Extra footage from this interview: kzfaq.info/get/bejne/prCpfrBenK_LcYk.html 3blue1brown video on the shadow a cube: kzfaq.info/get/bejne/otp8iMSV06ndmWQ.html

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

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

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]".😅

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.

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

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

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

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

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

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.

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!

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...

This collab instantly improved my day

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."

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.

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.

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!

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

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

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.

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!

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 KZfaq 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!

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.

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

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

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?

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.

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

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.

Love this team up! MOAR!!!!

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

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.

This is officially my favorite numberphile video

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

Another one with grant! Excellent job as well again.

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.

Thanks for bringing him

Just a amazing explanation,I was looking for it last 3-4 months,and finally found it.....Superb......❤️❤️❤️❤️❤️

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.

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.

lovely collab again as always!

I knew you guys were friends - love both channels - content, style and personalities.

Seeing Grant in person is always a great time!

3b1b + numberfile = instant click

This is great! Need more like this!

Two videos from Grant this morning! 🎉🍾

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.

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

Grant is revolutionizing maths understanding among peers.

Really nice intro video to the paradox! THX!!

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.

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

Hi grant, I am a huge fan of your math videos and I have started learning manim, in the process of understanding the codes. Beautiful problems of maths.

Awesome video. Out of curiosity, I tried making chords by connecting two random points in the circle or two random points in the bounding square. After finding the line between these points (and discarding those that don't intersect the circle when applicable), it was interesting to see that this yielded 75% greater than the equilateral triangle and 66% respectively. Toss these into the ring as examples of students with somewhat reasonable solutions.

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.

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

i love this channel n i love a lot 3blue 1 brown.... absolutly awesome collab

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!

I love how this relates directly to sizes of infinitys

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

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.

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

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'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.

Wow! That was a marvelous treat.

okay, I've been watching 3b1b for a while and this is the most excited I've seen him. lol

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

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.

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.

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

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.

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.

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.

This is a great example! Cristal clear explanation.

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.

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

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

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. 🧐

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

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].

This is a best of video for sure. The maths here are fascinating.

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?

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.

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

Stunning presentation!

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.

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.

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

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.

This is why I love and hate maths.

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.

Perfect circles, here we go. Great video. Nice work.

i love your videos numberphile!

after seeing him draw I can tell why he uses software for visualizations on his channel

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.

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

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.