Alright, Brady so I've got for you a puzzle. It's about darts. It involves some higher dimensional geometry and a couple of the most famous numbers in math So I'll describe the game. We want to hit as many bullseyes as we can. So right now the bullseye is this tiny red thing, but that's too small, we're going to start with a giant bullseye that fills up the entirety of the board. I have a proper compass so I'm gonna go ahead and cut out a piece, just a piece of paper to indicate our enlarged bullseye. [laughs]
- (Brady: That'll do) Not the greatest circle I've ever drawn. All you do is you try to hit the bullseye. Each time we hit it though, based on where our shot is the bullseye's gonna shrink. So it's gonna get harder and harder as we go.
- (Brady: You're not gonna hit the wall are you?) I'm not *that* bad! Oh, okay
- (Brady: Oh my goodness, hang on. That's amazing!)
- Thank you very much, that is amazing. And you're kind of rewarded for a good shot. So let me tell you what the rule is for how we shrink it. So step one, I'm gonna draw a line from the center to where I hit. And maybe for the future I'm just gonna call that distance H. So that's the distance of my hit and then what I'm gonna do is draw a chord of the circle that's perpendicular to this line. That's our chord. And our rule is that the length of this chord is the new diameter of the circle, which in this case because I had a good shot - right? The shot was nice and close to the center, I'm rewarded by the fact that it's not going to shrink that much. (Brady: Is the center of the new circle where your dart hit or where the original - ?)
- Great question. It's where the original centre was. All right so I'll draw a new circle where the original center was. Actually wish my shot was worse now because this is kind of too close to the edge, I mean you can kind of see it's, it's barely smaller.
- (Brady: Oh, yeah, so that's *just* trimming that circle down a bit) So I'm just gonna be giving it a little trim. This is like getting your hair cut, you know once every week or something like that. And presumably this means that the next shot is gonna be similarly difficult. This is actually fascinating because when I animate it then it's like oh boom the computer does it. And I'm not messing with the physical dartboard in any way. I think I'll stick to the animated style. For the sake of illustration I'm gonna try to make this a worse shot? Okay.
- (Brady: Oh my goodness!) So if I do the algorithm again, like this cord is a little bit shorter because we trimmed off the edges, right? And so it'll also be just a tiny bit smaller. So we kind of could run that again, but let me let me try to get a worse shot. (Brady: You can't even do a bad one when you try!) Yeah, consistency is only a virtue if you're not a screw-up. So this one's nice and far away from the centre, and I'm going to be punished more because it's far away. So we'll do the same game. We'll take this off. We're going to make some cuts. Distance is h1. And so it'll be a chord of the big circle, perpendicular to that line. The chord defines the new diameter, but it's kind of easier to think of it as half the cord as defining the new radius, certainly if you're drawing with a compass. So I'll set my new radius, but this time because I was closer to the edge of my you know, quote-unquote bullseye that cord ends up smaller. So I'm punished which means that the new bullseye is meaningfully smaller. Smaller bullseye, play the same game. I'm going to set it up so that the centre is at the centre of the board. Now I'm gonna actually try to hit closer to the centre
because I want more shots for a higher score. And I - I didn't hit close to the centre. (Brady: Ooh dear, you are going to get punished for that) So at this point, maybe you can already see how much that's gonna punish me, the idea that it's close to the edge means that the chord defining the new radius is going to be quite a bit smaller. Put it on the board...
- (Brady: Now we see how good you are)
- So far I've hit three, this is for a fourth. All right, I miss. (Brady: Oh dear, game over.) Game over.
- (Brady: So what's - so your score for that game is three, because you hit three?)
- I hit three I will say that would be the natural way to score it, based on what I want the answer to look like at the end here, you get one point just for playing. Okay? So you get one point just for playing and then one point for each bullseye that you hit. So in this case I got a score of four. You can say maybe you're - what we're counting is how many darts you throw in total? So I threw four darts and the fourth one happened to be a miss, whatever you want to do to artificially add one. You'll see why we like that at the end. All right, so I've defined the game, I haven't defined the puzzle yet. And so for the puzzle, let me just draw our board again, or at least draw a circle. All right, so we're gonna have a random darts player. So he throws at the board with some random distribution, we all do.
- (Brady: But there's no skill here, there's no - there is absolutely no waiting for the skill) (it could - it's as likely to be anything?) This will be a very bad - in fact, so he throws it in a weird way. We're gonna think of the square that encompasses the board, and this is a little unrealistic because any real dart player, their distribution is probably rotationally symmetric, right? I mean, why would it care about the diagonal directions? But basically he hits it at a random point anywhere within a square encompassing, whatever our first circle was. Okay, so maybe you imagine that the whole dartboard is a square and then there'll be shrinking within that. That's my X coordinates, Y, and maybe let's say that this happens to be radius of 1 for the original circle because why not. The x-coordinate for this person is always going to be something between negative 1 and 1. And it'll just be completely uniform within that. So this would be the simplest thing to model on a computer. To do it you just choose two different numbers, each one between negative 1 and 1, where every one of those is equally probable. So every point in this whole square is equally probable. And that square doesn't shrink it's only the circle that shrinks. And the question is what's the expected score? Okay, so we want to know the expected score for this thrower. (Brady: So Grant, after you throw your first dart and the circle shrinks the square doesn't shrink with it?) It does not shrink, no, so the the distribution for the thrower is the same. You might imagine the dart bored itself which is fixed is square, and the bullseye is shrinking. Okay? Because otherwise it's gonna be the same probability for each shot and some viewers might be able to guess what that probability is, and it wouldn't be as interesting a problem. Another way to think about the expected score, imagine this random player plays a thousand games, right? Just what's his average score for those thousand games? Another way to think about it, is you say it's going to be the probability that the score is equal to 1 times 1, for that score, plus the probability - because remember the lowest score you can get is 1, because you get that 1 point just for playing - the probability that the score is equal to 2, times 2, plus the probability that the score is equal to 3 times 3, and so on. So what we need to figure out is all of these probabilities. In principle you could have a perfect game, right? If you hit in the center every single time and it never shrinks, you're hitting a perfect game. What we'll find is that, that's there's a probability of zero that you hit a bullseye infinitely many times, but it's also interesting how quickly does it shrink? Would a score of 10 be absurd or would that be maybe expected? (Brady: What is the probability of getting an exact bullseye? 'Cause an exact bull's eye is like) (presumably an infinitesimally small point?)
- It's probability 0 Yeah, so like one of those paradoxical facts about probability is you can have events that are possible, it's just probability 0,right? It's possible to hit an exact bullseye, it's just probability 0. You know if you choose a random number from the number line it's possible that it's rational but its probability 0 and this - a lot of people kind of try to come to philosophical terms with like - how can it be possible but probability 0 the probability would have to be slightly bigger. But it's just one of those things of math. All right, so anytime you're solving a puzzle that's hard and this is hard, you see if you can ask an easier version of it or some easy sub problem within it, what's the easiest thing you can get a foothold in, right? And maybe in this case the easiest question to start that's not totally trivial is what's the probability of hitting the first shot? And this isn't that your score is exactly equal to one, if you hit a first shot your score is greater than or equal to one. Or no actually because you do get that point for free, this is the probability that your score is greater than one. Actually maybe you can tell me, right? What's the probability that on this first shot, with this random distribution within the square, it hits somewhere inside this whole circle? (Brady: I'm imagining we we need the area of the circle and the area of the square?)
- Yep, the area of the circle, this is pi r squared and that radius starts off as 1. So that ends up just being pi. And the area of the square is equal to, well the square side length is 2 times that radius right? So it's 2 by 2. So it's 4 which means the probability that you hit that first bullseye, that your score will be greater than 1 because you get that one point for you just for playing, is gonna be the area of that circle divided by the area of that square.
- (Brady: Grant, what happens if the dart lands on the line?)
- That's probability 0. Don't worry about it, it won't mess with the problem. So now a much, a much harder question is okay, you've hit that first shot, what's the probability that you hit at least 2? And this is where it starts to become interesting. Because this depends on where your first shot was. I'm gonna start writing a little bit of a chart, where I'm gonna keep track of the radius of each bullseye and I'm gonna keep track of just the hit length, at each point. And so the radius starts off as 1 and that first hit, it's, it's a it's a random point. Let's just give it a name for now Let's say it goes there. That's H_0. So it will be H_0 and by the Pythagorean theorem, if we want to describe what this is in terms of x and y, which will be helpful for later on to be thinking of it in terms of that, it's the square root of x squared, plus y squared if those are the coordinates of your first shot. And because we're going to have other shots I'm also going to give these guys a subscript. So now it's a geometry question to ask what's the new radius? Well, maybe we call this our zeroth radius, right? And I want to know what's the next radius after that initial shot? And if we remember our rule, we say you draw the chord and the chord was defined to be perpendicular to that. So when you know that it's defined to be perpendicular you know that you're probably going to use that fact in your solution to the puzzle. So the new radius is this length here, right? It's half of that chord. So what we might do, and I recognise I'll sort of be drawing over myself a bit here, because we have a right triangle here and that's what defined it we're gonna use the Pythagorean theorem. Because this was our old radius, radius one. That's the new hypotenuse so we know that the old hypotenuse, 1 squared, is equal to one of the legs which was H naught squared, plus the new radius squared plus r_1 squared. So what that means for what r1 is, is it's the square root of 1 squared minus h naught squared and remember the only way that you're even getting to this point is this h naught was less than 1, if you've got that first one in the bullseye. So this will be nice and well defined, no imaginary numbers or anything like that. Similarly now, we're gonna have a new shot it's determined by two new random coordinates, whatever the coordinates of your second shot happen to be or the coordinates of the next shot, maybe I should say. And this is an interesting question, what's the probability that you get that second one in? Because it depends on four different numbers, right? So I think it's helpful to just write out what the, what the actual requirement is. h1, the hit in that second shot, is less than r1. And I can go ahead and rewrite that by saying h1 squared is less than r1 squared and the reason is we have a whole bunch of squares going on, maybe that's going to be helpful for us. Right because I can expand out what h1 squared is, that's x1 squared plus y1 squared and that has to be less than, well what is r1 squared? That's 1 squared minus h naught squared. And then I can also expand out h naught so I'll be writing this as x1 squared plus y1 squared is less than 1 squared minus x-naught squared plus y naught squared. So all I'm doing is saying I want to explicitly write out what the requirement is for the four coordinates of my initial random shot. And now I have a bit of an interesting question, I might rearrange this so that all of my x's and y's are on one side and the 1 is just on the other. What I have is that xnaught squared plus y naught squared, you know bringing these two over to the other side, plus x1 squared plus y1 squared is less than 1. What's the probability that this happens when all these four numbers were chosen randomly within the range negative 1 to 1? So at this point that's kind of a hard question right? I tell you choose four random numbers. We're gonna add up their squares. What's the probability that they're less than 1? And you think to yourself oh, I don't know, you know if they were all less than 0.5 I guess that would happen? But if one of them was 0.9 that doesn't throw everything off because the others could be quite small. And to think about this let's just think about the very first situation where instead of doing it for h1 and r1; when we're over here doing it for what seems like a much simpler situation and it was, the probability that h naught is less than r naught which is just saying that x naught squared plus y naught squared is less than 1. We have a purely analytic statement, choose two numbers out of a hat based on this rule, what's the probability that the sum of their squares is less than 1? You answered it, geometrically. You immediately knew oh, we need to use pi. We're thinking of the area of the circle, right? And it's actually quite subtle what's happening there, you have a question about a pair of numbers and you're gaining intuition and using facts that we've discovered in math, like the area of a circle. By thinking of that not as two separate entities but as a single point in 2-dimensional space. That might sound obvious, the reason I'm hammering down on it is look at what we have over here. Four separate numbers, what you're asking for is a probabilistic property of these four numbers. It might be natural to think of them as being a single point x naught, y naught. x1y1 inside 4 dimensional space. And it's basically a space that has four real number coordinates. And so now Brady let me ask you, what's the analogous question? In the same way that asking what's the probability that the sum of two random numbers is less than one, begs the question of what the area of a circle is. When we ask what's the probability that the sum of four squares is less than 1. What do you want to know? (Brady: Well I was how do you think we were gonna end up using a) (cube and a sphere, but I feel like we've jumped past - ) We've jumped! We've jumped! The whole idea that, hey if I choose three random numbers x squared, y squared, z squared, what's the probability that they're less than 1? If they're chosen by the same rule what you're going to be asking for is the volume of a sphere, because what it means for the sum of their squares to be less than 1 is that they sit inside a unit sphere. The volume of a sphere is 4 thirds pi r cubed. r is just going to equal 1. So that's 4/3 pi. That's the volume of a sphere. The volume of the cube which is 2 by 2 by 2, is going to be 2 cubed, which is 8. So if we were asking an analogous question where 3 dimensions happened to pop up you would just divide these two numbers. You would have 4/3 pi, divided by 8. Which would be pi sixths. You'll see it's actually quite helpful that we're gonna skip over odd numbered dimensions here. And so here we went straight from 2 to 4, the question that we want to ask right now is what's the volume of a 4 dimensional sphere? Or the 4-dimensional equivalent of volume. Instead of saying hyper volume or coming up with a new word, they just say volume. You could fancifully say measure, but we're just gonna say volume of a 4d sphere. And again if you wanted to be pedantic, the mathematician might actually say ball? Because when they use the word sphere they typically refer to the boundary right? It's just the peel of the orange. It's not the contents of the orange. So what we really want is the ball. The other thing that we need is the volume of a 4d cube. The cube is the easy one. Just like in the other cases the side lengths are all 2, 2, by 2, by 2, by 2. So that volume is 16. Volume of the sphere, there's a whole interesting discussion that we could have about the volumes of higher dimensional spheres, the surface areas, or the analogs of surface areas of higher dimensional spheres, and where they come from. I'm not going to give you that full story now, maybe at some other point. It happens to be the case that it is pi squared halves times r to the fourth. So it picks up an extra pi. Which is kind of interesting, when you go from 2 to 3, you don't pick up that pi but when you go from 3 to 4 you pick up that extra pi. And because our radius is 1, in this case that'll just be pi squared halves. And let's remember what this is. This isn't the probability of hitting exactly two shots, right? It's that you hit those first two shots, so it's the probability that your score is at least two. Maybe it also happens to be three or four, so we can write our answer here. The probability that our score is greater than or equal to two is equal to pi squared, and I'm going to actually write it as being over like this, 16 as 2 to the fourth, just to remember where it came from. So this was kind of the, the cube, or the hypercube. And then times 2. Where that was the denominator that we had from our volume of the 4d ball. And now we just keep doing this for all infinitely many different scores that we might have. Very similarly when we're asking about the probability that our score is at least 3 or that it's bigger than 3, what this comes down to is asking - we're going to have 6 different coordinates: x naught, y naught, x1, y1, x2, y2; and that the sum of their squares is less than 1. (Brady: I don't know how to calculate the volume of a six dimensional ball)
- No! Luckily mathematicians have figured it out for us. And again, it could be a whole interesting story, but if you go to Wikipedia, you could see a chart, there's quite a nice pattern that we're gonna - about to point out for, where that comes from. But the volume of a 6d ball happens to be pi cubed, over six times r to the 6. And then the volume of the 6d cube is of course 2 to the 6, and so the answer to the probability of hitting those first three shots, is going to be pi cubed divided by that 2 to the 6 and then what its denominator was, which was the 6. Because that r ends up being 1. And now I'll tell you the more general fact, which is pretty mind-blowing. Is that the volume of a 2n ball is equal to pi to the n, so half of the dimension, right? So in 2 dimensions you see a pi, in 4 dimensions you see a pi squared, and six you see pi cubed. So pi to half the number of dimensions, divided by n factorial. Very clean, very beautiful. Very reminiscent of another celebrity real number that a couple people might be familiar with now and that might be poking in their minds. But if it's not we'll get to it in just a moment. Given this as a fact that we'll just hand down from on high, let's figure out the expected score. (Brady: Please, that's what I'm here for!) Great. First let me get a cup of water. [Hold music] Alright, are you ready for the grand finale?
- (Brady: Yeah!)
- The finishing touch. It's, it seems like it'll be chaotic but it'll collapse in the most wonderful of ways. So the expected score. Expected score, now remember what this means, we're gonna have - it's going to be 1 times the probability that your score is 1, plus 2 times the probability that your score is 2, plus 3 times the probability that your score is 3, and so on. Now all of our expressions are greater than things. They're not saying exactly equal to, but that's completely fine. Because if I wanted to say for example what's the probability that your score is precisely equal to 2? So this will be your probability that your score is greater than 1 right? Minus the probability that the score is greater than 2. Because for it to equal 2 it's greater than 1, but it's not greater than 2 and this entirely encompasses the possibilities, right? If you're probably if you get a score of greater than 1, that encompasses all the possibilities of getting a score greater than 2, so it's okay to just subtract this off. We don't have to do anything more than that. So I'm gonna expand all of these out just a little bit before things collapse nicely. So this would be 1 times the probability that the score is bigger than 0, minus probability that the score is bigger than 1, plus 2 times probability that the score is bigger than 1 minus the probability that the score is bigger than 2. And just one more for good measure, probability that the score that's bigger than 2 minus the probability that the score is bigger than 3. So at this point we have a lot of cancellation that happens. So we've got this probability that your score is bigger than 0 that just stays on its own. For the probability that it's bigger than 1 we're subtracting one here but then we're adding two of them over here. So we've got probability that the score is greater than 1 and then similarly for score bigger than 2 we're subtracting 2 here but we're adding 3 back here. And so we're just going to be adding probability for the score is bigger than 2 And in general all we're doing is adding up all of the numbers that we just saw. All of those fun things involving pi. Because with the 3 we're subtracting 3, but in the next one we'll be adding 4, and so on. So every puzzle that we just had, every micro puzzle, we just add up all of their answers. So the probability that your score is at least zero, that's 1, because definitely your score is bigger than zero because you get a point just for playing. Probability that it's bigger than 1, that's the one that you answered for me first. Which was saying that it's pi divided by 4. The next one, so I'm actually going to write this slightly differently because we always have some power of pi divided by some power of two. I'm going to write this as being PI fourths squared divided by 2. Is that all right? So that 4 squared came from - that's the same as 2 to the fourth and then the pi squared just went in there. And so I'm going to write it here as pi fourths squared times 1 over 2. And then the next one we had up here. And again, I'm gonna write it in terms of pi fourths. Rather than writing pi cubed over 2 to the sixth, which is the same as pi cubed over 4 cubed, I'm gonna write it as pi over 4 cubed. And 6 is really 3 factorial - like the reason that we see the 2 and then the 6, it's coming from this general pattern of hanging, of having a factorial. So over here we've got pi over 4 cubed times 1 over 3 factorial, this is really two factorial. And maybe you see where this is going or maybe at this point you're saying you promised me this would look simple. It actually looks exceedingly complicated, you've got factorials, you're grouping the pis together for some reason. So there's some viewers right now, especially if they're maybe a little bit fresh off of a calculus class, or if they're particularly ingratiated to the number e, who will just see something screaming in their head right now. There's something known as the Taylor series for e to the x. Which is that it's 1 plus x and all right, it's x to the 1 over 1 factorial. x squared over 2 factorial, x cubed over 3 factorial and so on. Where you evaluate it as an infinite polynomial where each one of the terms is 1 over n factorial. It's not just that e to the x happens to equal this, I actually think the healthier way to view the exponential function and what e to the x is - this defines it, right? This is the thing that should pop into your head when you think of exponential growth and e to the x is this particular infinite series. This particular polynomial. This is where it will come up, especially in probability, it lends itself to an easier interpretation of why it is its own derivative, there's all sorts of nice things, it extends, it makes it easier to understand things like e to the (pi)(i); all of that. This is the healthy way to think of e to the x. So if you looked at this and you weren't thinking of e already, what it means is that there is a healthier relationship with e waiting for you in the future.
- (Brady: Or if you didn't see it you have an unhealthy relationship) You currently have an unhealthy relationship with e, that's absolutely true. If you didn't see it. Or you have no relationship with e. But if we compare this to the series, which I've unhelpfully drawn kind of far away, but you can match them up pretty easily, the thing that's playing the role of x right now is pi fourths. So all this is, is e to the pi fourths. Which, if we go and we plug it into a calculator, approximately equals two point one nine three two, on, on, on. Which is to say if you are a terrible random darts player, who unrealistically hits within a square with a uniform probability, which you wouldn't because it would be rotationally symmetric - but whatever you're hitting within a square. And you keep going like this then you play a thousand games, on average your score would be 2.1932. And remember my score was 4. (Brady: So you're you're not quite twice as good as someone who has absolutely no skill whatsoever?)
- I couldn't ask for a better recommendation Yeah, that's that's about right. So there's a couple things I like about this puzzle, right, the reason that I want to relay it. Hopefully what was the more unexpected or mind-blowing component is that we're even talking about higher dimensional geometry. You found yourself naturally asking, in the middle, what is the volume of a six dimensional ball? Right? And nowhere in that was asking, is the universe six dimensional? You know, like the string theory invokes 6 - no, no, no, like that's not why mathematicians necessarily care about higher dimensions. What was happening is you had six numbers and you were encoding a property of those six numbers with something that we like to describe geometrically. Rather than saying the sum of their squares is less than 1, you say it shows up inside a six dimensional ball. That also means it saved you some work, I put out this question to like some channel supporters as a early teaser of things, and one of them got back to me saying you know, I've been working on it and I, it's just like really hard. I've been like working through these integrals. What's very strange is even in the case with the second dart I'm getting pi squared for some reason. So he's working through all these integrals, under the hood what's happening is he's rediscovering the volume of a four dimensional ball, right? So it gives a more universal language for people to talk with on that. It came about from two-dimensional geometry. Nothing about the dartboard is four dimensional and I think this is just a common misconception that people listening to mathematicians describe things like manifolds in four dimensions or the Poincaré conjecture has been answered for everything except four dimensions. They don't actually care about like a thing where you can move in four dimensions, it's about encoding quadruplets at points. And the second one is that I think it helps build a healthier relationship with e. Because this this series is much more important than the number itself. I might, I might even just ask the puzzle and then we won't answer it, but it's a good thing to end on. (Brady: Yeah) So here we were choosing these random numbers between negative 1 and 1 and what were we asked - what we were asking is something about when you sum their squares. When is that less than 1? You can play a much simpler probability game where let's say I'm just gonna choose numbers from 0 to 1 with uniform probability. And I'm gonna keep going until the sum of the numbers that I've chosen ends up being bigger than 1. so for example, if the first number you choose is 0.3 and then the second one is 0.6, their sum is 0.9. And if the next one you choose is 0.5, that's the point when you go over. And so the question you can ask is what's the expected number of samples you need to take before it over blows one? e shows up in the answer, right? And the way it shows up is actually quite similar here and it's a way that's distilled because it doesn't involve circles so there's not the confusing factor of pi. You see much more pristinely the factorials. Very important is that I mention where this puzzle comes from. I saw this on Twitter, I think it was Greg Egan. He specifically designed a puzzle such that the answer would look like adding up the volumes of higher-dimensional balls, because we have this wonderful formula for the volume of higher dimensional balls, when the number of dimensions is even. And everyone who has a healthy relationship with e looks at this, a thing to the power n divided by n factorial, they scream in their mind the exponential function with e to the x, right? And it it sort of asks you to add them. Which is a weird thing to do, why would you do that? It's very strange. What does - how do you even interpret the sum of the area of a circle to the volume of a four-dimensional sphere? So he specifically designed a puzzle such that this would be the answer, which I think is beautiful and clever. Here's a very interesting thing. We're adding up all of these volumes, right, it converges. And what that means is that higher and higher out, the volume of a high dimensional sphere is quite small, right? In fact, we could compute it out. If I said hey, what's the volume of a 100 dimensional ball? Well, that would be pi to the 50th, divided by 50 factorial. Well, the thing is with, with 50 factorial the numbers you're multiplying in, it's 1 times 2 times 3 times 4 times 5 times 6 - you're, you're multiplying bigger and bigger numbers. Pi to the 50, you're always multiplying a pi, right? So the denominator starts winning out because by the end you've added an extra pi to the top, but an extra 50 to the bottom. So it shrinks back quite a lot. Around 2.368 times 10 to the negative 40. If you think about what that means, if you're in a hundred dimensions, right, and you look at a 100 dimensional cube and you say let's look at the sphere that's touching every single edge of that cube, what proportion of the square does that sphere occupy? It's, it's around 10 to the negative 40, right? Which - a lot of people I might quite counterintuitive, right, because if you think of a two-dimensional circle it fills up most of the square. Or a 3-dimensional sphere it fills up most of the cube. But if you think of what it's actually asking - (Brady: Hang on, are you doing what you just told people off for doing though, and imagining these as real spaces?) (Rather than - )
- Well they are! So - sorry. When I say real, I mean, they're exactly as real as numbers are right? Like, the real number line you're not gonna find that in nature. You're not hiking through the woods and there's the real number line. So in the same way like 100 dimensional space, it's a useful abstraction. What I, what I would suggest people not hinge on is asking is the universe 100 dimensional? And is the only way that it's meaningful to ask questions about 10 dimensions, if the universe has that wiggle room? But people find this quite counterintuitive how small in a sense small balls are up in higher dimensions. But if you interpret it much more literally by saying choose 100 numbers, okay, all of them chosen uniformly between negative 1 and 1, add up their squares, what's the probability that the sum of all those squares is less than 1? Like, oh, we got 100 of them - of course is gonna be bigger than 1, there's just so many numbers. Same thing that's going on. There's so many dimensions. So this is why it's helpful to have a back and forth between analytic thinking and geometric thinking. If you haven't already, make sure you check out Grant's incredible math channel: 3blue1brown. And if you'd like to hear more from Grant on the Numberphile podcast, check out the links on the screen and down in the video description
He had been teasing this on his Patreon for a while and god damn it nerd sniped me... what I took away from this video is that independence is a damn nice property to have.
I love my Mathematical Conan O'Brien
Or, my favorite representation of the answer, sqrt(i-i )
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I thought it was an interesting problem with a nice solution with a bit of discussion on what it means to be thinking in higher dimensions (nothing massively controversial here I'd imagine, but nice to have in a video anyway). Also, 3Blue1Brown is always great, and there was a little puzzle at the end:
If I choose numbers uniformly from 0 to 1, what's the expected number required before it becomes greater than 1?
He's a better presenter/teacher than most professors on numberphile.
I now have a less unhealthy relationship with ex
I wish I could have a natural skill at explaining/learning probability. Takes me forever and I need to read it 3-4x and if I’m trying to explain something I need my notes with me. He just makes it look like clockwork
Wow this was such a great video. Great illustration of why we care about n dimensions (something that honestly I've never had a good answer for myself!)
Honestly the best math presenter on YT, and do nice he leaves us another problem to play with at the end too lol!