Darts in Higher Dimensions (with 3blue1brown) - Numberphile

高维空间中的飞镖问题（与 3blue1brown 合作）- Numberphile Grant Sanderson 提出了一个飞镖靶谜题：每次成功击中后，圆形靶心都会沿着由击中点到中心距离所确定的垂直弦发生缩小。在对限定正方形区域内的完全均匀随机投掷者建模时，确定连续投掷的存活概率，便转化为计算坐标元组是否落在更高偶数维的单位球内。通过裂项求和计算期望得分，这些超球体的体积直接对应于指数函数的泰勒级数展开项。这得出了一个简洁的闭式期望得分 e^(π/4) ≈ 2.1932，展示了高维几何如何作为多变量代数条件的直观编码工具。 Grant Sanderson 提出了一个几何飞镖谜题：目标最初是一个单位圆，每次击中后，通过将新半径设置为垂直于击中点的半弦长来缩小。 (0:00) 对于在固定边界正方形 [-1, 1] × [-1, 1] 内均匀随机投掷的选手，存活 n 轮在几何上等同于一个点落在 2n 维单位球内。 (9:59) 对期望值总和进行裂项求和，可将概率直接映射到 x = π/4 时的 e^x 泰勒级数展开式上，得出精确的期望总分为 e^(π/4) ≈ 2.1932。 (20:59) 该谜题由科幻小说作家 Greg Egan 逆向设计，他注意到偶数维单位球的体积公式 π^n / n! 正好对应于指数泰勒级数的各项。 (28:49)

Grant Sanderson demonstrates how a shrinking bullseye dart game naturally encodes higher-dimensional Euclidean geometry, revealing that an unskilled, uniform random thrower achieves an expected score of exactly $e^{\pi/4}$. The Shrinking Target and Pythagorean Reduction (0:00) The game begins with a unit circular bullseye of radius $R_0 = 1$ centered on a dartboard, surrounded by a fixed outer square $[-1, 1] \times [-1, 1]$ of area 4. Whenever a dart lands at distance $H$ from the center, a chord is constructed perpendicular to the line segment connecting the center to the hit point. The full length of this chord defines the diameter of the next bullseye, keeping its center pinned to the board's origin. By the Pythagorean theorem, the previous radius serves as the hypotenuse, establishing the new radius as $r = \sqrt{R^2 - H^2}$. A shot close to the center removes only a thin margin, whereas a shot landing near the perimeter sharply contracts the target. Mapping Sequential Throws to Even Dimensions (9:59) To analyze the expected score, Sanderson models an untalented player whose coordinates $(x, y)$ are chosen uniformly and independently across the bounding square $[-1, 1] \times [-1, 1]$. Surviving the first throw simply requires landing inside the starting unit circle, which occurs with probability equal to the circle-to-square area ratio, $\pi / 4$. For the second throw to hit the shrunken target, its squared distance from the center must satisfy $x_1^2 + y_1^2 < 1 - (x_0^2 + y_0^2)$, rearranging into the single condition $x_0^2 + y_0^2 + x_1^2 + y_1^2 < 1$. Rather than resolving four separate random variables analytically, Sanderson interprets $(x_0, y_0, x_1, y_1)$ as a single point inside a four-dimensional hypercube of volume $2^4 = 16$. The success condition describes the interior of a four-dimensional unit ball, which has volume $\pi^2 / 2$. This pattern extends to any arbitrary round $n$: surviving $n$ consecutive throws requires $2n$ random coordinates to lie within a $2n$-dimensional unit ball. Using the general volume formula $V_{2n} = \pi^n / n!$ and dividing by the hypercube volume $2^{2n}$, the probability of achieving a score of at least $n$ simplifies to $(1 / n!) (\pi / 4)^n$. Telescoping Series and the Emergence of e (20:59) The game's scoring awards 1 point for participating plus 1 point for each successful hit, meaning the score equals the total number of darts thrown until the first miss. The expected score is defined as the sum of all possible scores weighted by their exact probabilities: $\sum_{k=1}^{\infty} k \cdot P(\text{Score} = k)$. Because geometric ball volumes yield cumulative survival probabilities $P(\text{Score} \ge k)$ rather than isolated probabilities, Sanderson expands the expectation sum using telescoping differences, where $P(\text{Score} = k) = P(\text{Score} \ge k) - P(\text{Score} \ge k + 1)$. When expanded, the terms collapse completely: each intermediate subtraction of $k$ copies is offset by the addition of $k+1$ copies in the adjacent bracket, leaving a simple sum of the survival probabilities. This yields $1 + (\pi / 4)^1 / 1! + (\pi / 4)^2 / 2! + (\pi / 4)^3 / 3! + \dots$. Sanderson connects this infinite series directly to the Taylor series expansion for the exponential function, $e^x = \sum_{n=0}^{\infty} x^n / n!$. Evaluating the polynomial at $x = \pi / 4$ produces an exact expected score of $e^{\pi / 4} \approx 2.1932$. Geometry as an Encoding of Multi-Variable Data (28:49) Sanderson attributes the puzzle to author and mathematician Greg Egan, who noticed that the volumes of even-dimensional unit balls, $V_{2n} = \pi^n / n!$, mirror the terms of $e^\pi$ and reverse-engineered a probabilistic game to sum them. Because the series converges, the volumes of unit hyperballs diminish rapidly in high dimensions. For example, a 100-dimensional unit ball has a volume of $\pi^{50} / 50! \approx 2.368 \times 10^{-40}$, filling an infinitesimal fraction of its surrounding $2^{100}$ bounding hypercube. Sanderson emphasizes that higher-dimensional geometry serves as an intuitive spatial language for lists of numbers rather than a claim about physical spatial dimensions. While the disappearance of volume in high-dimensional spheres seems geometrically counterintuitive, it is algebraically obvious: choosing 100 independent numbers between $-1$ and $1$ and summing their squares almost guarantees a total exceeding 1, making it exceedingly rare for a point to remain inside the unit sphere. The puzzle demonstrates how translating multi-variable algebraic constraints into higher-dimensional geometry can resolve seemingly complex conditional probabilities. Sanderson concludes that moving between geometric hypervolumes and analytic sums offers clear intuition for abstract spaces, showing that higher dimensions function as practical mathematical tools rather than physical realities.

Grant Sanderson在Numberphile节目中探讨了一个圆盘不断缩小的飞镖概率谜题，展示了二维平面上的随机投掷如何自然映射到偶数维欧几里得球体的体积计算，并最终收敛为自然常数的高斯指数 $e^{\pi / 4}$。 几何收缩规则与计分机制 (0:00) 该飞镖游戏始于一个半径为 $R_0 = 1$ 的单位圆盘（靶心），靶盘放置在范围为 $[-1, 1] \times [-1, 1]$ 的固定正方形内。玩家若命中靶心，设落点距靶心距离为 $H$，则过落点作垂直于该半径的弦，该垂直弦的全长成为下一轮新靶心的直径，即新半径满足勾股定理 $r = \sqrt{R^2 - H^2}$，且新靶心始终锚定在原点。这种几何机制对靠近边缘的较差击中施加严厉惩罚，使后续目标急剧收缩，而精准接近圆心的投掷仅使靶心轻微减小。 针对完全无技巧的纯随机投掷者（坐标在正方形内独立均匀分布），正方形大小在每轮投掷中保持不变。计分规则规定玩家只要参与游戏即自动获得1分基准分，其后每次成功命中靶心加1分，直至首次脱靶游戏结束，因此总得分等价于首发失误前的累计投掷次数。连续命中无限次或恰好命中无穷小圆心点的概率在测度意义下均为严格的零，该问题的核心在于计算该随机投掷者的期望得分。 多轮命中映射偶数维球体 (9:59) 分析单次投掷时，首发命中靶心的概率即单位圆面积与正方形面积之比，即 $\pi / 4$。进入第二轮投掷时，落点坐标 $(x_1, y_1)$ 命中新靶心 $r_1$ 的条件是 $x_1^2 + y_1^2 < r_1^2$。将第一轮的勾股收缩关系 $r_1^2 = 1 - (x_0^2 + y_0^2)$ 代入后，不等式转化为四维坐标约束 $x_0^2 + y_0^2 + x_1^2 + y_1^2 < 1$。这意味着连续命中两镖的概率，等价于四维超立方体 $[-1, 1]^4$（体积为 $2^4 = 16$）内四维单位球体（体积为 $\pi^2 / 2$）的占比，即 $\frac{\pi^2 / 2}{16} = \frac{\pi^2}{32}$。 将该逻辑推广至第 $n$ 轮投掷，连续命中 $n$ 镖的条件对应 $2n$ 维空间中坐标平方和小于1的区域，即 $2n$ 维单位球体。根据数学上已知的偶数维球体体积公式 $V_{2n} = \frac{\pi^n}{n!}$，在体积为 $2^{2n}$ 的超立方体中，连续命中至少 $n$ 轮的尾部概率为 $\frac{\pi^n / n!}{2^{2n}} = \frac{1}{n!} (\frac{\pi}{4})^n$。投掷步数的增加直接跳过了奇数维度，把二维几何概率连续提升至 $2n$ 维超空间体积计算。 裂项求和与泰勒级数收敛 (20:59) 期望得分的常规定义为各得分与其概率加权求和 $\mathbb{E}[\text{Score}] = \sum_{k=1}^\infty k \cdot P(\text{Score} = k)$。通过单点概率向尾部累积概率转化，$P(\text{Score} = k) = P(\text{Score} \ge k) - P(\text{Score} \ge k+1)$，该展开式呈现明显的裂项相消结构：每一项概率在前面被减去 $k-1$ 次，而在后一项被加上 $k$ 次，展开重组后所有复杂系数完全抵消，简化为各项尾部概率的直接求和：$P(\text{Score} > 0) + P(\text{Score} > 1) + P(\text{Score} > 2) + \dots$。 由于参与基准分对应 $P(\text{Score} > 0) = 1$，第 $n$ 轮概率形式为 $\frac{1}{n!} (\frac{\pi}{4})^n$，整个无穷级数精准对应指数函数 $e^x$ 的泰勒展开式 $\sum_{n=0}^\infty \frac{x^n}{n!}$，其中变量 $x = \frac{\pi}{4}$。因此，随机投掷者的数学期望总分精确等于 $e^{\pi / 4} \approx 2.1932$。这表明即便完全依靠随机乱掷，玩家平均也能获得略高于2分的成绩。 高维几何抽象与反向构题 (28:49) Sanderson说明，该谜题最初由科幻作家兼数学家Greg Egan在Twitter上逆向设计。Egan注意到偶数维单位球体体积序列 $\frac{\pi^n}{n!}$ 与 $e^\pi$ 的泰勒展开项完全契合，因此构造了这一连环缩小的飞镖投掷模型，使得累加各维度球体体积具有实际概率意义。这解释了为何纯暴力积分求解者也会反复推导得出偶数维超球体的体积公式。 从高维收敛性质来看，由于分母阶乘 $n!$ 的增长速度远超分子 $\pi^n$，高维单位球体的体积随维度攀升迅速归零。例如100维单位球体的体积公式为 $V_{100} = \frac{\pi^{50}}{50!} \approx 2.368 \times 10^{-40}$，在棱长为2的100维超立方体中几乎只占无限小的一点微量。若单从高维视觉去想象此现象会极度违背直觉，但从代数角度看，在 $[-1, 1]$ 之间独立抽取100个随机数，其平方和小于1的概率极其罕见，代数与几何视角的相互转换能够彻底阐明这一反直觉现象。 Sanderson强调，数学家研究高维空间并非将其视为物理实体，而是作为编码多元数值列表与约束条件的几何语言。该谜题展示了代数分析与几何空间之间相互转换的力量，但分析未涉及正方形同步缩小或非均匀分布等更贴近真实物理投掷情形的扩展讨论。

[2:05] Grant Sanderson: Consistency is only a virtue if you're not a screw-up. [4:27] Grant Sanderson: One of those paradoxical facts about probability is you can have events that are possible, it's just probability 0.

[0:00] Alright, Brady so I've got for you a puzzle. It's about darts. [0:04] It involves some higher dimensional geometry and a couple of the most famous numbers in math [0:10] So I'll describe the game. We want to hit as many bullseyes as we can. [0:13] So right now the bullseye is this tiny red thing, but that's too small, [0:16] we're going to start with a giant bullseye that fills up the entirety of the board. [0:19] I have a proper compass [0:20] so I'm gonna go ahead and cut out a piece, just a piece of paper to indicate our enlarged bullseye. [0:28] [laughs] - (Brady: That'll do) [0:29] Not the greatest circle I've ever drawn. All you do is you try to hit the bullseye. Each time [0:34] we hit it though, based on where our shot is the bullseye's gonna shrink. [0:37] So it's gonna get harder and harder as we go. - (Brady: You're not gonna hit the wall are you?) [0:41] I'm not *that* bad! [0:44] 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. [0:51] So let me tell you what the rule is for how we shrink it. So step one, [0:55] 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. [1:02] So that's the distance of my hit and then what I'm gonna do is draw a chord of the circle [1:07] that's perpendicular to this line. That's our chord. [1:09] And our rule is [1:11] 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 [1:17] shot was nice and close to the center, I'm rewarded by the fact that it's not going to shrink that much. [1:22] (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. [1:29] All right [1:30] so I'll draw a new circle where the original center was. [1:32] 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, [1:38] it's barely smaller. - (Brady: Oh, yeah, so that's *just* trimming that circle down a bit) [1:42] So I'm just gonna be giving it a little trim. [1:44] This is like getting your hair cut, [1:46] you know once every week or something like that. And presumably this means that the next shot is gonna be [1:52] similarly difficult. [1:54] This is actually fascinating because when I animate it then it's like oh boom the computer does it. [1:58] 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 [2:05] I'm gonna try to make this a worse shot? [2:10] Okay. - (Brady: Oh my goodness!) [2:13] So if I do the algorithm again, [2:15] like this cord is a little bit shorter because we trimmed off the edges, right? [2:18] And so it'll also be just a tiny bit smaller. So we kind of could run that again, but let me [2:23] let me try to get a worse shot. [2:25] (Brady: You can't even do a bad one when you try!) [2:28] Yeah, consistency is only a virtue if you're not a screw-up. [2:32] So this one's nice and far away from the centre, and I'm going to be punished more because it's far away. [2:37] So we'll do the same game. We'll take this off. We're going to make some cuts. [2:41] Distance is h1. And so it'll be a chord of the big circle, [2:46] perpendicular to that line. [2:50] The chord defines the new diameter, but it's kind of easier to think of it as half [2:54] the cord as defining the new radius, certainly if you're drawing with a compass. So I'll set my new radius, [3:00] but this time because I was closer to the edge of my you know, quote-unquote bullseye that cord ends up smaller. [3:05] So I'm punished which means that the new bullseye is meaningfully smaller. [3:10] Smaller bullseye, play the same game. I'm going to set it up so that the centre is at the [3:16] 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. [3:24] And I - I didn't hit close to the centre. [3:26] (Brady: Ooh dear, you are going to get punished for that) [3:28] So at this point, maybe you can already see how much that's gonna punish me, [3:33] 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. [3:43] Put it on the board... - (Brady: Now we see how good you are) - So far I've hit three, this is for a fourth. [3:51] All right, I miss. [3:53] (Brady: Oh dear, game over.) [3:57] Game over. - (Brady: So what's - so your score for that game is three, because you hit three?) - I hit three [4:02] I will say that would be the natural way to score it, [4:05] based on what I want the answer to look like at the end here, you get one point just for playing. [4:10] Okay? So you get one point just for playing and then one point for each bullseye that you hit. [4:14] So in this case [4:14] I got a score of four. You can say maybe you're - what we're counting is how many darts you throw in total? [4:20] So I threw four darts and the fourth one happened to be a miss, [4:23] whatever you want to do to artificially add one. You'll see why we like that at the end. [4:27] All right, so I've defined the game, I haven't defined the puzzle yet. [4:30] And so for the puzzle, let me just draw our board again, or at least draw a circle. [4:35] All right, so we're gonna have a random darts player. So he throws at the board with some random distribution, [4:39] we all do. - (Brady: But there's no skill here, there's no - there is absolutely no waiting for the skill) [4:44] (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. [4:49] 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 [4:57] rotationally symmetric, right? I mean, why would it care about the diagonal directions? [5:01] But basically he hits it at a random point anywhere within a square encompassing, whatever our first circle was. [5:08] 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, [5:14] Y, and maybe let's say that this happens to be [5:18] radius of 1 for the original circle because why not. The [5:21] x-coordinate for this person is always going to be something between negative 1 and 1. And [5:26] it'll just be completely uniform within that. So this would be the simplest thing to model on a computer. To do it [5:32] you just choose two different numbers, each one between negative 1 and 1, where every one of those is equally probable. [5:38] So every point in this whole square is equally probable. And that square doesn't shrink [5:42] it's only the circle that shrinks. And the question is what's the expected score? Okay, so we want to know the expected [5:50] score [5:52] for this thrower. [5:54] (Brady: So Grant, after you throw your first dart and the circle shrinks the square doesn't shrink with it?) [6:02] It does not shrink, [6:02] no, so the the distribution for the thrower is the same. You might imagine the dart bored itself [6:07] which is fixed is square, and the bullseye is shrinking. Okay? [6:11] Because otherwise [6:12] it's gonna be the same probability for each shot and some viewers might be able to guess what that probability is, [6:17] and it wouldn't be as interesting a problem. Another way to think about the expected score, [6:22] imagine this random player plays a thousand games, right? Just what's his average score for those thousand games? [6:29] Another way to think about it, is you say it's going to be the probability that the score is equal to 1 [6:37] times 1, for that score, plus the probability - [6:41] because remember the lowest score you can get is 1, because you get that 1 point just for playing - [6:45] the probability that the score is equal to 2, [6:48] times 2, [6:49] plus the probability that the score is equal to 3 times 3, and so on. [6:54] So what we need to figure out is all of these probabilities. In principle [6:57] 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. [7:05] What we'll find is that, that's there's a probability of zero that you hit a [7:10] bullseye infinitely many times, [7:11] but it's also interesting how quickly does it shrink? Would a score of 10 be absurd or would that be maybe expected? [7:17] (Brady: What is the probability of getting an exact bullseye? 'Cause an exact bull's eye is like) [7:23] (presumably an infinitesimally small point?) - It's probability 0 [7:27] Yeah, so like one of those paradoxical facts about probability is you can have events that are possible, [7:32] it's just probability 0,right? It's possible to hit an exact bullseye, [7:36] it's just probability 0. You know if you choose a random number from the number line [7:40] it's possible that it's rational but its probability 0 and this - a lot of people kind of [7:44] try to come to philosophical terms with like - how can it be possible but probability 0 the probability would have to be slightly bigger. [7:50] But it's just one of those things of math. [7:54] All right, so anytime you're solving a puzzle that's hard and this is hard, [7:57] you see if you can ask an easier version of it or some easy sub problem within it, [8:01] what's the easiest thing you can get a foothold in, right? And maybe in this case the easiest question to start [8:07] that's not totally trivial is what's the probability of hitting the first shot? [8:12] And this isn't that your score is exactly equal to one, [8:14] if you hit a first shot your score is greater than or equal to one. Or no [8:18] actually because you do get that point for free, this is the probability that your score is greater than one. Actually [8:23] maybe you can tell me, right? [8:24] What's the probability that on this first shot, with this random distribution within the square, it hits somewhere inside this whole circle? [8:30] (Brady: I'm imagining we we need the area of the circle and the area of the square?) - Yep, the area of the circle, [8:36] this is pi r squared and that radius starts off as 1. [8:40] So that ends up just being pi. And the area of the square is equal to, well [8:45] the square side length is 2 times that radius right? So it's 2 by 2. [8:49] So it's 4 which means the probability that you hit [8:53] that first bullseye, that your score will be greater than 1 because you get that one point for you just for playing, [8:58] is gonna be the area of that circle [9:01] divided by the area of that square. - (Brady: Grant, what happens if the dart lands on the line?) - That's probability 0. [9:07] Don't worry about it, it won't mess with the problem. So now a much, a much harder question is [9:12] okay, you've hit that first shot, [9:13] what's the probability that you hit at least 2? And this is where it starts to become [9:17] interesting. Because this depends on where your first shot was. [9:20] I'm gonna start writing a little bit of a chart, [9:22] where I'm gonna keep track of the radius of each bullseye and I'm gonna keep track of just the hit length, [9:28] 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. [9:35] Let's just give it a name for now [9:38] Let's say it goes there. That's H_0. So it will be H_0 and [9:44] by the Pythagorean theorem, if [9:46] 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 [9:50] in terms of that, [9:52] it's the square root of x squared, plus y squared if those are the coordinates of your first shot. And [9:59] 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 [10:05] what's the new radius? [10:07] Well, maybe we call this our zeroth radius, right? [10:09] And I want to know what's the next radius after that initial shot? [10:13] And if we remember our rule, we say you draw the chord and the chord was defined to be perpendicular to that. [10:20] So when you know that it's defined to be perpendicular [10:23] you know that you're probably going to use that fact in your solution to the puzzle. [10:28] So the new radius is this length here, right? It's half of that chord. [10:33] So what we might do, and I recognise I'll sort of be drawing over myself a bit here, [10:39] because we have a right triangle here and that's what defined it we're gonna use the [10:44] Pythagorean theorem. Because this was our old radius, radius one. That's the new hypotenuse so we know that [10:51] the old hypotenuse, 1 squared, is equal to one of the legs which was H naught squared, [10:59] plus the new radius squared plus [11:02] r_1 squared. So what that means for what r1 is, is [11:07] it's the square root of [11:09] 1 squared minus h naught squared and [11:12] remember [11:12] 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 [11:17] 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 [11:24] it's determined by two new random coordinates, whatever the coordinates of your second shot [11:29] happen to be or the coordinates of the next shot, maybe I should say. And this is an interesting question, [11:34] what's the probability that you get that second one in? Because it depends on four different numbers, right? [11:39] So I think it's helpful to just write out what the, what the actual requirement is. [11:43] h1, the hit in that second shot, is less than r1. And I can go ahead and rewrite that by saying [11:49] h1 squared is less than r1 squared and the reason is we have a whole bunch of squares going on, [11:55] maybe that's going to be helpful for us. Right because I can expand out what h1 squared is, that's x1 squared plus [12:02] y1 squared and that has to be less than, well what is r1 squared? [12:07] That's 1 squared minus [12:10] h naught squared. And then I can also expand out h naught [12:14] so I'll be writing this as x1 squared plus [12:17] y1 squared is less than 1 squared minus [12:22] x-naught squared [12:24] plus y naught squared. [12:26] So all I'm doing is saying I want to explicitly write out what the requirement is for the four coordinates of my [12:34] initial random shot. And now I have a bit of an interesting question, [12:37] 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. [12:42] What I have is that xnaught squared [12:44] plus y naught squared, you know bringing these two over to the other side, [12:49] plus x1 squared plus y1 squared is [12:53] less than 1. What's the probability that this happens [12:57] when all these four numbers were chosen randomly within the range negative 1 to 1? [13:02] So at this point that's kind of a hard question right? I tell you [13:06] 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 [13:12] oh, I don't know, you know if they were all less than 0.5 I guess that would happen? [13:16] 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 [13:22] let's just think about the very first situation where instead of doing it for h1 and r1; [13:27] when we're over here doing it for what seems like a much simpler situation and it was, [13:31] the probability that h naught is less than r naught which is just saying that x naught squared [13:37] plus y naught squared is less than 1. [13:40] We have a purely analytic statement, choose two numbers out of a hat based on this rule, [13:44] what's the probability that the sum of their squares is less than 1? You answered it, geometrically. You immediately knew [13:50] oh, we need to use pi. We're thinking of the area of the circle, right? [13:54] And it's actually quite subtle what's happening there, [13:56] you have a question about a pair of numbers and you're gaining [13:59] 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 [14:05] but as a single point in [14:08] 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, [14:14] what you're asking for is a probabilistic property of these four numbers. [14:19] It might be natural to think of them as being a single point x naught, y naught. [14:28] x1y1 inside 4 dimensional space. [14:34] And it's basically a space that has four real number coordinates. And so now Brady [14:38] let me ask you, what's the analogous question? In the same way that asking what's the probability that the sum of two [14:43] random numbers is less than one, [14:46] begs the question of what the area of a circle is. [14:49] When we ask what's the probability that the sum of four [14:52] squares is less than 1. What do you want to know? [14:55] (Brady: Well I was how do you think we were gonna end up using a) [14:58] (cube and a sphere, but I feel like we've jumped past - ) [15:00] We've jumped! We've jumped! The whole idea that, hey if I choose three random numbers x squared, y squared, z squared, [15:07] what's the probability that they're less than 1? If they're chosen by the same rule [15:11] what you're going to be asking for is the volume of a sphere, [15:14] because what it means for the sum of their squares to be less than 1 is that they sit inside a unit sphere. [15:23] The volume of a sphere is [15:25] 4 thirds [15:27] pi r [15:28] cubed. r is just going to equal 1. So that's 4/3 pi. That's the [15:33] volume of a sphere. The volume of the cube which [15:37] is 2 by 2 by 2, is going to be 2 cubed, [15:41] which is 8. [15:43] 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 [15:49] 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 [15:56] here. And so here we went straight from 2 to 4, [15:59] the question that we want to ask right now is what's the volume of a 4 dimensional sphere? Or the 4-dimensional [16:04] equivalent of volume. Instead of saying hyper volume or coming up with a new word, they just say volume. [16:08] You could fancifully say measure, but we're just gonna say volume of a [16:13] 4d sphere. And again [16:15] if you wanted to be pedantic, the mathematician might actually say ball? Because when they use the word sphere [16:20] they typically refer to the boundary right? It's just the peel of the orange. It's not the contents of the orange. [16:25] So what we really want is the ball. The other thing that we need is the volume of a 4d cube. [16:30] The cube is the easy one. Just like in the other cases the side lengths are all 2, 2, by 2, by 2, [16:36] by 2. [16:37] So that volume is 16. Volume of the sphere, [16:40] there's a whole interesting discussion that we could have about the volumes of higher dimensional spheres, [16:44] the surface areas, or the analogs of surface areas of higher dimensional spheres, and where they come from. [16:48] I'm not going to give you that full story now, maybe at some other point. [16:52] It happens to be the case that it is pi squared halves [16:57] times r to the fourth. So it picks up an extra pi. [17:01] Which is kind of interesting, when you go from 2 to 3, you don't pick up that pi [17:04] but when you go from 3 to 4 you pick up that extra pi. And because our radius is 1, in this case [17:09] that'll just be pi squared halves. And let's remember what this is. [17:13] This isn't the probability of hitting exactly two shots, right? It's that you hit those first two shots, [17:18] so it's the probability that your score is at least two. Maybe it also happens to be three or four, [17:23] so we can write our answer here. The probability that our score is greater than or equal to two is equal to [17:29] 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. [17:38] So this was kind of the, the [17:39] cube, or the hypercube. And then times 2. [17:43] Where that was the denominator that we had from our volume of the 4d ball. And now we just keep doing this [17:48] for all infinitely many different scores that we might have. Very similarly when we're asking about [17:53] the probability that our score is at least 3 or that it's bigger than 3, [17:59] what this comes down to is asking - we're going to have 6 different coordinates: x naught, y naught, [18:06] x1, y1, x2, y2; and that the sum of their squares is less than 1. [18:21] (Brady: I don't know how to calculate the volume of a six dimensional ball) - No! Luckily [18:27] mathematicians have figured it out for us. And again, it could be a whole interesting story, [18:30] but if you go to Wikipedia, you could see a chart, [18:32] there's quite a nice pattern that we're gonna - about to point out for, where that comes from. [18:35] But the volume of a 6d ball happens to be pi cubed, over [18:42] six times r [18:44] to the 6. And then the volume of the 6d cube is of course 2 to the 6, and [18:51] so the answer to the probability of hitting those first three shots, is going to be pi cubed [18:58] divided by that 2 to the 6 and [19:00] 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, [19:06] which is pretty mind-blowing. Is that the volume of a 2n ball is equal to [19:13] pi to the n, so half of the dimension, right? So in 2 dimensions you see a pi, in 4 dimensions [19:18] you see a pi squared, and six you see pi cubed. [19:21] So pi to half the number of dimensions, divided by n factorial. Very clean, [19:26] very beautiful. Very reminiscent of another celebrity real number that a couple people might be familiar with now and that might be [19:33] poking in their minds. [19:34] 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, [19:41] let's figure out the expected score. [19:43] (Brady: Please, that's what I'm here for!) [19:44] Great. First let me get a cup of water. [19:47] [Hold music] [19:52] Alright, are you ready for the grand finale? - (Brady: Yeah!) - The finishing touch. [19:56] 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, [20:03] now remember what this means, we're gonna have - it's going to be 1 times the probability that your score is 1, [20:10] plus 2 times the probability that your score is 2, [20:15] plus 3 times the probability that your score is 3, and so on. [20:21] Now all of our expressions are greater than things. [20:24] They're not saying exactly equal to, but that's completely fine. Because if I wanted to say for example [20:29] what's the probability that your score is precisely equal to 2? [20:34] So this will be your probability that your score is greater than 1 right? [20:38] Minus the probability that the score is [20:42] greater than 2. Because for it to equal 2 it's greater than 1, but it's not greater than 2 and this entirely [20:48] encompasses the possibilities, right? [20:50] If you're probably if you get a score of greater than 1, that [20:53] 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 [20:59] do anything more than that. [21:00] So I'm gonna expand all of these out just a little bit before things collapse nicely. So this would be 1 [21:08] times [21:09] the probability [21:11] that the score is bigger than 0, [21:14] minus probability that the score is bigger than 1, [21:18] plus 2 [21:20] times [21:21] probability that the score is bigger than 1 [21:25] minus the probability that the score is bigger than 2. And just one more for good measure, [21:32] probability that the score that's bigger than 2 [21:35] minus the probability that the score is bigger than 3. So at this point we have a lot of cancellation that happens. [21:42] So we've got this probability that [21:43] 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 [21:49] but then we're adding two of them over here. So we've got [21:53] probability that the score is greater than 1 and then similarly for score bigger than 2 we're subtracting 2 here [22:01] but we're adding 3 back here. And so we're just going to be adding [22:06] probability for the score is bigger than 2 And in general [22:09] 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 [22:15] we're subtracting 3, but in the next one we'll be adding 4, and so on. [22:19] So every puzzle that we just had, every micro puzzle, we just add up all of their answers. [22:25] So the probability that your score is at least zero, [22:28] that's 1, because definitely your score is bigger than zero because you get a point just for playing. [22:33] Probability that it's bigger than 1, that's the one that you answered for me first. [22:36] Which was saying that it's pi divided by 4. The next one, [22:40] so I'm actually going to write this slightly differently because we always have some power of pi divided by some power of two. [22:47] I'm going to write this as being [22:49] PI fourths [22:51] squared [22:53] divided by 2. Is that all right? So that 4 squared came from - that's the same as 2 to the fourth [22:59] and then the pi squared just went in there. And so I'm going to write it here as [23:04] pi [23:05] fourths [23:06] squared [23:08] 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. [23:17] Rather than writing pi cubed over 2 to the sixth, which is the same as pi cubed over 4 cubed, [23:22] I'm gonna write it as pi over 4 cubed. And [23:26] 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 [23:33] hanging, of having a factorial. So over here we've got [23:37] pi over 4 [23:39] cubed times 1 over 3 factorial, this is really two factorial. And maybe you see where this is going [23:45] or maybe at this point you're saying you promised me this would look simple. It actually looks exceedingly complicated, [23:49] you've got factorials, you're grouping the pis together for some reason. [23:52] 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 [23:59] ingratiated to the number e, who will just see something screaming in their head right now. [24:03] There's something known as the Taylor series for e to the x. Which is that it's 1 plus [24:10] x and all right, it's x to the 1 over 1 factorial. [24:14] x squared over 2 factorial, [24:17] x cubed over 3 factorial and so on. [24:23] Where you evaluate it as an infinite polynomial [24:27] where each one of the terms is 1 over n factorial. [24:30] It's not just that e to the x happens to equal this, [24:33] I actually think the healthier way to view the exponential function and what e to the x is - this defines it, right? [24:39] This is the thing that should pop into your head when you think of exponential growth and e to the x is this [24:43] particular infinite series. This particular polynomial. This is where it will come up, especially in probability, [24:48] it lends itself to an easier interpretation of why it is its own derivative, there's all sorts of nice things, [24:54] 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 [24:59] of e to the x. So if you looked at this and you weren't thinking of e already, what it means is that [25:05] 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) [25:11] You currently have an unhealthy relationship with e, that's absolutely true. [25:14] If you didn't see it. Or you have no relationship with e. But if we compare this to the series, [25:19] which I've unhelpfully drawn kind of far away, but you can match them up pretty easily, [25:25] the thing that's playing the role of x right now is pi fourths. [25:29] So all this is, is e to the pi fourths. Which, if we go and we plug it into a calculator, [25:36] approximately equals two point one nine three [25:41] two, on, on, on. Which is to say if you are a terrible random darts player, [25:47] who unrealistically hits within a square with a uniform probability, which you wouldn't because it would be rotationally symmetric - [25:53] but whatever you're hitting within a square. And you keep going like this [25:56] then you play a thousand games, on average your score would be 2.1932. And remember my score was 4. [26:06] (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 [26:14] Yeah, that's that's about right. [26:18] 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 [26:24] unexpected or mind-blowing component is that we're even talking about higher dimensional geometry. You found yourself naturally asking, in the middle, [26:31] what is the volume of a six dimensional ball? [26:33] Right? And nowhere in that was asking, is the universe six dimensional? You know, like the string theory invokes 6 - [26:39] no, no, no, like that's not why mathematicians necessarily care about higher dimensions. [26:44] 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 [26:51] geometrically. Rather than saying the sum of their squares is less than 1, you say it shows up inside a six dimensional ball. [26:57] That also means it saved you some work, [26:59] 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 [27:05] you know, I've been working on it and I, it's just like really hard. I've been like working through these integrals. [27:09] What's very strange is even in the case with the second dart I'm getting pi squared for some reason. [27:14] So he's working through all these integrals, [27:17] under the hood what's happening is he's rediscovering the volume of a four dimensional ball, right? [27:21] So it gives a more universal language for people to talk with on that. It came about from two-dimensional geometry. [27:26] Nothing about the dartboard is four dimensional and I think this is just a common misconception that people [27:31] listening to mathematicians describe things like manifolds in four dimensions or the Poincaré conjecture has been answered for everything except four dimensions. [27:38] They don't actually care about like a thing where you can move in four dimensions, [27:42] it's about encoding quadruplets at points. And the second one is that I think it helps build a healthier relationship with e. [27:48] Because this this series is much more important than the number itself. [27:51] 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. [27:56] (Brady: Yeah) [27:56] So here we were [27:58] 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 [28:04] squares. [28:05] When is that less than 1? You can play a much simpler probability game where let's say I'm just gonna choose numbers [28:13] from 0 to 1 with uniform probability. And I'm gonna keep going until the sum of the numbers that I've chosen [28:20] ends up being bigger than 1. so for example, if the first number you choose is 0.3 and then the second one is [28:27] 0.6, their sum is 0.9. And if the next one you choose is 0.5, [28:30] that's the point when you go over. And so the question you can ask is [28:34] what's the expected number of samples you need to take before it over blows one? [28:38] e shows up in the answer, [28:39] 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 [28:45] so there's not the confusing factor of pi. You see much more pristinely the factorials. [28:49] Very important is that I mention where this puzzle comes from. I saw this on Twitter, I think it was Greg Egan. [28:54] He specifically designed a puzzle such that the answer would look like adding up the volumes [29:01] of higher-dimensional balls, because [29:03] we have this wonderful formula for the volume of higher dimensional balls, when the number of dimensions is even. And [29:10] everyone who has a healthy relationship with e looks at this, a thing to the power n divided by n factorial, [29:15] they scream in their mind the exponential function with e to the x, [29:19] right? And it it sort of asks you to add them. Which is a weird thing to do, why would you do that? [29:24] 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? [29:29] So he specifically designed a puzzle such that this would be the answer, which I think is beautiful and clever. [29:35] Here's a very interesting thing. We're adding up all of these volumes, right, it converges. And what that means [29:41] is that higher and higher out, the volume of a high dimensional sphere is quite small, right? [29:47] In fact, we could compute it out. If I said hey, what's the volume of a 100 dimensional ball? [29:53] Well, that would be pi to the 50th, [29:57] divided by 50 factorial. Well, the thing is with, with 50 factorial the numbers you're multiplying in, [30:02] 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, [30:07] you're always multiplying a pi, right? [30:09] 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. [30:14] So it shrinks back quite a lot. Around 2.368 times 10 to the negative [30:19] 40. If you think about what that means, if you're in a hundred dimensions, [30:23] right, and you look at a 100 dimensional cube [30:26] and you say let's look at the sphere that's touching every single edge of that cube, [30:30] what proportion of the square does that sphere occupy? [30:34] It's, it's around 10 to the negative 40, right? Which - a lot of people I might quite counterintuitive, [30:40] right, because if you think of a two-dimensional [30:42] circle it fills up most of the square. Or a 3-dimensional sphere it fills up most of the cube. [30:46] But if you think of what it's actually asking - [30:49] (Brady: Hang on, are you doing what you just told people off for doing though, and imagining these as real spaces?) [30:53] (Rather than - ) - Well they are! [30:55] So - sorry. When I say real, I mean, they're exactly as real as numbers are right? Like, the real number line [31:00] you're not gonna find that in nature. You're not hiking through the woods and there's the real number line. [31:03] So in the same way like 100 dimensional space, it's a useful abstraction. [31:08] What I, what I would suggest people not hinge on is asking is the universe 100 dimensional? [31:14] And is the only way that it's meaningful to ask questions about 10 dimensions, if the universe has that wiggle room? [31:20] But people find this quite counterintuitive how small in a sense small balls are up in higher dimensions. [31:26] But if you interpret it much more literally by saying choose 100 numbers, [31:29] okay, all of them chosen uniformly between negative 1 and 1, add up their squares, [31:34] what's the probability that the sum of all those squares is less than 1? Like, oh, [31:38] we got 100 of them - of course is gonna be bigger than 1, there's just so many numbers. [31:42] Same thing that's going on. There's so many dimensions. [31:44] So this is why it's helpful to have a back and forth between analytic thinking and geometric thinking. [31:52] If you haven't already, make sure you check out Grant's incredible math channel: 3blue1brown. [31:58] 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