitfall of deciding not to use rejection sampling because it seems wasteful and using polar coordinates instead (which leads to the inverse of the problem shown in THIS video; it will cause a bias of samples toward the center of the circle). (It also goes over two other approaches, definitely worth a watch) Personally I think Grant's answer in the follow-up video as to which one he would choose as 'the answer' (spoiler: p=0.5 from imagining the circle traversing a field of lines) was ultimately the one he intuitively went with very near the beginning of this video even if he didn't quite realize it. He states a chord is a line connecting any two points on the circumference of the circle.. but he never just connects the two dots he draws.. he just draws a line through them that extends beyond the perimeter.

[@DanMusceac]

You are perfectly right:the distribution of the middle point on the radius is not uniform.

[@frostyusername5011]

The most anticipated Cross over episode of all time! Brady and Grant together!!

[@Pheonix1328]

I remember watching a programming video and they were were going over different methods of randomly picking a point in a circle, or something like that. As it turns out you don't always get a uniform distribution. One of the methods produced (I believe) a concentration of point at the center.

[@jamesknapp64]

Correct if you just say its distance from center is 0 to 1 and angle is 0 to 2 pi this over concentrates points towards the center of the circle. Using this polar form you have to make the probability of distance from 0 to 1 be linear increasing not uniform.

[@f5673-t1h]

Wasn't that one of the Summer of Math video submissions? (the 3blue1brown event)

[@NoNameAtAll2]

the paradox is that *it's still uniform* just not uniform in the way _you_ think of

[@martinepstein9826]

@@NoNameAtAll2 In probability theory there is no ambiguity about choosing a _point_ uniformly in a circle. The uniform distribution is the unique distribution such that for any two regions A1 and A2 of equal area in the circle the probability that the point is in A1 equals the probability that it's in A2. Choosing a _chord_ uniformly is a different story.

[@garethdean6382]

Yeah. The simplest method you just pick points in a unit square and reject any outside the circle. That's a nice, even density , but of course this means less points near the center of the circle (Since there's less area there.) Sometimes though you WANT to have it so that in any collection of points as many are within a half radius circle as without and then you need to get tricky.

[@kai-xuanyao4666]

In the first minute of the video I was thinking "how do you define randomly choosing a chord?" Turns out that's essentially the gist of this paradox.

[@klosty2]

9:38 Did Grant just yada yada a math problem?

[@zesalesjt7797]

This was fun to watch. It reminds me of a thought I'd had the other day where the probability arrangement is the circumstance of the circle instead of a number line. When you approach zero from left using infinity, it seems impossible until the result is achieved. Likewise, approaching from the 1 turns a sure thing into a zero. So, it's like the 1/0 wave on a continuum where the critical point is not a limit but a gateway between infinite and infinitesimal possibility.

[@zesalesjt7797]

May have reversed orientation in the original, but the idea should still be clear.

[@noahwelke5231]

I love how this relates directly to sizes of infinitys

[@daniel.watching]

I don't think it does. Do you mean countable vs uncountable infinities? Because that's not what's going on here. The reason they get different answers is because they're all assuming a uniform distribution across their "coordinate system". But the "fix" just means adding some probability weighting curve and integrating over that. That calculus is the only place infinity comes in and I can't see how sizes of infinity are involved. But I'm still wrapping my head around it, maybe I've missed some nuance.

[@manswind3417]

​@@daniel.watching It indeed does, perhaps you need to look deeper into the equivalences drawn. But yes, as you rightly inferred, it's not countable vs uncountable coz that's just irrelevant here. Your conclusions are indeed correct and I think they should've led you to the sizes of infinities' argument: the probability is essentially a ratio of cardinality of 2 infinite sets, i.e. chords of size under rt(3) and total number of them. The numerator remains the same in each counting, it's the denominator that varies as per the exact interpretation of the word random/uniform. All the quantities being discussed here are of infinite size btw.

[@wohlrajh666]

I have a fourth solution for the problem: (2-sqrt(3))/2. -pick a point on the perimeter of the circle -pick an orientation (left/right) -pick a length between 0 and 2. You have defined a unique cord this way and have randomly sampled amongst all the possible cords. By design, the probability of the length of that cord to be bigger than sqrt(3) is (2-sqrt(3))/2 PS: you actually don't need the orientation and can assume you always make a cord on the right direction. It s the same when you pick two points A and B, you either have the cord (A,B) or (B,A). PPS: With what I wrote above, you in fact are twice likely to pick a diameter or a chord of length zero than any other chord. So you would have to in fact - pick an orientation - if left, pick a length within [0,2[ ; if right, pick a length within ]0, 2].

[@benweieneth1103]

I think you lost a factor of 2 in your final solution. 2 - sqrt(3) is the size of the "successful" interval, but you need to divide by the size of the original interval.

[@wohlrajh666]

@@benweieneth1103 you are right, I edited my comment !

[@slitfidgetspinnerdabbodmod9953]

I'd like to propose a 4th method: a chord is a line that passes through the circle, and a line is uniquely defined by 2 points. Therefore we can choose 1 point on the circumference and another within the circle according to uniform distributions. With this method the probability becomes 1/3 + sqrt(3)/2pi or ~60.9%. If the 2nd point is instead defined as being anywhere in space then the probability drops back to 1/3. I initially considered picking 2 points randomly within the circle (or maybe even randomly in space, discarding if the line doesn't intersect the circle) but that sounded like a pain to work out so I simplified to 1 point being on the circumference. I would be interested to know the result of these methods though.

[@mikki429]

My first thought was the third method. Wild that the distributions are both so distinct for this problem and give such nice answers.

[@jishcatg]

I think the problem with the 3rd method as it is, is that you can't give equal probability to every position down the length of the radius. This is because points close to the center are sweeping a smaller circle as the angle of the radius changes than at the end closer to the circle edge. So the distribution of points on the line should be weighted towards the circle end, otherwise if you just draw the points, you'll find you're favoring the center of the circle for point density. I'm not sure what the distribution needs to be (some trig function), but it can't be uniform.' EDIT: The correct distribution on the line is x^(1/2) or the square root of the uniform random point on the line. So if you pick a random point on the line uniformly, then take the square-root of that for the actual length on the line, your point density will be constant throughout the circle. I don't know if the chords would then be as well, but I feel like that's closer to the right answer.

[@wetbadger2174]

You could make a list of all possible random selection methods and randomly pick a method from that list. Assuming these are the only three, you would get a 13/36 chance of a chord being shorter.

[@rudranil-c]

I believe 13/36 chance of the chord being longer.

[@wallywutsizface6346]

He mentions this in the follow up video. He says something about how he doesn’t like this conclusion, since, like you said, it assumes we can’t think of any more ways to choose a chord

[@ninjafruitchilled]

Only if you evenly weight all three methods. And actually there are infinitely many methods so it doesn't help anyway :).

[@TechnocratiK]

Numberphile's brown paper with the dulcet tones of Grant's baritone narration? More please!

[@JohnWaylandHarper]

I think the core assumptions of the second and third methods are actually just flawed. Chords are not uniquely defined by their midpoint, because there's one midpoint (0,0, the center of the circle) that is shared by infinitely many chords (the chords that are also diameters). And even if you arrive at the point 0,0 by picking an angle and a distance, that gives two chances to generate what is functionally the same chord: one for a given angle and one for that angle +180 degrees. The first method is the only one that assigns an equal slice of probability to every possible chord.

[@wolfcatcombo5859]

Thank you! I was trying to figure out the last cause for this. I realized the difference for the second method biasing chord plotting differently but I wasn't accounting for duplicate chords due to these examples!

[@godowskygodowsky1155]

Having a chord pass through the center is a measure zero event in the first distribution. The issue is that you need to fix a distribution before doing any probability. You like the first distribution because it respects the symmetry of the Euclidean plane. When you say that it assigns an equal probability to each chord slice, you are actually making an interpretation. There is a unique measure (Haar measure) on the space of affine lines that is invariant under the action of the isometry group of the plane, and you are identifying chords with the Euclidean lines they belong to. This is one reasonable notion, but all of these distributions are reasonable in a some way, and there are many more that are just as reasonable.

[@PhilfreezeCH]

@@godowskygodowsky1155 I don‘t really understand anything after „measure zero event“ but I think his issue is mostly with that identifying chords by their mid-point has a uniqueness/identifiability issue. With both definitions the event of a chord going exactly through the middle if a measure zero event but only in the first one is such a chord uniquely identifiable, in the second definition the probability of it happening is still measure zero but the assigned ‚value‘/‚weight‘ (probably not the right word) is infinitely large which makes me feel like there is some Dirac-pulse shenanigans involved here. Also kinda unrelated. Personally I think the first and second definitions basically use a different view of this circle to define a chord. In the first one the circle is seen as a part of a larger plane and there are random lines on the plane, some of them just happen to cross the circle (identified by entry and exit point). On the other hand the second definitions to me basically assumes only the circle exists and everything outside it is nothingness, it places us inside the circle and defined a chord from our perspective.

[@godowskygodowsky1155]

@@PhilfreezeCH Your unrelated point is essentially what I was trying to get at. How do you know that the circle is part of a Euclidean plane with Euclidean notions of distance? This is somewhat of an arbitrary decision. For instance, it could instead have had hyperbolic distance. Also, you can rest assured that there are no Dirac pulse shenanigans going on. That only happens when one measure is singular with respect to the other. If you really want to be pedantic about it, you can swap out the underlying topological space of the configuration space, and the identification issue goes away. Essentially, imagine the same distribution as the third, except you proactively reweight the probability of choosing r so that it's proportional to r.

[@JohnWaylandHarper]

@@godowskygodowsky1155 Thank you for trying to help me with this, but I'm afraid I don't know how a lot of these terms apply to this situation. Probably my level of understanding is lower than yours. What I do think I understand is this: In order to have an equal chance of n things occurring, each thing should occupy an equal slice on a hypothetical dart board, even if that dart board has infinite slices, as it does here. The first method, choosing two points, does this. Each chord is represented once and only once, by an unordered pair of points. The second method does not do this. Most chords are represented by a single point, with their orientation determined automatically, but when that point is the center of the circle, their orientation can't be determined automatically, so such chords can't be chosen. They don't show up on the dart board at all. (I think this is one reason why the example resulting from the second method has a big hole in the middle). The third method also does not do this. Every chord that crosses the center of the circle has a paired chord, which it identical. This gives it two chances to be generated every time the program pick new chord, making diameters more likely than non-diameters. And, I really have no idea what I'm talking about regarding higher dimensions, but I'm not sure what Euclid has to do with any of this. Shouldn't this basic principle of probability hold true regardless of the curvature of space, or am I missing something fundamental?

[@Sciguy95]

Videos like this show that math can be a lot more fun than it seems when you're in school

[@mickschilder3633]

3b1b + numberfile = instant click

[@lenskihe]

Choose a random number between 0 and 2 uniformly. This will determine the length of the chord. Then choose a random radial line. Now, similarly to the third method, choose a point on that line such that the length of the perpendicular chord at that point is equal to the chosen number. Now P(ℓ>S) = (2-√3)/2 ≈ 0.134

[@SG2048-meta]

Another one with grant! Excellent job as well again.

[@loganjones75]

This is officially my favorite numberphile video

[@galloway_joseph]

Team Student #1 for the win.

[@TheAcer4666]

While I love Grant's animation + voiceover video style, I also love to see his cheeky little smile as he talks about maths

[@peter4928]

Seeing Grant in person is always a great time!

[@zlac]

Here's mine: 1. Choose a random angle. 2. Make two tangents with this angle - touching the small circle from both sides. 3. Choose a random point in the big circle. 4. If a random point is in between two segments - it's bigger, and if it's in any of the segments - it's smaller. 5. You have 60.9% chance of having a line bigger than a triangle side. 6. Now you have 4 results -> 25%, 33.3%, 50% and 60.9%!

[@jonathancohen2351]

That's a great video, but I don't see how infinity is involved directly. In this case it is, but that's not the crux of the problem. The same would occur in a discrete case where there were different notions of what uniformly distributed meant. I learned about this problem in the 70s as a part of bayesian inference, where people take a uniform distribution as a natural a priori distribution. The problem is that you get difference answers depending on which uniform distribution you choose.

[@SnijtraM]

The real problem is the incompleteness of the definition of the problem itself. The moment he started to define a distribution of midpoints uniformly on their radius to the circle center, I thought "hey! we're talking a *different* question here", no surprise at all when we'll be having a different answer. And, indeed ... what surprises me is that, how is this a paradox at all, doesn't everybody else immediately see that we're comparing apples and oranges?

[@Rodhern]

@@SnijtraM I agree with you. Like so many of these puzzle kind questions, the information is presented 'out of order'; first we are told to choose a random line segment of a particular kind, and only later we are told that it means choosing a 'second point' along the circumference of a circle in a fairly well-defined uniform way. That is great, we can usually deal with the information presented 'out of order'. But then the rules change, now we should choose a line segment by some other algorithm. It is a bit weird then, after the rules change, to go "Oh, now the result changed too!".

[@Rodhern]

Maybe infinity is meant like the smoke and mirrors of a magic trick, that it is an important part of the misdirection? Certainly I cannot see there should be a difference either, would the 'paradox' not be the same if you used a finely grained but finite lattice to choose points(?).

[@Rodhern]

@Joji Joestar Thank you for the teaser to watch the extra footage :) Maybe lattice was not the correct turn. In the first instance you should choose one of N points arranged with regular spacing on the circumference. In the second instance one of N points in a grid (cropped/restricted to) inside the circle. In the third instance one of N points arranged by regular 'bearing and distance' from the center. The point is that for N large you won't see a significant difference by 'going all the way' to a continuum. Edit: Ah, now I think I see what you are saying. If there was 'only' N line segment you stand a better chance that the question asker will list them all for you in advance in the first place. Is that it?

[@jonathancohen2351]

@Joji Joestar 1/N is just another claim for the meaning of randomly choosen. You could still imagine a situation where it made sense to choose two points on the edge vs a point in the middle and then one on the edge.

[@egnos]

I mean, you can't watch this video without wanting to watch the follow up. Well done!

[@runderdfrech3560]

I think the first one (1/3) is correct. It's the only calculation where the propability of every point is the same. We can say, the first point doesn't matter, it will only determine how the triangle is orientated. It makes sense to simply calculate the propability that the second point is in the third of the circle between the two other points of the triangle. In the third attemp, with the drawing of the second, you could traw a line from the centre of the circle through the intersection point of the red circle and the red triangle to the circle itself and traw a line rectangle to it from any point of that line. The legth of the part of the great circle that would lead to a midpoint outside the red circle is 1/3, the leght of the parts of the circle that would lead to a midpoint inside the red circle is 1/2 - 1/3 = 1/6. So the propability of a midpoint inside the red circle is (1/6) / (1/2) = 1/3. In the second attemp you should traw a circle touching the great circle on a corner of the triangle and its centre. It represents all the midpoints of the pairs of points when one point is the triangles corner. Then connect the 2 intersection points of that circle and the triangle. You have a smaller equilateral triangle with the centre beeing the centre of the small circle and you can easily see, that a third of that circle lays inside the great triangle meaning the two points are further away from each other than a side of the great triangle.

[@slickytail]

It's not a problem with a correct answer. The set of chords on a circle is not a priori a metric space, and so we can't infer what the distribution of the chords should be assumed to be.

[@runderdfrech3560]

@@slickytail. Why shoudn't it be a metric space? You take totaly RANDOMLY two points of the circle. You could say the points can be from 0 to pi, the computer chooses randomly the first digit, then the second one, then the third one and so on. And then you connect the two points. The other shown ways don't lead to a equaly distribution of points and make some configurations more likely than others. And if not told we have to assume that the possibilities are equaly likely. Otherwise you could also say ,,It's 90% probable that the result on roulette with one throw is the number 5" ... yes, that's a lobsided table where the ball always stops on the number 5. No, this doesn't make sense.

[@slickytail]

@@runderdfrech3560 (I meant measure space, but the point still stands) Because you're imbuing it with a specific measure by declaring that the uniform distribution on it is induced by the map from (S1)^2 that takes two points and gives you the chord between them. That's a reasonable measure to use, but you can't a priori assume that that is the distribution on the space. As we saw in this video, there are at least three easily-describable measures on the set of chords of the unit circle - if a problem asked you to consider a random chord and didn't specify which distribution to give the set of chords, then the problem would be underspecified.

[@DonPaliPalacios]

Joseph Bertrand has two other paradoxes named after him: one in economics and another one in probability theory ("Bertrand's box paradox"). The latter would be a fantastic subject for another Numberphile episode!

[@addymant]

The problem isn't anything to do with infinity, it's that the problem is poorly defined.

[@GGenoce]

Yea, I was thinking the same. At the beginning of the video my first thought was "does random line mean picking 2 random points, or randomizing the angle+offset, or what?". At that point I was wondering if these would give different distributions (and as shown in the video, they do). It's just that there's many ways to generate a "random" line going through a circle, and different random functions give different answers. Since the question doesn't define which way you should use, literally all 3 answers here are correct - they're all just answering different questions, since the actual question didn't give enough information. So the actual takeaway from the video is that one should always make sure they're giving enough details when writing a question related to "picking something at random".

[@ophello]

The question of random chords being selected should be well-defined before the probability is calculated.

[@matthewbertrand4139]

ah yes, how fitting that a man bearing my family name came up with a paradox in which you can never be sure that what you're doing is the correct thing and you are doomed to paranoia forever

[@palleppalsson]

Change your name to Alzheimer and forget all about it.

[@ericcooke2661]

I posted this on the second video also where the symmetry part comes from: I think the answer to Bertrand's Paradox is more about the way the chord was chosen was through a multi-space probability rather than the symmetry of the object. Choosing a chord through finding points on a circle is a probabilistic space of {CxC}, where C is the probabilistic space of points on the edge of a circle. When choosing a chord by finding a midpoint of a chord parallel to the midpoint of a circle, one effectively uses bounded {XxY} probability space bounded by the circle, where the choice of Y is dependent on X, if X is the first choice when picking a random point. Finally, choosing a point on a line extending from the center of a circle then choosing a point on the line is in a probability space of {CxX}, where X is bounded by the radius of the circle. The reason why all this matters in when finding the probability is that C is not a topological space of bounded X, even though it is common practice to do such a thing, because when transforming C->X the circular order property is lost thus the distribution of points of C must be transformed to "fit" the points within bounded X. I should mention there are ways to mathematically model a circular distribution into a line. First, one can just use an unbounded line but that isn't used in the problem. Second, one can remove one point from the circle which it can be modeled onto any line segment. Both of these models deal with the circular order of a circle, one lets infinity be a number and equal to negative infinity, where the unbounded opposite ends of the line connect, the other removes a single point to remove the one point that would have to mapped onto both ends of the line segment.

[@choco_jack7016]

the 3rd one is clearly wrong. the points in the center would be chosen more often, which means you should have a non-equal distribution

[@johnboyer144]

I was thinking the same thing, but mostly because he is always picking a chord that is perpendicular to the line from the center, when it doesn't have to be.

[@choco_jack7016]

@@johnboyer144 if you think about it all chords are perpendicular to the center of the circle

[@gamerdio2503]

@@johnboyer144 Yes it does. All chords are perpendicular to the radius

[@roblaquiere8220]

What I think is happening is the algorithm for selecting random chords is skewing the results because not all chords have equal weight to be selected in a given selection algorithm.

[@Kriscor01]

@@roblaquiere8220 thats the point of the video

[@anurag_verma_youtube]

6:31 "There is no reason to prefer any one of those" Avoiding pi vs tau argument 😅😅😅

[@drumset09]

This is why I love and hate maths.

[@bigcountrymower4263]

I love both Numberphile and 3b1b. The audio on this video is way better than on 3b1b. Normally Grant's voice has a grating, ASMR like quality to it. In this video he sounds like a normal guy.

[@BrunsterCoelho]

For the second method as explained in the first video, can someone help me understand why a point on the circle uniquely identifies a cord? Doesn't the center of the circle have infinite many cord passing through it? I thought the overall angle would be maintained (e.g. only horizontal cords, then argue by symmetry) but then that falls into the third example.

[@Cyrinil142]

I suppose the origin of the circle doesn't uniquely describe a particular chord, but for the sake of this problem does describe a unique length as any chord that passes through it is just a diameter. Though the proportion of the infinite chords that are a diameter is zero I suppose.

[@diniaadil6154]

Any point besides the center uniquely defines a chord given that it is the midpoint of the chord. The circle however defines infinitely many chords (diameters). This is kind of a reason why we get a different answer : all diameters are "packed" into one single instance of the distribution , which lowers the probability since diameters are the largest among all chords.

[@RexxSchneider]

@@diniaadil6154 I'm assuming you meant "The centre however defines infinitely many chords (diameters)", and that's a valid issue to raise. I should also point out that all points on the circumference also define chords of zero length, and there are an infinite number of those (which are the smallest chords, of course), but it raises the issue of whether the infinite number of points making up the circumference is the same infinity as the infinite number of points in the annulus of the circle you use when considering whether the chord is longer than the side of the inscribed equilateral triangle.

[@albertrenshaw4252]

The origin does have infinitely many chords but they’re all the same length so it’s moot. A more valid wording of the property would be for any midpoint it will only produce one chord-length then you can resume with his implementation as is, no confliction

[@nathank3721]

@@albertrenshaw4252 Right, but therein lies the bias -- in averaging the chord lengths given by each point, you're implying that any set of chords defined by a shared point is weighted equivalent to any other set, as opposed to each individual chord being weighted equally. You're packing the distribution of chords non-uniformly.

[@DestroManiak]

Bertrand Paradox isnt about "infinity" it is also about "uniform distribution". This is a very important point. The paradox is that different ways of defining "uniform" can result in wildly different probability distributions.

[@nifets]

so basically it's about the underlying measure you consider

[@jessejordache1869]

I walk around with this implicit assumption that "everyman"'s understanding of numbers grows along side mine. So when something like the Monty Hall problem comes up, and someone starts to argue with "1/2", my reaction is always "still? Why don't you understand it yet?" when it's probably someone encountering it for the first time.

[@TampaCEO]

Numberphile teaming up with Grant?!?! It's like watching Superman teaming up with Batman. You guys could conquer the universe together. Definitely happy to see Grant on your channel!!!

[@fergalcarling9274]

I’m quite sure the answer is 1/3: A chord is a straight line between two points on the circumference of a circle. The two variables for making a chord are therefore the positions of the two points the chord is drawn between. A chord then also has attributes dependant on these two variables, such as: angle, midpoint position, and length. In this question we are concerned with the distribution of the length attribute of a ‘random’ set of chords. The first method uses a uniform distribution of the position of the two points on the circle to create the chords. This will then define the distributions of the chords’ attributes e.g. length. Notice how this method does not pre-set the distribution of its attributes. The second and third methods use a uniform distribution of the attributes (midpoint position) of the chords to define the chords, which causes an altered distribution of other attributes. These are false methods, as they are not using the definition of a chord to decide which properties of a chord are the creator variables and therefore to be uniformly randomised.

[@catinthebox9400]

I know this is theoretically supposed to be unsolveable, but I think I came up with another answer and more reasoning for 1/3 If a chord occurs then it occurs infinitely. This means every length of chord is equally likely. To start, have a point fixed on the edge of a circle and a point of both the chord and triangle fixed on that point. Then trace a line from that point to every point on the edge of the circle. at this point you have a double of every possible length except for 1 and 0. As you have a copy of one half of the circle on the other half, you can simply do with a half circle, it makes no difference, I’m just lazy. Now, at this point you have an infinite amount of points with probabilities of 0, so this cannot work, however, an infinite amount of points in 1d space is a line, so if we measure the length of the line and the ratio of it in the critical length we get ⅓. However if the lengths themselves are equally likely than the probability is approximately ⅙. Here's what I got down.

[@guepardiez]

Fourth method: choose two random points inside the circle and draw the cord that passes through them. What's the probability in that case?

[@diegomo1413]

If you choose points like that, there’s an infinite amount of pairs of points that will give you the same chord, so you might as well define that whole set of pair of points by the end points of the chord they make, which is basically the first method. But maybe I’m wrong.

[@MegaYouNiko]

This feels more difficult to solve as you have two random variables. I would guess that this gives you the highest likelihood of ending up with a longer line, compared to the methods of the video.

[@MegaYouNiko]

Diego Mo the issue is that you change the distribution: when you chose completely random points, and you happen to end up on the edge with the first point, for the second point the overwhelming majority of possible locations is within the area of the triangle and behind it, whereas in the first method the edge segment on the far side only accounts for a third of the entire edge.

[@landsgevaer]

How do you pick the two random points? Random x and y, but inside the circle; or random r and phi, inside the circle; or... 😉

[@MegaYouNiko]

@@landsgevaer Well I would argue that here you clearly should pick uniformly from the circle's area, so either random x and y, or random sqrt(r) and phi, which yields the same distribution.

[@YiamiYo]

I intuitively arrived to 1/2 as an answer before even listening to the possibilities, because since the side of the equilateral triangle divides the half-circle in half, the outer half chords are smaller and the chords closer to the center are larger.

[@curtaustin8119]

Well, I'm no expert, but the moment I heard "random chord" I wondered "What's a random chord?". If I were Bertrand's student, I'd have insisted he give us the procedure he had in mine for generating them. Are the three discussed here the complete set?

[@frogkabobs]

They are certainly not a complete set, just three “natural” ways of picking a random chord, but otherwise arbitrary. There is an infinitude of possible ways to choose a random chord.

[@MCredstoningnstuff]

"Choose n random sets that don't contain themselves. What is the probability that you chose the set that you just made?" ~ The Bertrand-Russel Paradox

[@fahrenheit2101]

Haven't seen the rest of the video yet (just reached the second student's idea), and my current guess is that for some reason, using a randomly chosen midpoint, doesn't correspond to a randomly chosen point on the circumference, i.e. the distributions are different. Not quite sure why this would be the case, but this seems most likely to me so far. Edit: I'm at the third person's idea, and it looks like it clears things up - the second person's midpoint was biased away from the centre, since the area of a small central area is very small in comparison to the remaining area, though I'm not sure how to express such an idea mathematically (probably some calculus involved). The third person's idea seems to solve this with their uniform distribution along a radius, so I'd expect it to work out to 1/3. Edit 2: Ok, I seriously didn't expect that turn of events.

[@harishganesan3575]

It is a paradox, in the sense, that the question is not well defined.. All 3 of them are reasonable types of interpretation.