Bertrand's Paradox (with 3blue1brown) - Numberphile
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- Опубликовано: 20 сен 2024
- Featuring Grant Sanderson, creator of 3blue1brown.
Extra footage from this interview: • More on Bertrand's Par...
3blue1brown video on the shadow a cube: • A tale of two problem ...
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Special thanks to our friend Jeff for the accommodation and filming space.
Extra footage from this interview: ruclips.net/video/pJyKM-7IgAU/видео.html
3blue1brown video on the shadow a cube: ruclips.net/video/ltLUadnCyi0/видео.html
🤓 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. ❤
@@LightVortexMatrixStudy can you please elaborate what you are trying to express gentle sir
@@PluetoeInc. Seek and Find... That is our way.
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.
Hi there✌️Would you mind if I translate this video to my native language (Russian)? May I upload this translated video to my RUclips channel?
It must be nice to collab with Grant since he did his own animation
Next vid, Grant teaches Numberphile how to use Manim
@@Xingchen_Yan No, I'm sure he's keeping his tricks a secret for himself. Like a magician.
@@vigilantcosmicpenguin8721 It's open source tho
@@jayd2279 news to me now I'm gonna have so much power
@@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.
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]".😅
For anyone who was confused when he said one 4th -- he meant, as you say, 1/4 [of the outer circle's area].
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.
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]".
For anyone who is still confused still, if the area of the circle is A, then one fourth means A/4
For anyone still confused, go learn some English first
Every other person: This is because of my bad drawing.
Grant: Thats a "function" of my Bad Drawing.
7:22
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.
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.
@@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.
@@MultiPleaser Random points in *_*2 Dimensions._* You can’t have 3 dimensions in a circle.
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
@@user-he1cx9my2xWhat is the ambiguity? 🤔
This ended in such a cliffhanger, never felt so compelled to see the extra footage
Have you watched the extra video? Do they present a solution to this paradox?
@@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.
@@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
Me tooooo lolol xD
The problem lays in the defined rules. Stricter rules about how the lines can be drawn.
A triple cheers for bringing this important example up and making it known to the greater public! And so wonderfully rendered and explained, too!
*triple
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.
What do you mean?
@@michaellinner7772 Huh? (
I totally get it!
Blue sharpies were invented to solve difficult math problems on certain kinds of brown paper.
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.
mind = blown
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?
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?
@@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.
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?
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!
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.
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.
@@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.
This collab is a better Christmas gift than anything I have ever got from my family
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!
🥶
Damn... that's rough. I hope life gets better for you soon
This is a beautiful video. The style of numberphile mixed with Grant and his animations makes for a wonderful combination
Back to back videos by Grant. He just released one video on his channel 3b1b. What an amazing day!
And he talks about that at 9:55 when he mentions a problem with choosing a random orientation 🙂
I didn't even realize before reading your comment that this was numberphile. I was so confused why there was brown paper.
Taking the average of these results gets you 13/36.
but, I got 26/72
@@Redditard 26/72 simplifies to 13/36.
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
but how would this problem be fixed? what’s right?
Same, I was at about 4:40 in the video and I realized both the paradox and the solution.
@@WolfgangGalilei there is no right answer. You have to just rigorously define your chord selection.
In both the 2nd and 3rd options it's not very random to have that point as the centre of the cord.
@@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.
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...
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.
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.
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.
@@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.
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.
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."
someone="Donald Knuth"
mathematicians have allways the best quotes
@@gabitheancient7664 Physicists like Heisenberg would feel "uncertain" about your assertion.
@@jatinsoni1979 Are you sure? When I google the quote, everywhere it says that it's by Robert R. Coveyou.
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.
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.
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.
@@lukeearthcrawler896 It is a veridical paradox.
I feel like there is a _degrees of freedom_ issue here.
This collab instantly improved my day
6:31 "There is no reason to prefer any one of those"
Avoiding pi vs tau argument 😅😅😅
Couldn't expect less form a collab from my two favorite math channels. Awesome stuff!
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!
"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?
Anybody who's ever, against better judgement, rolled for stats in DnD, would be familiar with gaussian distributions.
@@wymarsane7305 But gausian distribution is not uniform. The video implies that all 3 distributions are uniform on their own ways.
@@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.
@@wymarsane7305 "Against better judgement"? The expectation for 4d6 drop lowest is higher than the average score using point buy.
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
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!
So which method is the most correct?
@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.
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
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!
Yes but how "random" was this book you were reading in college? ;)
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)
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 RUclips 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!
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
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.
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. 🧐
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
For those who didn't understand, the point of the video is that **all/none** of the answers are correct! It depends on what probability distribution you choose:
If the "random chord" is generated by randomly and independently choosing 2 random points on the circumference and drawing a line between them, the probability is 1/3.
If the "random chord" is generated by choosing the midpoint to be a random point inside the circle, the probability is 1/4.
If the "random chord" is generated by how it divides the radius, the probability is 1/2.
The problem is with the question: the term "random chord" is not well defined!
So it may be obvious to some people that OF COURSE choosing different ways of constructing the chord causes the distribution to change, but for a lot of us the question is still "wait, why does that happen? In every example you construct a chord by selecting a random point? What's different?"
In the first example, you select the two points by choosing two random points on the outer perimeter of the circle. Since the first point you select isn't important (you can always rotate the circle to align it with a fixed point), the random chord you select is constrained by choosing a single point on the perimeter.
In the second example, you only select one point inside the area of the circle. The random chord you select is constrained by choosing a single point in the circle's area--which is a very different distribution than the circle's perimeter.
In the third example you select a point on the perimeter and THEN select a random point on the radius you drop from it. Since, again, the point on the outside of the circle isn't important, the random chord you select is constrained only by choosing a random point on the circle's radius.
So, yeah. That's why this distribution isn't well-defined. You are, technically, dealing with several infinities here: there are infinite points in a shape's area, and there are infinite points on its perimeter, and infinite points on the shape's derivative lines. Any one of them will give you an infinite amount of points to choose from, which may trick you into thinking your first idea is the only correct one. But not all infinities are equal.
This is the best explanation I’ve read on how the probability distributions for these methods of choosing a chord differ. I couldn’t quite put a finger on it myself, but when Grant brought up choosing a random point on a sphere I could immediately see what the distribution problems with that would be.
That was brilliantly explained. I think I can see where the contradictions originate, though, helped a lot by those spiderweb diagrams he made.
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.
I should learn python.Jpg
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.
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].
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.
@@benweieneth1103 you are right, I edited my comment !
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.
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
The three approaches have different results because they start from different assumptions on what events should be equally likely
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%!
Yay, 3Blue1Brown is now a regular Numberphile guest !!!
What is the frequency of periodicity that makes it regular?
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!
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.
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
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.
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.
Shouldn't the third one be a solid helix?
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.
That can't be right
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.
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.
@@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.
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.
You are perfectly right:the distribution of the middle point on the radius is not uniform.
Answer: consider the set of all possible chords. The largest possible chord (diameter) has a length of 2, while the smallest has a length arbitrarily close to 0, or 0. There are infinitely many of each length of chord, making the probability distribution of selecting any length of chord at random a uniform distribution. So the probability of selecting a chord from [0,2] of length >√3 = (2-√3)/2 = 1-√3/2≈13.4%
My "favorite" thing about sampling infinite spaces it that you can't uniformly sample from the Naturals (or the Integers).
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!
@@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.
@@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.
@@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.
@@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.
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.
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
No, none of the answers are correct! The question just isn't well-defined. There is no single way to select a "random chord".
@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.
@@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.
How would this solution work for a line that has the length of the diameter ?
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.
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.
Yeah, like given that we know any rational number must either terminate or repeat, what is the probability that it does either?
@@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.
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.
Thats exactly what he mentions at the end of the video...
@@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.
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!
"They're all valid"
But only the third one gives you the true answer
@@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.
@@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.
@@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.
@@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?
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.
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.
@@invenblocker now I got it Thanks
@@umaresrar717 Glad I could be of help.
Without loss of generality are the four sweetest works in the English language.
Thank you for your job! It’s always interesting to see something new at this channel 😃
Another solution to this is to simplify the problem down to a line rather than even considering the different orientations of the circle and triangle.
We know the length of a side of an equilateral triangle inscribed in a circle of radius 1 is √(3), and we're focused on all chords longer than that.
If you arrange the entire circle to be infinitely many vertical chords, and separate the circle into 3 regions: a segment on the left with all vertical chords shorter than √(3), a segment in the middle with all vertical chords longer than √(3), and a segment on the right with all vertical chords shorter than √(3).
The solution becomes the width of the middle segment of the circle, as it gives you the proportion of chords that are longer than √(3) compared to chords shorter than √(3).
The sagitta of the left and right segments of this circle would each be of length 1/2, which leaves the width of the middle segment to be of length 1.
So, our "line" (the width of all 3 segments of the circle added together) is of length 1/2 + 1 + 1/2 = 2. Meaning the proportion of the middle segment (chords longer than √(3)) compared to the proportion of the left and right segments together (chords shorter than √(3)) is 1/2, or 50%. This matches the solution in the video.
It doesn't matter which way we orient the chords (vertical in my solution), because we're looking at a circle. If you compare ALL chords of any orientation to all other chords of that same orientation, the proportion of chords shorter than √(3) to chords longer than √(3) would always be 1/2 or 50%. You can consider ALL infinite orientations of the chords within the circle this way, and the proportion would always be 50%.
Edit: After writing out my solution and looking back at the video's solution, they're basically the same idea, but looked at from two different perspectives.
The problem here is that in all 3 cases they didn't choose a random chord. They chose a random point on a circle, point inside the area of a circle and point on a line respectively, from which they CONSTRUCTED the chord.
Correct me if I'm wrong, but to construct a truly random chord, take an infinite 2D plane, draw your circle on that plane and draw an infinite amount of line segments with random starting and ending coordinates (with X and Y both being real values for those coordinates). Prune all the line segments that don't touch or intersect the circle and now determine the proportion of all lines left of which the chord inside the circle is longer than s. Sure this isn't feasible to actually perform as an experiment, but you might be able to approach this answer using a large, but finite plane and by using double precision floating point numbers, rather than real numbers. Then again, you can argue that this is not a random chord, but a chord constructed from two random points on a plane. The term "random chord" is just too ambiguous.
You should watch the second video. Grant shows that the third construction is the same as what you describe. But from my perspective, all three methods of picking a chord are "random" chords. You just have different distributions as noted by the density of what 1000 chords looks like in the end.
I think you could compute this efficiently by picking a random point ON the circle and then picking a random angle for the chord.
A fourth method:
Choose a random point P on the circle.
Choose a random radius r between 0 and 2.
Draw a circle around P with radius r.
Randomly choose one of the two intersection points between the two circles and connect it with P to get your random chord.
You can get each cord with this method, so it's as legit as the other three.
The chord you get is longer than the side of the triangle iff r>sqrt(3), so it's probability is (2-sqrt(3))/2 which is about 0.134.
3b1b + numberfile = instant click
This is why we can't assume that a distribution is defined the way which we assume that it is. If your criteria for placing a chord are different, then you can different results in the way that they're placed. We can see this demonstrated in how the web plots are visibly different in where the lines are distributed.
I think you could come up with even more answers if you get creative in how you define the distribution of chords. Here's one I thought of:
Any chord in a unit circle has a value between 0 and 2. Any two chords of equal length can be rotated to become the same as each other, so we can eliminate congruent chords by planting one endpoint in a stationary spot, and moving the other endpoint around the circumference of the circle. If we only draw chords on one side of the diameter line, then each chord we draw from our starting point will be the only chord that with a given length between 0 and 2. If we decide to plot our chords uniformly by their length, then we can just divide our desired cutoff length by the range, which is 2. In this plot, half of the chords are between 0-1 and half are between 1-2. One quarter are between 0 and 0.5, etc. So if we want the number between sqrt(3) and the maximum value of 2, we subtract sqrt(3) from 2, then divide the difference by 2. This gives an answer of (1-[sqrt(3)/2]), which is approximately 0.134.
This is a weird distribution and doesn't make much sense, but the answer depends on how your chords are distributed.
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.
Could you describe the paradox :)
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.
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.
@@cmyk8964 Where you said that the probability is 1/3, I think you meant 3/4
@@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.
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.
so basically it's about the underlying measure you consider
Team Student #1 for the win.
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.
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.
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!
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.
@@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.
@@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.
@@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?
**cries**
**takes another bong hit**
**chillaxingly ready to watch extra footage**
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.
" In this article we resolve Bertrand’s probability paradox and show
why it has perplexed researchers for 120 years." Wang, J., & Jackson, R. (2011). Resolving Bertrand’s probability paradox. Int. J. Open Problems Compt. Math, 4(3), 73-103. - the solution comes at the end of the article ;)
The problem isn't anything to do with infinity, it's that the problem is poorly defined.
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".
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.
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...
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.
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.
Wasn't that one of the Summer of Math video submissions? (the 3blue1brown event)
the paradox is that *it's still uniform*
just not uniform in the way _you_ think of
@@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.
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.
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.
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.
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.
@@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.
@@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.
The most anticipated Cross over episode of all time! Brady and Grant together!!
I love how this relates directly to sizes of infinitys
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.
@@daniel.lupton 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.
I am at 1:30 and before I look at the answer, here is my intuitive idea: look at where the "first point" of the chord is. Draw the triangle in such a way where one corner of the triangle touches that point. Then pick another point on the circle at random, that will be the "second point" of the chord. Since the circle is divided into 3 equal 120 degree arcs, there is a 1/3 chance that the second point will be between the other two points of the triangle, hence making it longer.
Therefore my answer before I watch the video is: the chance is 1/3. Let's see if I'm right.
You are 1/3 right
Another one with grant! Excellent job as well again.
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.
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
Change your name to Alzheimer and forget all about it.
To my schock I found that I was not subscribed to the channel. Well, the Algorithm did not seem to care. I hope you reach 4 million subs before 2022!!
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
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.
@@johnboyer144 if you think about it all chords are perpendicular to the center of the circle
@@johnboyer144 Yes it does. All chords are perpendicular to the radius
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.
@@roblaquiere8220 thats the point of the video
The answer to this is simple circular geometry. You categorize all chords by length and you you count how many chords there are. You know there will be an infinite amount of chords so you are dividing infinities into 2 groups. The chords that are longer and the chords that are shorter. You can atleast say that the the ratio will be defined by the comparison of the 2 arc lengths. Then ontop of that you would say that theoretically, a perfectly random chord drawing machine would have to pick 2 random points on the circle and draw a line between them. As a result, the painting of the circle would be weighted away from the centre because we're dealing with chords which generally have a higher chance of not being through the centre and more towards the sides. You would then merge this probability distrobution with the cut off set by the ratio set by the chord lengths and compare your 2 new sets and the ratio between the 2 of them to get your answer. You definitely need a geometric understanding of infinities to know this. One of the answers he wrote is actually quite close to the true answer.
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.
I believe 13/36 chance of the chord being longer.
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
Only if you evenly weight all three methods. And actually there are infinitely many methods so it doesn't help anyway :).
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.
May have reversed orientation in the original, but the idea should still be clear.
Fourth method: choose two random points inside the circle and draw the cord that passes through them. What's the probability in that case?
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.
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.
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.
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...
😉
@@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.
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.
Seeing Grant in person is always a great time!
Thank you google for making this the next video in autoplay after a meme compilation, i actually lerned something new instead of wasting more of my lifetime.
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.
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.
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.
@@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.
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
@@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.
I believe this demonstrates how ambiguous mathematics or numbers can be.
I ended my University life by accepting all we are taught are very limiting, giving us very little freedom to ask questions and reach unique but still relevant answers.
What we are not taught is that 'This is one way of doing this' and we are left more ignorant than wise.
I was a little perplexed at first, but then I realized that Bertrand Russell´s paradox is of course referred to as "Russell´s paradox". So I managed to have at least something sorted out.
Grant is revolutionizing maths understanding among peers.
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.
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?
@@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!".
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(?).
In the extra footage it does confirm that infinity isn't the problem.
@@Rodhern with finely grained lattice points there is only 1 answer. Because you can then say if there are N total points that the probability of picking a point is 1/N. That would be a uniform distribution and is the standard meaning of picking randomly from finitely many things.
"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
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?
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.
Methods 1 and 3 actually give you the same results when handled carefully. Method 1 is assigning a uniform distribution to points on the circumference, but changing those to the line segment of method 3 no longers yields a uniform distribution and when computed, the probabilities are the same 1/3
You could, however, assign a uniform distribution to points on the line segment in method 3. Then when transformed to points in the circumference both will give a probability of 1/2.
There is no inconsistency, just a problem that is not well defined by the original statement and freedom to choose what we assign the uniform distribution to.
I think "1/2" is right. Here's why:
Imagine 2 equilateral triangles inscribed in the circle (creating a Star of David shape). [When calculating the side lengths of radius 1, each triangle side is square-root(3).]
If we begin by drawing all possible horizontal chords (starting from the bottom most point of the star to the upper most point), we can see that each chord will be unique (though an infinite number of them). Approximately 1/2 of them will be longer than the square-root(3), and the other half will all be shorter (there will always be exactly 2 that are equal to).
Now if we rotate the circle (but not the star), we can see that each infinitesimal angle change will give us more brand new (unique) chords, until we have done so for every possible rotation, until we have gone through 180 degrees (no need for 360, because they all begin to repeat after 180, or pi). That would indicate that all of the probability still comes out to 1/2.
Regarding the first proof (the 1/3 one), if you consider all possible unique chords that pass through just one single point of the triangle, you indeed are then finding 1/3 that are longer than the triangle's side length. BUT! When you find ALL possible unique chords in total, you must begin to move to the next infinitesimal distance away point. Then you must take into account that there is OVERLAP with the previous set of unique chords-implying that this infinitesimal shift has caused one single chord of overlap. As you begin moving around the circumference, you see more and more overlap-fewer and fewer UNIQUE chords. When subtracting out the overlapping chords, you again arrive at 1/2.
I don't have quite as clever a way of bringing the "1/4" answer into the 1/2 range, but that is partly because I'm not really familiar with how a chord could be defined by its midpoint.
I made my reply, then saw yours and we basically constructed the same solution! (for the correct "1/2" solution)
->the correct answer is 1/3. to choose a totally random cord, one must choose 1 random point, then another random point completely independently from the first. This will be a truly random cord created. This will result in a cord less than √3, 1/3 of the time.
->The third example with 50% is incorrect since the first point picked influences the position of the second point. The second is mirror image of the first point, creating not a truly random cord. This study gives this result "given any point on the circle, a point on the opposite side of a predetermined axis dividing the circle in half, will create a cord that is less than √3 long 50% of the time"
->the 2nd example is flawed since there are MANY cords created passing through the center point. We count all these as 1 case and all the other points all around as just as likely to create a cord. The odds of falling into the inner area is 1/4 only, but on the unlikely chance we hit the center, we get a "jackpot" and get infinite cords creates. Any random point on the circle will create 1 unique cord, but the center point will create infinite cords.
so case 2= "a point on the circle has a 1/4 chance of generating a cord with less than √3 length, regardless of the quantity of cords generated", what we are looking for is "a random, unique, independent cord created on this circle, what is chance it is less than √3 long" which is 1/3