More on Bertrand's Paradox (with 3blue1brown) - Numberphile

[@toughnerd]

Brady: "I'm going to burn your house down if you don't tell me the answer" Grant: "That's a great question."

[@jamaljohnson2414]

That literally had me laughing out loud. ...and I'm a trucker.

[@gerrybaker7055]

@@jamaljohnson2414 how does being a trucker change your ability/willingnrss to laugh though?

[@jamaljohnson2414]

@@gerrybaker7055 It's not about being a trucker. Or that being a trucker would cause anything. It's about reaching a trucker, as part of your result, after the interviewer chose a less clinical approach.

[@minikretz1]

I read this before they said that and I thought it was funny but way over the top. Then they had that exact conversation lol

[@snickers10m]

His complete deadpan seriousness at considering the problem had me in stitches. Easily my new all-time favorite numberphile moment.

[@jajohnek]

Brady's certified method to finding answers to unanswerable mathematical questions: find a mathematician and threaten them to give you the answer. Love it :D

[@linuskerr]

News headline in 2022: "Mathematics has been solved! One Austrailian man flew across the world and threatened every mathematician to give him all the answers."

[@alkylperchlorate388]

“That’s a great question” -grant

[@IceMetalPunk]

@@linuskerr "...either 1/4, 1/3, or 1/2 of all mathematicians' houses were burned to the ground. More on this during the news at 8, 9, or 10 o'clock."

[@heartles_xyz]

@@linuskerr P != NP proved! by way of a Ruger 57

[@damamoot2291]

**AN* answer

[@Rubrickety]

I'm in awe of Grant's ability to speak, unscripted, with perfect clarity. (Both in the sense of what he's saying, and incidentally, his flawless diction.)

[@TheGreatAtario]

I feel like this is a skill you end up with by being a teacher, or else you soon stop being a teacher

[@stephenbeck7222]

Yes but remember this is a question he has thought deeply about and also clearly read much about. So it’s not so much unscripted.

[@starpawsy]

ANd no ums and arhs, either. Very unusual. Agree wtih you compltely.

[@pawelkorzeniewski4897]

He he. Tele prompter goes brrrr. Seriously though impressive

[@DynestiGTI]

The more you know about the topic, the easier it is to talk about it.

[@iamnorwegian]

15:04 The lack of even a chuckle at the premise of that question tells me that Grant 100% has laid awake at night thinking about Bertrand's paradox.

[@ThePondermatic]

You can see his eyes light up and his mind start organizing his thoughts. You only respond that way if you have unironically laid awake at night thinking about something.

[@sender1496]

Rewatching that part after reading this comment was the funniest thing ever

[@Salchipapafied]

I've already been a fan of 3blue1brown's videos, but seeing him explain/answer/clarify all his points in "real time" was nothing short of amazing. In his videos, you assume preparation, practice, etc., but here he's just talking to someone else and making a whole bunch of sense on the fly.

[@andreybashkin9030]

Watch his interview to Lex Fridman, then.

[@Daiwie44]

I am just disappointed he doesn't actually have the eyes of his logo

[@NottoriousGG]

In all honesty, it is reasonable to assume that if you script yourself very many times, and reason with yourself through your own arguments, you're speach seemingly innately acquires an aura of congruence and logic to it. Albeit, all of his talent we witness, we also see much much practice without a shadow of doubt.

[@miscellany_media]

Daiwie actually, he does! he has sectoral heterochromia in his right eye, though i guess that would still be the *eye* of his logo and not the eyes.

[@Fred-tz7hs]

it's still edited

[@dojelnotmyrealname4018]

I think this paradox can be partially summed up with saying "Math can answer questions for you, but it can't ask your questions for you."

[@nahblue]

Grant speaks on that topic at 15:15 i think

[@samuela-aegisdottir]

"What's the answer?" "42" "What's the question then?"

[@ismailshtewi8560]

@@samuela-aegisdottir 7x6 😉

[@666Tomato666]

@@samuela-aegisdottir The Ultimate Question!

[@mjj29]

@@ismailshtewi8560 Canonically the question is 'what is 6 times 9'

[@willhastings731]

Actually got chills when Grant so quickly and eloquently explained the failings of Brady’s proposed method.

[@finnwestergren8670]

yeah it's inspiring. as someone who is so bad at making sense I really admire this.

[@phmfob]

Him explaining the random dice analogy felt like I was back in school ignorantly asking a silly question

[@peterhaugen887]

@@phmfob you mean picking the low hanging fruit on your knowledge tree?

[@xenonchikmaxxx]

yes, Its always pleasure to spectate really smart people arguing

[@aapjeaaron]

I kind of thought he was too hand wave-y. Because there is an elegant way to assigning a random number to each possible cord. If we pick a random direction then assign every single cord parallel to that direction a number in between [0;1[. Then we do it again but we rotate the direction infinitesimally small amount and assigning them [1;2[. Then we do this for the infinity many directions. It's the same as solving it like solution 3. Random direction, random length. So the answer would also be 1/2.

[@tyranneous]

Grant's on-the-fly analogy of the multi-sided die really got me there with this paradox. Absolutely nailed it!

[@lb5299]

This guy is brilliant, this was a perfect analogy

[@Wheat_Thinn]

That analogy really interests me! If it's completely random what the die that I pull from the box looks like, there's infinitely many outcomes where the die I have doesn't even go up to 5. Because of this I'm tempted to say that rolling a 5 is possible but has probability zero.

[@Wheat_Thinn]

What about in our chord problem? What if our circle is floating in a 1 by 1 by 1 cube and only lines coming from the edges of the cube are considered? There must be infinitely many more unspecified possibilities that should maybe be left to probability. I'm reaching the idea that any problem without enough specifications might have probability zero for *any* specific outcome

[@nathanepstein4079]

@@Wheat_Thinn But what is a room full of monkeys were rolling dice forever? What would the probability be then?

[@Wheat_Thinn]

@@nathanepstein4079 As the dice are rolled repeatedly forever, I still think that a 5 would be rolled approaching 0% of the time. We truly have no idea what dice are in the box so we must assume any and all dice are possible. I would wager to say that a vast majority of things that can be considered dice have a small chance of rolling a 5. Like a 100 sided die or a 4 sided die. Whatever the case, it must be a very small number

[@martijn3151]

18:55 it's just amazing how quickly Grant grasps the definition Brady provides and is able to flawlessly lay bare its weaknesses.

[@sujals7108]

IKR, I was completely in awe of how quickly he could find the flaw and point it out so clearly

[@SilverLining1]

The uncountably infinite part is obvious to mathematicians, that was an easy correction, and the part about how you label them affects the probability is obvious to staticians, since every continuous distribution over an infinite (and specifically compact) space is necessarily "uniform" in the sense that the probability of picking any one element is equal: 0. That is, to even look at a continuous distribution in such a space one must consider the neighborhoods of each element. That said, his response about labeling a bunch of small chords with most of the unit interval was an incredibly succinct counterexample!

[@Tony-cm8lg]

@@SilverLining1 Yup, I agree with that. I think it’s difficult for laymen to understand that this is the language mathematicians use. Especially such fundamental concepts like cardinalities of sets and functions between them, these are basically the ABC’s of mathematics and Grant of course is deeply familiar with them.

[@elidrissii]

I love Grant tbh. I wish there were more people like him in the world.

[@jamesonskalinski6910]

I liked the idea posed and tried to analyze it. I'm pretty sure it would tend towards half. Sure there are infinitely many, but if you use the third approach, and use an infinitesimal delta on the radian, it is still obviously true half of them will be longer than the side of the triangle. Make the radian the full diameter instead and now you have every possible chord on this diameter. Still you will have a half probability. Now, change the radian infinitesimally, and do this for every possible orientation. On every orientation, half of the chords are longer, half are shorter. Now every possible chord is drawn, half of them and the ratio is 50/50.

[@JaapvanderVelde]

I think Grant is amazing as well, for all the reasons in the top comments here - but I think Brady deserves an enormous amount of credit as an interviewer of mathematicians. He always succeeds in asking the questions (be it scripted or on the fly) I want to hear asked, as well as a couple I didn't even think of, but which sends the interviewee on exactly the tangent they need to be on for a great video. This one was epic.

[@jvcmarc]

"When you lie in bed at night, and think about Bertrand's Paradox..." has got to be one of the best quotes from numberphile

[@PetWolverine]

And of course Grant doesn't argue with the premise. It seems perfectly natural to him that one would lie in bed at night and think about a paradox like this.

[@madisondampier3389]

@@PetWolverine I would argue that lying in bed at night is precisely the only time someone might try to solve paradoxes. They're the stuff of nightmares! Infinite possibilities!

[@johanlopez655]

It seems like the meme "dad can you tell me a paradox before sleep" i really laught in that moment of the video.

[@JohnMeacham]

I think a whole lot of people are going to be doing that after watching this video. I know I will.

[@SoleaGalilei]

I am literally lying in bed at night thinking about Bertrand's Paradox right now!

[@Henrix1998]

This should be part 2 on the main channel, too important to miss

[@mahxylim7983]

true

[@LeonardEisen]

100%

[@macronencer]

Grant is a uniquely gifted communicator. I can't think of anybody with greater ability to explain mathematics clearly. Absolutely love this guy.

[@skilz8098]

If it's not him then I'd say Ben Eater is another great communicator.

[@otonanoC]

And his saucy voice too.

[@sabinrawr]

I think it comes from his incoming perspective on mathematical topics. Many teachers focus on logic and procedure given to us from on high in ancient times. Grant doesn't do that. He finds ways to make these concepts feel intuitive, discoverable, and relatable. He doesn't just give us a problem and a solution, he gives us a story and a journey.

[@macronencer]

@@sabinrawr Well said - I think that's very accurate!

[@realchrisward]

This reminds me of the problem of choosing a random location using latitude and longitude. If you don't correct the distribution used for selecting latitude, the polar regions are over sampled

[@6612770]

Please explain...

[@highviewbarbell]

​@@6612770I'm not a mathematician so this will be worded wrong, but, imagine looking top down at the earth and each latitude line is a slice. The circumference of each slice is smaller as you go farther north (or south), but because they are pieces of a sphere, each one is still broken into the same number of sections. A grid square between two lat and long lines at the poles is much smaller than the same unit at the equator, but they would be represented equally in a list of coordinates

[@jukmifggugghposer]

I’m pretty sure this is exactly the reason that a multiple of sinθ (θ being the up/down angle) shows up when integrating over spherical coordinates. With the way we define the coordinates, there’s just not as much volume in sections of high and low θ as there are near the center, so the sinθ term corrects for that.

[@FuzzyLogic0]

Doesn't the same 'correction' here change the first example from a third to a half? If instead you choose your chord by going from a random point on the circumference at a random angle. Due to symmetry you only need to consider half the circle and angles less than 90. If you correct for over sampling as you describe using sin then it results that half the chords longer. I'll look up more about what you describe but that may additionally convince me of the 0.5 interpretation.

[@jimsmedley234]

Our friend, Danish mathematician Piet Hein, offers a take on this: "When you're desperately trying to make up your mind and bothered by not having any; you'll find that the simplest solution by far is to simply try spinning a penny. No, not that chance should decide the event while you're passively standing there moping; but once the penny is up in the air you'll suddenly know what you're hoping" Aaaah his powerful Gruks.

[@Idorise]

"I will go before thee, and make the crooked places straight: I will break in pieces the gates of brass, and cut in sunder the bars of iron".

[@LeeClemmer]

This is a powerful but subtle way to dig in and find out what you prefer when trying to make a decision between two things. I've used with friends and family who say "I can't decide." Once they see the coin moving, or see the result, they know what way they wanted it to land.

[@aenetanthony]

I do this all the time, where I assign two different values to each side of a coin, flip it, and choose whichever I'd prefer, not even taking into account what it lands on. It's funny how well it works.

[@Idorise]

You have a Book of Lazlo Meró on this subject.

[@mibber121]

I use this all the time on my boyfriend. If he cant decide between two things for us to do, I flip a coin and depending on his reaction to the result i choose what to do

[@BowlOfRed]

This is probably the first time I've seen an "extras" bit that someone put on their second channel that I enjoyed more than the main presentation. The discussion of the symmetries and the choices was fabulous. Thank you for taking the time for putting this out there (and for your curiosity in asking these questions)

[@vell0cet517]

I agree. The original video was awesome, and this one was even better.

[@kevinsmith9385]

100% agree. I was captivated through the whole video, both parts. I really appreciated Grant's thoughts about the limitations that do exist for math and being ok with that.

[@sam5992]

omg yes, I was dying to hear whether or not he'd mention infinite chords.

[@krish4288]

15:26 "One of the biggest misconceptions is that maths shows us truths, but it doesn't. It tells you 'given certain assumptions, what are the necessary links to consequences'" - Grant Sanderson

[@alynames7171]

An extremely important point that, at least in my neck of the woods, a bunch of tech bros with more confidence and venture capital than sense need to have explained to them. I'm very glad to learn Grant doesn't seem to fall into that common intellectual ditch.

[@KT-dj4iy]

@@alynames7171, isn't it frustrating that it often feels like the relationship between level of confidence and amount of venture funds on the one hand, and sense on the other, may be one of _inverse_ proportion? 🤓

[@ExplosiveBrohoof]

Mathematics is nothing more than the hunt for interesting tautologies.

[@rainzhao2000]

I absolutely loved your interactions with Grant in this video. Each time you probed him with questions, the discussions and explanations became more and more elaborate, helping me understand the crux of this paradox. Especially when Grant went back and forth with you to clarify your definition of a random selection of the chord and how that implicitly assigns a distribution related to some symmetry.

[@dominicveconi4301]

Possibly another (equivalent but mathier) way of explaining the “paradox”: When you talk about “random cords”, doing the calculation requires a *procedure* for constructing random cords. That procedure is essentially a measurable function from a parameter space (such as the space of pairs of points on the circle, or the space of angles and midpoints) into the space of cords. If you assume a uniform distribution on your parameter space, the measurable function into the space of cords carries that distribution forward-but there’s no reason to assume that different measurable functions from different parameter spaces will carry forward to the same distribution in the space of cords.

[@bonezman5]

This is the most satisfying explanation I've seen. Most of us are used to thinking about measurable spaces where there's only one natural measure, and I think this paradox is simply demonstrating that there can be more than one natural interpretation for measure on the space of chords.

[@LK90512]

ftw

[@oshuao414]

Grant's reaction after Brady saying "draw every possible chord, each one is numbered... just wonderful

[@zachrodan7543]

"you're a math person" "what emotion does this paradox invoke in you? delight? frustration?" um... as a math person myself, I consider delight and frustration to be indistinguishable as it pertains to math problems: the frustration produced by a problem is precisely what makes it a fun problem to contemplate

[@madisondampier3389]

It fills you with determination!

The challenge is what makes math enticing. How creative can you be with the approach, and how many corners can you cut? In the video I would have loved for them to go a bit more into the details of topology and why the problem is ill defined, and maybe setting up some examples where each of the different interpretations make sense. That's what I think would help people the most in grasping it.

[@JohnMeacham]

I feel like there should be a word for this.

[@zachrodan7543]

@@JohnMeacham If it were up to me... "mathocistic pleasure?"

[@Merthalophor]

But the frustration isn't that you can't solve a mathematical problem; it's that the problem is unsolvable. It's a problem prior to math, it's a problem of philosophy.

[@daniellambert6207]

I love how Brady pokes to inject emotion into the discussion. A true master of the craft. 15:03, 21:05

[@frankjohnson123]

I almost enjoyed this more than the main video because the discussion was great. I love Brady's way of asking the prying questions to get more insight.

[@_ilsegugio_]

also it's impressive how quick he can reasonably answer Brady's questions being thrown at him

[@JP.Q]

I think one can articulate the 3rd solution in a way that makes it sound less artificial: Since we have no preferred direction, we can without loss of generality assume the chord is vertical. Now visualize all vertical line segments covering the disk. "Clearly" half of them are within distance 1/4 of the circle's center, and those are precisely the long chords. Of course this is no deep insight, just a rephrasing of the same argument, but the choices of radius and midpoint (which Grant made very explicit) are more well-hidden.

[@Sarimae23]

nope, sorry, it's more easy the Radius has the midpoint on it's half. so, if you go with the corde in direction of the midpoint of the circle, it's longer , and if you go towards the rim, it's longer. There are only 2 possibilities. The game is rigged by putting this system into play ;)

[@jameswood8021]

If all the chords are vertical, wouldn’t that mean you have a preferred direction?

[@thecommexokid]

@@jameswood8021 Choose the chord (in any direction) first, then define your axes afterward such that the chosen chord is parallel to the y-axis.

[@victorquesada7530]

Thank you, that was a helpful insight! I got that the random radius and random chord with a midpoint on that radius as an appropriate step, but you clarified things greatly!

[@watchm4ker]

@@jameswood8021 You have it backwards. If there's no preferred direction, all the lines in a single direction should have the same distribution as lines in random directions. So we can analyze one case knowing that it is already generalized to the other cases.

[@malcolmw513]

I really like that he is able to engage with Brady’s questions in a sensible way. A lot of people would deflect Brady’s “burn your house down” question trying to be clever, but Grant actually gives a thoughtful response to its intent.

[@asamenechbayissa553]

"Probability is the extension of logic" This line will stay with me

[@digitig]

All of mathematics is an extension of logic.

[@marsulgumapu2010]

Thus was born the quantum computer

[@digitig]

@@marsulgumapu2010 Or not. Until someone observes it.

[@alxjones]

There's actually a formalization of this exact idea called fuzzy logic.

[@columbus8myhw]

"Infinity comes with its own little first-aid kit." That's a great quote!

[@yashrawat9409]

Following it up as soon as I finished the first video

[@supu8599]

Me too 😀

[@MrDiscoTube]

I love the way Grant smiles throughout - he is having such fun, and so are we!

[@Yupppi]

I'm just so impressed by the animations. Moving the circle with triangle and circle inside over the lines and actively changing the colors based on if they are longer or shorter.

[@soymilkassassin]

You'll probably like Grant's videos on his channel...

[@zygoloid]

I was definitely the kind of student who'd see that the distribution matters and then pick a weird distribution on purpose, eg: a (proper) chord is uniquely identified by a point on the circle plus a chord length in (0,2r), where the chord runs clockwise from the chosen point. Selecting those uniformly gives P(l > s) = 1 - √3/2.

[@sheridanvespo1269]

Wouldn't that cover only half the circle? So I suggest going from -2r to 2r to cover both halves. Don't know what probability that works out to.

[@matthewhubka6350]

@@sheridanvespo1269 I presume it’s the same as what he gives

[@columbus8myhw]

@@sheridanvespo1269 You could get the other half by choosing a different point for your first point. (Every chord has a "left" side.) (Well, excepting diameters, but they're measure 0. I hope.)

[@ancientswordrage]

I wonder what the symmetry there is

[@addy9961]

@@sheridanvespo1269 I think the "missed" chords would be counted by picking the supposed missed end point as your new start point and using the same chord length. But if it changes the distribution then idk that's weird

[@proxyprox]

21:09 This is the real proof by intimidation

[@Lemon_Inspector]

Alternative title: The 7 Stages of Jaynes "I don't like Jaynes. His smug aura mocks me." ... "So in conclusion, I agree."

[@verdigris9742]

I was really nervous until he picked 1/2 as the "don't burn my house down" answer. Whew! I think 1/2 makes the most sense if you just consider the set of all parallel lines with a particular angle. (E.g., all horizontal lines.) Throw out the ones that don't intersect the circle at all, and if you choose uniformly from the ones that remain, you get 1/2... Excellent video, love to hear Grant and Brady go down those rabbit holes!

[@ericbright1742]

The best way I see to look at this problem is this: Pick a direction. Fill the circle with chords in that one direction, evenly spaced. Every possible chord would be created in this case, via rotational translation. Calculating the ratio would give the 1/2 solution.

[@TristanCunhasprofile]

Actually, you could pick any number of directions, and add any number of chord (including infinite) pointing in those directions and you'd get the same answer, right? Because every direction would give a 1/2 answer, and adding more directions wouldn't change the answer for the chords from other directions? I suspect that some methods are counting some chords more than once as the number of chords approaches infinity, and that's giving biased results? For example, if you start at a point and draw random chords to any other point on the circle, the answer is 1/3 for the first point. If you then pick another point slightly clockwise and repeat the random picks, you have some small chance of picking a chord that you've already counted. As you keep repeating with more and more chords from more and more spots, the chances that you'll count a chord you've already counted already increases. Taken to the limit, the chances of getting a duplicate chord when you've started at 90% of the points on the circle is 90%, and most of those duplicate chords will be shorter. It seems like this method has to be biased towards duplicating the pick of a shorter chord more often. So if we deleted the duplicated chords counted, the answer would have to be greater than 1/3?

[@dominicveconi4301]

@@TristanCunhasprofile The problem isn't that they're counting some cords more than once as the number of cords goes to infinity (the second method really does generate all cords without repetition and gives a probability of 1/4). The problem is what Grant said towards the end of the video: if you try "counting up cords", you run into problems because the number of distinct cords is uncountably infinite. You can't avoid making a choice about a probability distribution. In your example construction, you're using a countable process to generate all of the cords. That's not possible. In fact, if you just go point by point around the circle, you're picking points discretely, not continuously. In other words, you've picked a countable collection of points. Even if it's countably infinite, even if it's dense, it's not 90% of the circle, it's 0%. (Consider that if you randomly choose a point between 0 and 1, assuming a uniform probability distribution, the probability of picking a rational number turns out to be 0%, because the rational numbers are countable.)

[@gleedads]

This (and the video that it is an extension of) are one of the best math videos I've watched in a very long time! I suppose in the same sense that we need to precisely define what we mean by "pick a random chord" I should specify what I mean by "best". I'm a statistical physicist. This video will actually impact my research because it has revealed a way of thinking about the consequences of picking initial conditions for molecular dynamics simulations which hadn't occurred to me before. For quite a while I've been bothered by how some researchers initialize their simulations in a particular way while others initialize them in another way. I think this has showed me some better ways to think about the differences between those two ways.

[@ricobarth]

That's cool. I was actually wondering how Grant's simulations were actually coded, because small changes of "initial conditions" there would affect the outcome.

[@jounik]

That 1/2 probability seems to come from defining chords as a circle sampling a uniform distribution of lines with uniform distribution of orientations, so that the lines intersecting the circle at two points are included in the set of chords as a kind of Bayesian prior. The construction given with a random point on a random radius as the midpoint gives the same set more by a coincidence than by design. The other constructions discount the existence of "failed" chords entirely so they end up with a set that is denser in short chords and thus a lower probability of longer ones. It's interesting that if one started the question with "Pick a random line. If it forms a chord with the circle" instead of "Pick a random chord of the circle" the answer can change.

[@iurigrang]

The funny thing about language is that if it was merely “pick a random line that is a chord” it would probably make people lean a lot more to 1/2 than “pick a random chord”, even though “a line that is a chord” and “a chord” are the same thing. The book was also kinda missleading when using that method to find 1/2. 1/2 is not as artificial as it seems from the video, but that particular way of finding it is not natural at all.

[@pxlated2366]

I had a very similar way of approaching this problem. I drew a rough circle on graph paper since the uniformity of graph paper implied randomness by “eliminating” it. Very quickly I was able to make the same assumption that the third method did. This solution makes the most sense to me but, unfortunately, it’s still all in the way a chord is defined.

[@buttonasas]

"uniform distribution of lines" is not a thing unless you say _where_. The _where_ affects its symmetries. A plane has different symmetries compared to a sphere's surface, for example (for points, at least, which is what they are comprised of).

[@jounik]

@@buttonasas I'll admit that the setup of the question didn't _explicitly_ state that the circle itself - and thus any chord it has - is embedded in a plane but that's what all the discussion is about. Besides, while the comment is true, that wouldn't change the answer since such a line on a curved surface won't form a chord with a planar circle.

[@buttonasas]

@@jounik You've just said it yourself: "planar circle". That means it's on a plane. There are other kinds of circles not on planes nor spheres but instead spaces of digital values, spaces of energy and mass readings, spaces of analog state, etc.

[@jacemandt]

Here's another reason that the 1/2 answer makes intuitive sense to me: When we say "choose a random angle for the chord" in that chord-selection method, we're noting that it shouldn't change the probability if we just fix the angle, since that doesn't affect the length. So without loss of generality, assume that this is the unit circle on the plane, and that all the chords are vertical (since angle doesn't matter), distributed with uniform density across the interval (0,1), since we really only have to consider half the circle. This way of thinking of it seems to unify Grant's original description of chord-selection with the idea of "consider all lines on the plane and place the circle randomly".

[@LucenProject]

I think I'm still lost. I had the idea to make them all vertical lines, but then thought it was a simple case of seeing "How often are these vertical lines longer that radius*SQRT(3)?" But I still ended up getting that that would be 1/3 of the time rather than 1/2.

[@timothybexon6171]

My problem with the 1/2 method is that it is the probability of a 1 dimensional line in a 2 dimensional shape. This sits uneasily with me. I understand and agreed with the logic though.

[@Khaim.m]

You're right that angle doesn't affect the length, and so it's perfectly fine to arbitrary pick an angle. The issue is the probability distribution over the resulting 1-D space. You get a different answer depending on how you define that distribution. Even if you want to insist on it being uniform, you can get different answers from different ways to _define_ the 1-D space over which you're sampling. For example: radius of chord center, or length of chord.

[@bendotc]

@@LucenProject The trick is that if you define the chord distribution in terms of the angle to the location on the circle where they cross, (e.g. solving (√3)/2 = sin(θ)) then you end up with the 1/3 answer, and the density of chords in your distribution increases with the slope of the circle. Instead you can define chords uniformly along the x axis regardless of the circle. By using the equation of a circle 1 = x² + y², you can consider just one quarter of the circle (again, because of symmetry) and solve for (√3)/2 = √(1 - x²). Doing this, you end up with the cross-over point at x = 0.5 -- half the range.

[@Joshman0131]

Grant does an excellent thing where he does not stop if your question leads to a blunt, short, answer, but recognizes the principal behind what you asked. When Brady asks to label every chord with a natural and you wouldn't be able to, Grant continues with the natural order of questions about labelling them with reals, how many reals, etc. I'm sure his students are ecstatic to feel heard when he answers their questions in class

[@joemonster55]

Grant's explanation of Frequentist vs Bayesian stats was so clear and concise! He's got a real gift for teaching. I'd love to hear him apply this framework to the Copenhagen Interpretation vs the Pilot Wave theory. It seems very analogous to me.

[@momerathe]

Suppose you want to generate a random point on a sphere, and you do so by independently generating lat/long pairs. You'll end up with the areal density of points being higher towards the poles. So it seems reasonable to me that there are "better" ways of generating random things. And not just because I guessed 1/2 at the start of the first video ;)

[@huckthatdish]

We have a transitive symmetry in the points on a sphere case. That method is not invariant over that transitive symmetry.

[@adelarscheidt]

I'd plot radial lines off its center with randomly generated alpha/beta/gama angles between 0°-360° - that should do it I think

[@finn8518]

there was a great SoME1 video about the best way of getting a random dispersion of points on a circle, that was amazingly explained and made me know the answer to this one already :)

[@lonestarr1490]

That's not how you (should) randomly select two points on a sphere. Better start with one point and apply a random rotation, two times in a row.

[@evanrosenlieb8819]

But that would be an artifact of how we define latitude and longitude, you could use a different coordinate system to define points on a sphere -- like one that arbitrarily has two perpendicular parallels instead of a parallel and a meridian.

[@linkspring1287]

17:12 Probability of getting a 5 in this event(said by Grant), explained me the whole point of this paradox instantly.

[@andrewharrison8436]

Actually Grant has an extensive colllection of D4 dice.

[@QuantumConundrum]

@@andrewharrison8436 that is a genius level trap for a street performer right there.

[@ten.seconds]

I think the 1/2 makes the most sense the more I think about it, and here's how it can be rationalized: 1. Choosing an angle is just choosing an orientation of the line. 2. Choosing from 0 to r is just translating the line around it's normal, and limiting it to r limits it to lines that are chords (that touches the circle) And there you go, it feels natural that this is analogous to the "every possible line, plop a circle down" approach.

[@okuno54]

That's exactly what I was thinking as Grant complained about that method's artificiality! It's really all in the presentation isn't it?

[@JohnMeacham]

Yeah, the 2 choices exactly line up with the 2 symmetries. Makes sense. Cool.

[@glenmatthes8839]

I was imagining the last method not as a bunch of random lines, but as a plane filled with infinitely many parallel lines. You can translate the circle through this plane and the distribution stays the same. You also get every single length of chord through (on?) the circle of which half are longer and half are shorter than sqrt(3). Choosing the angle is the same as rotating the circle on this plane. But rotating the circle doesn't affect anything. You can't tell when a circle has been rotated. Adding more infinite sets of parallel lines at different angles doesn't change the distribution either. Add an infinite number of these infinite sets and you still end up with the same 50% probability.

[@LeonardEisen]

Yup, makes sense

[@jamesflames6987]

If you think you've found a special method that "feels like it makes the most sense" you've completely missed the point of the video.

[@aajpeter]

3 different flavors of chord spaces, three answers. I love it. The paradox is that it only feels paradoxical before going through the explanation of each approach, as by the end you have embedded the precise methodology of each into the original underspecified problem statement.

[@jukmifggugghposer]

the 1/2 option feels pretty natural as someone who works with spherical and plane-polar coordinates a lot. Points within the circle are defined by two coordinates, call them θ and r, and you’re choosing a random value for each of them. My first choice for choosing a chord would be the two points on the perimeter option, but knowing that a point within the circle uniquely defines a chord, I would definitely go with the third option over the second.

[@jorgechavesfilho]

This situation in itself raises another problem: for every number between 0 and 1, is there a way to model a solution such that the probability found is this number? We already have solution modelings for 1/2, 1/3, and 1/4.

[@franciscofernandez8183]

Yes there most definitely is. The hard part is to find ones that came "naturally" from a simple symmetry recognizable by humans like the 3 exposed in the video.

[@Bovineprogrammer]

What if you choose your chord by randomly choosing two points WITHIN the circle? What if these points are both chosen by the radial method? What if you generate two chords using any combination of other methods, and then connect the midpoints of those chords to create your chosen chord? Intuitively I expect any rational number between 0 and 1 to be attainable, but perhaps not every real number.

[@mzg147]

This is true. There is a paper showing that, but I cannot find it now :(

[@beartankoperator7950]

fantastic question

[@Aaron-th7xx]

Sure, an easy way is with the third method. Instead of taking the uniform distribution on the radius, you can take any distribution that you want on the interval [0,1]. The problem is, this usually won't feel as naturally symmetric as the others.

[@kaidenschmidt157]

Brady: “I’m gonna burn your house down” Grant: “ok fair question”

[@wiseSYW]

the 1/2 solution is the most correct one for me. divide a circle into vertical strips. when you choose a chord, you in effect choose one of these strips, since the length are the same just rotated.

[@dougaltolan3017]

You have a reduced version of Jaynes approach. Instead of filling space with all lines and placing a circle on it, you can consider filling space with parallel lines, since the result for any set of parallel lines will be the same (circles are rotationally symmetrical)

[@jezennalakomne]

Ha one of very rare examples where gray matter between shoulders is used! Excelent Right solution is only one and it is 1/2. First and second example are double counting same lines. It is only third that it doesn't. There is even more simple example that proves 1/2 is right. Circle with triangle in it. Now make one side of triangle bold. Rotate triangle for 180°. Now all lines between bold lines are longer, while all others are shorter. And this split circle diameter on 1/4, 1/2 and 1/4 while both 1/4 are shorter, we are left with 1/2. Since circle is symmetrical this is valid for all orientations. First example in video has divided area on three sectors each counting for 1/3 of the lines. But if triangle is slightly rotated in any direction, new set of the lines starts covering already existing ones. This double counting happens to be one of the two (left and right sectors) that is double counted. After 120° rotation 1/3 of the lines is double counted. And double counting is not allowed. Therefore you are left only with two valid 1/3. One presenting longer lines while other shorter. So you are left with 1/2. Second example is even more bizarre because at that instance lines are triple counted, therefore two 1/4 should be counted. So half of four 1/4 is 1/2 again. Enjoy

[@artsmith1347]

The random point on a radius also parallels the "best" way to randomly distribute points on a sphere: pick a random point on a radius and project it at a random angle to the surface of the sphere.

[@insightfool]

I wish he would have just started with what he said at minute 17:53 at the very beginning instead of doing all this math gymnastics without motivating the problem. This two videos were so frustrating! 😖 ...but, to be honest, seeing the simulation really helps me understand the different methods better than when he was describing it. And I understand the problem better than I ever have before, so... Thanks!

[@eranbernstein7972]

Amazing video! I love the question "how does this make you feel at night?" and I love to hear Grant's philosophy about math, it's restrictions and relation to life. The example with the bag of random cubes deeply impressed me. Amazing video :)

[@protocol6]

The differing interpretations that Grant is pointing out (frequentist vs bayesian) is the source of some interpretational issues in particle physics and gives rise to a lot of quantum woo.

[@hens0w]

Copenhagen being bayesians and many worlds being frequentist? If I understand you right and thing for about 15 seconds then it seems to me the cat is the best argument against bayesianism I even heard. (Beeting out my usual "disposition" of its hard but they get the same answers)

[@protocol6]

@@hens0w No, Copenhagen utterly fails to realize that a knowledge update doesn't actually change anything other than your knowledge. Many worlds doesn't really improve the situation, It just moves the goal post beyond an event horizon where it's less obvious. You have to move toward hidden variable interpretations like deBroglie-Bohm or others that sidestep Bell to get closer to Bayesian.

[@lachlancooke]

quantum woo 😂

[@recompile]

Never listen to anyone who uses the term 'woo'. They're easily suckered by con-men like James Randi.

[@jessejordache1869]

@@hens0w Many worlds and Copenhagen are in agreement. I think the non Copenhagen model is the pilot bubbles theory. If that's the same thing as deBroglie-Bohm, then, "what Transmission Control said". The Copenhagen model is the limit concept of probability, and the pilot bubbles model is the Bayesian model (there is a predictable answer of what the final state of the quantum wave will be, but we lack enough information to predict it).

[@zacfubar1]

This was absolutely the best anecdote for the difference between the Frequentist and Bayesian philosophies that I've ever seen (and a confirmation of my opinion that Bayesian thinking is more practically effective).

[@Evan490BC]

I love the Bayesian approach (I did my doctorate on it) but by itself doesn't solve Bertant's "paradox".

[@GoodBrownBear]

They are not contradictory approaches but complementary. No one will disagree on the calculation of an expected value but Bayesian approach takes a practical point of view when faced with uncertainty. In this case the uncertainty is the definition of random chord. It took the position that random chords should result in uniform density.

[@stevescott2819]

The third method seems a little obscure, but there is another way to describe the third method that makes it seem the most natural: The third method is basically describing the circle intersecting all lines parallel to one side of the triangle. It’s easy to imagine that the circle is completely carpeted by a uniform distribution of non-overlapping parallel lines (unlike in the first method, where picking all chords passing through a single point intuitively seems to cover the region around the point more than the region far from the given point; and unlike the second method, where the computer simulation clearly shows a non uniform distribution.). By the argument given for the third method, the probability is 1/2. Although this is only for the infinite number of lines that are parallel to the one side of the triangle, intuitively, it is clear that there is a rotational symmetry that keeps the probability at 1/2 when you start considering all of the infinite parallel lines from some other orientation. I think if the third method had been explained first in a more intuitive sense, there would be much less debate, or interest, in this problem. But the warning regarding how to think about such things is important, and appreciated.

[@fruitshuit]

Thank you, phrased like that it does sound much more natural!

[@Aleksandr011]

What I like about the third one is it's constructing lines randomly by using polar coordinates. That feels the most intuitive to me.

[@randomnobody660]

But the premise of the question is that we are drawing chords in a circle. A chord is, according to some, "a line segment joining two points on any curve". Does a random "chord" not just mean the line segment joining 2 such random "points" on the circle then?

[@fwiffo]

The problem here is that you're creating a bias toward the center of the circle. Within a circle, there are fewer points near the center than near the edge, but this is not the case on a line segment. It would be like choosing a random point on a sphere by choosing a random point uniformly from its shadow. You'd have an over-concentration at the poles.

[@Aleksandr011]

@@fwiffo I disagree that there is a bias. It stands to reason that any line that randomly intersects with a circle will have a 50% chance of intersecting with a concentric circle with half the radius, which is exactly what happens.

[@Kenneally151]

Picking arbitrary point within circle is rotationally invariant P(l>s) = 1/4 Picking arbitrary point on boundary of circle is rotationally, scale invariant P(l>s) = 1/3 Picking arbitrary point along radius of circle is rotationally, scale, and translationally invariant P(l>s) = 1/2

[@russellthorburn9297]

I was somewhat lost until 17:32 and the analogy with the dice. Well done Grant.

[@Nikception]

I think another way to think about the 1/2 answer is to imagine that you're picking a random chord from the set of all perfectly vertical chords. This feels like a reasonable thing to do. It's like you've got all your ties of various lengths fitting perfectly in your circular closet and you want to pick one at random. Then you just imagine that the chords are at a random orientation, and you get the 1/2 answer.

[@andrewharrison8436]

Ahh yes, but when picking one tie aren't you more likely to pick the longer ones more often than the shorter ones (I can never find my 1 cm tie)? So I think you are right picking a tie (numbered 1 to 100) by choosing a number with equal probability has a right answer of 1% each. A different problem is to muddle all the ties in a pile then chose a piece of material and pull out the tie it is attached to which gives a different distribution. A clearly defined choosing technique gives a clearly defined answer - the problem with the ties and the chords is the lack of definition. (Half my ties are hung neatly, the rest have slipped into a heap at the bottom of the cupboard - I never select a tie at random).

[@kasuha]

Ambiguous specification of the task is the source of so many so called paradoxes in mathematics. Declaring single "unambiguous" interpretation for such cases is in my opinion even worse than recognizing and treating them as ambiguous, and asking for unambiguous specification instead.

[@FManga18]

Yes if the single unambiguity is called on a case by case base, But if you decide that the shifting has always to be true whenever you can apply it then I'm fine with it

[@blacklistnr1]

To add to that, I have yet to encounter a paradox which is not formed by omitting some ideas. They all read like "Outside is sunny. Outside is rainy. Ha! I have omitted that those were two different days."

[@Varksterable]

@@blacklistnr1 But it is perfectly valid for it to be sunny and rainy at the same time. 🌈 (And no, this isn't just pedantry; it relates to an interpretation of the problem under discussion.)

[@frechjo]

@@blacklistnr1 For Russell's paradox (the one about the set of all sets that don't include themselves), what information would you say it omits?

[@blacklistnr1]

@@frechjo As far as I remember it, it omits the distinction of self, much like the barber who shaves everyone who don't shave themselves. As per my first reply, it reads: "I shave everyone [...]. I don't shave myself. Ha! I omitted that I talk about everyone who is not me".

[@rustymustard7798]

"Choose a random chord." Define choose, random, and chord.

[@Lemon_Inspector]

Define "define", "and", "a", and ","

[@tim40gabby25]

@@Lemon_Inspector So "." is unambiguous?

[@Lemon_Inspector]

@@tim40gabby25 Yes.

[@tim40gabby25]

@@Lemon_Inspector Interesting...

[@Lemon_Inspector]

@@tim40gabby25 No.

[@borisgrozev2289]

Fascinating. The third method seemed the most unintuitive to me, too. More than that, it's equivalent to selecting a single random point on the circle with a skewed distribution, which felt like the wrong thing to do. But selecting chords this way does have this special property. Really cool.

[@umbertorodrigez8213]

My fix to the solid state method. Bisect the circle on each axis. Number the two lines. Check. Double the number of lines, space them equidistant from parallel lines. Number them. Check the average. Repeat doubling process. Look for convergence.

[@fangjiunnewe3634]

Something to notice is that "choose a random chord where the length is longer than X" presupposes that the chord length has to have a "uniform" distribution, and only in the third case does the chord length have such a uniform, transitive symmetry. In the first case, it's choosing a random point on a circle relative to a fixed point, and the chord lengths are biased to be shorter, since the underlying distribution is the angle between the fixed and random point subtended by the center, and trig functions are non linear. In the second case, choosing a random point in the area of the circle, the chord lengths are also biased to be shorter because for a uniformly dense forest of midpoints the chord lengths are non uniformly shorter the further away from the center. By choosing both an angle and a height, the chord length can now be uniform because the calculation for length "cancels out" the choice of angle in a totally coincidental way. The chord forest loses density the further away from the center while also loses on chord length at the same rate. I need to emphasize that the fact that this works is a coincidence, Jayne's method really is the correct way to think about it. One perspective is to consider the common question in intro to stats classes for choosing a random point in a circle, where people tend to choose an angle and a distance because circular coordinates are easier, but it's completely wrong because the conversion from angular density to Cartesian density was forgotten. Basically the same thing happening here, I think it's not so ill defined as Grant puts it, there is definitely some non linear conversion of one measure to another that is being forgotten.

[@nestoreleuteriopaivabendo5415]

I just thought the same thing about the rotational symmetry. The distribution of the chords over one of the radii is not uniform, because the farther away from the center, the more points to choose from the circumference there are, based on the radius of the random point selected from the center of the circle, just because the circumferences are longer there. Hence the probability is biased towards the edge of the circle.

[@cheshire1]

The method of averaging all the approaches is very interesting. Basically you're assigning a probability distribution over all the possible meanings of the question. In this case it's a "uniform" distribution of 1/3, 1/3, 1/3. So you would probably run into the same problem again. Is there a symmetry in the space of all those approaches? What happens if you add more and more layers?

[@beetehotraroy3468]

What you are trying to do is assign a measure over all possible measures assigned over the space of chords... A metameasure, if you will.

[@Zinong-b3m]

Well said. I was a bit confused about that part and your interpretation helped a lot.

[@travismyers3396]

Every finite space of possibilities trivially has a symmetry associated with it that gives a uniform distribution: just label each possibility with a number 0 to N - 1 (where N is the number of possibilities), and then the symmetry operation transforms the label n to (n + 1) mod N.

[@sidsixseven]

I felt like the 1/2 answer is the most intuitive. Consider that half the points of any circle lie on the opposite side of that circle. You can change the orientation of the circle in any direction, but that fact remains the same. Given that, it seems natural that the answer would be 1/2.

[@strehlow]

And how can you prove, or can you prove, that you've enumerated and described all the possible methods of randomly choosing the chord? One needs to know unambiguously how many methods there are to find the mean of their probabilities.

[@TheBasikShow]

I feel like Grant is, probably without intending to, biasing the audience against the third distribution as the one that “makes sense”. He himself states that it feels the least natural. I think the problem is that his presentation of the third distribution is... odd. Maybe it’s what Bertrand initially did, but it’s not the version I first saw. Okay so, here’s the way I was taught: Remember how we set up the first distribution? We said, “Because the random chord is rotationally symmetric, we can rotate the circle so that one of the points is on top.” Then, one natural way to make it “uniform” is to place the other point of the chord uniformly anywhere else. (The other natural way to make it “uniform” would be to give it a uniformly chosen angle, but that turns out to produce the same result in this case. It would lead to a different, fourth distribution if we were using anything other than a regular triangle, but that’s not important here.) But, we could also use a different, equally natural first step: “because the random chord is rotationally symmetric, we can rotate the circle so that the chord is vertical.” Ta-da! Now we have to choose how different chords of the same orientation are distributed, and the most natural way to do that is to assume they cover the space uniformly-that is to say, they’re distributed evenly along the radius. This also gives an intuitive reason for why distribution 3 is so nice-looking: from this construction it is easy to see that each point in the circle is contained in a chord about as often as any other, and so the image made by overlaying thousands of chords looks uniformly shaded, instead of being brighter near the edges. This, in turn, explains why this distribution has translational AND scaling symmetry.

[@ricobarth]

I agree. I thought the third way was ridiculous until I reframed it myself. Rotate the triangle to make a side parallel with the random line, and drag that line across the entire circle. Which gives the insight that we're dealing with different infinities.

[@bendotc]

Yes! I actually took the time between watching part one and part two of this video and worked the problem myself and ended up taking this line of reasoning. It's got rotational symmetry, so I can simply consider "vertical" lines. Additionally, I can just consider a quarter circle, since symmetry means that the same distribution will hold for the others. So I take the equation for a circle (x² + y² = 1) and transform it into y = √(1 - x²), and only look at values of x between 0 and 1. Now, if I wanted to solve the original question of the average length of the chord, I could integrate that, but since we're just interested to know the likelihood of a chord being longer than the side of the inscribed equilateral triangle (√3), then we just need to know where the space between the quarter circle and the x axis crosses half the length of that line (half since we're only dealing with the top half of the circle). So to find that, we can subtract (√3)/2 from our function and find where it crosses zero; that root should tell us what proportion of the lines in the quarter circle are longer than the length of the equilateral triangle. And if you solve for 0 = √(1 - x²) - (√3)/2 for positive values of x, you find that the answer is 1/2. But the thing that inspired me to come here is the turn this logic can take that'll get you a different answer. Still thinking of using the rotational symmetry to only consider vertical lines on a half circle, we could forgo the equations in terms of x and y, and instead think of the portions of the circle in terms of angles. If I define each chord by where it touches the half circle, I can enumerate them by the angle θ. And using a similar logic to above, I can soon find the roots for the half circle: 0 = sin(θ) - (√3)/2 for values of θ between 0 and π. And it turns out that measured this way, the cords are shorter for values in the intervals [0,π/3) and (2π/3, π]. In other words, the chords are longer 1/3 of the time! However, the problem is the density of vertical chords in the second example is dependent on the curve of the circle. We end up with a greater density of chords considered where the slope is high. Although not materially a different argument from the one Grant makes, I found this way of thinking about it to more obviously favor the uniform distribution of lines that yields the 1/2 answer.

[@joeeeee8738]

@@bendotc great explanation

[@rafaelmarkos4489]

@@bendotc I believe the standard phrasing for describing the second part of your explaination is 'angle subtended by chord', i.e. the angle between the radial lines from the endpoints of the chord. To add to your analysis, if we're given a chord subtedning an angle θ between 0 and π, the length of the chord is 2sin(θ/2). Now, this function in the given interval varies between [0,2], and averages out to 4/π, and the length we are looking for is > √3. Thus, we can confirm that the length of a chord in terms of the angle subtended by it is a non-uniform distribution, and that we are looking past the average value of the distribution - thus, there are 2 reasons to expect a biased selection of chords. And yet, somehow 'choose 2 points on a circle' still appears the most natural method to me. Go figure.

[@Vacuon]

I also think the translatable answer to be the only correct one, but man that argument was so beautifully put. In my mind I was thinking "well obviously the one with 'uniform' density is the correct one", but I never would have thought about moving the circle on "preexisting" lines. This is such amazing outside of the box thinking, and 3blue1brown's channel is full of those amazingly intuitive explanations.

[@akhilsajeev1601]

Like you said when we say random we usually use uniform distribution. Here the key words are chord and circle. Not points and circle. How can we ensure a chord has uniform distribution. It has uniform distribution if there's a uniform in both the polar coordinates. Uniform distribution of the distance from center and angle wrt to some initial line. In the first method we the angular distribution is not uniform. The first 150 has 1/3 probability. The next 60 has 1/3rd and the last 150 has 1/3. So it heavily favours the chords which are longer. Second method has a uniform distribution when it comes to angle, but the probability of selecting a chord increases proportional to the distance from center. Hence it's also not uniform. Favours chords which are longer In the third method when we choose a random initial radial line we are also selecting the angle of our chord. Initial radial line is perpendicular to our chord. When we select a random point of that chord we are also selecting the distance. We are ensuring both the distributions are uniform here. I loved the idea of traversing the circle. Made it easier to grasp it. Also sorry for any grammatical mistakes.

[@limbridk]

Thank goodness for this follow up video. The first one was driving me crazy. But this one tied up so many lose ends. It's a relief. Great videos as a pair. Thanks

[@Tumbolisu]

The second method felt the most artificial to me, with the first and third being tied for most intuitive. But after hearing the argument about being able to move and scale the circle without changing the answere, it feels obvious that chosing 2 random points on the circle line is wrong. Like, what makes the circle line so special? A modification would be to first chose a random radius and draw a new circle, then chose 2 points on that new circle. If the line connecting the two points is outside of the original circle, we discard it. Now what are the chances? Well, smaller radiai are always going to make a valid cord, while larger radiai have a chance to make an invalid cord. So that immediately makes it obvious that we were actually missing out on a lot of long cords! In other words, the distribution might have looked uniform at first, but that is because our eyes aren't perfect. The first method actually is sparser at the center than at the edge.

[@passerby4507]

You missed the point where if the natural universe of consideration is the circle and only the circle, there would be no "other chords".

[@W0X42A]

"What makes the circle line so special?" The definition of chord: In plane geometry, a chord is the line segment joining two points on a curve.

[@Tumbolisu]

The idea that "lines outside the circle don't exist" just feels artificial to me. Why shouldn't they exist?

[@jpdemer5]

@@Tumbolisu You can make them part of your universe of chords, or you can choose not to do so. That's where the "paradox" arises - the universe of lines from which you're choosing isn't specified, and the "random distribution" of chords isn't defined in the absence of that information.

[@passerby4507]

@@Tumbolisu It's an abstract space, not the geometric one being drawn. An analogy is your clock. You don't think there's more numbers on the clock face other than on the circle do you?

[@ClydeCoulter]

My first inclination is to consider only half of the circle, where all chords are parallel to the line across the diameter. All other chords are either a mirror of or a rotated version of those.

[@jacksonbrim7359]

My thoughts exactly. And then it's just a number line problem. How many of those lines are longer than the other... 1/2. I'm having a hard time imagining how the answer can't be 1/2, tbh, and I don't think it requires the translational distribution bit.

[@martinhall1188]

Yes, but how do you select the chords "randomly". You can select "uniformly" along the line perpendicular to the selected diameter (which gives 1/2). Or you can select "uniformly" along the circumference, which gives the answer 1/3. [Not sure what is the selection process which gives 1/4, but it will exist :)].

[@tb45g]

Isn't that technically just the same as the third method?

[@ForsakenDAemon]

The explanation of the difficulties with assigning a number from 0 to 1 to every chord was such an elegant description of the issues with copulas

[@GoodBrownBear]

The first method quickly showed its flaw because the angle between chords are not uniformly spaced. Imagine we divide the circumference into 360 tick marks. Starting at any point A, you draw 359 chords by connecting A to the other 359 points. These 359 lines will not have the same angles between them. The first few lines and the last few lines will be incredibly close together while the 180th line is widely spaced from 179th and 181th.

[@thomaswalters7117]

1/2 was the most natural to me! It seemed the simplest way to solve, since you just remove everything "below" the line, (then mirror and rotate to show it fills "every" chord). It's also the simplest way to fill space with lines: just take one line, repeat it infinitely vertically, then repeat that infinitely through 180 degrees.

[@vitorbortolin6810]

This is an important question in the core of probability and statistics. In foundation of statistics a great book by Leonard Savage this topic is explored. Although this is still an unsolved problem and I think it will always be. Great Video!

[@cparks1000000]

It's not unsolved. It's undefined. You need to define WHAT random means. You have to define the distribution.

[@marsulgumapu2010]

I'd say there's a 50/50 chance it will be solved.

[@ricobarth]

@@cparks1000000 The problem is how do define it, and that problem remains unsolved.

[@KoHaGames_]

In my opinion this is relatively obvious. We cant tell the odds, if we look at it as a circle and as random lines through that circle - the question is: How *can* we tell the odds? The solution is, to simplify the given Model into something we can specify without doubt. In fact every line, no matter how chosen, is parallel to every other line, no matter how chosen, if you turn the circle the right way. It doesnt change the length of the line, nor the way you chose the line. What we get as a result is a bunch of lines which are all parallel to each other, randomly chosen and every line crosses the circle. We have 4 borders, that define the result of one line (all parallel to the other lines). One on the left side of the circle, one on the right side, and two equally long lines, that have the length of one side of the triangle. Every line you draw can be defined by the positioning relative to these borders, so the circle is obsolete. We dont need the circle, because it doesnt change anything about anything anymore. We have two borders. Between these two borders are all the lines. And then there are the two other borders, that define which line is longer or shorter than the side of the Triangle. 1: |(| |)| - 4 Borders, one Circle 2: | | | | - 4 Borders, no Circle The position of a randomly drawn line parallel to the borders, between the outer borders defines, if its longer or shorter than one side of a triangle, IF there was a circle. In conclusion we can chose a number between 0 and 1 If 0 < x < 0.25 its shorter If 0.25 < x < 0.75 its longer If 0.75 < x < 1 its shorter again. That gives the whole thing a probability of 1/2, without dealing with a circle, or anything. It's basically the 4th version, but in my head it makes so much more sense. Reasons why this is the right way to do it: 1. Every line that crosses the circle at least once *can* be defined by this 2. If you chose any other method of drawing lines, and you give them all the same direction, so they are parallel to each other, you can recognize more dense and less dense collections of lines, depending on the method. 3. It doesnt matter how big the circle is, or where its placed, its everywhere the same 4. The direction of the line isnt asked, it's the length. If you throw a coin with a "1" on both sides, and you want the probability of the "1" pointing into a specific direction, then you have to deal with the angle, or even with the position of the peak of the one, relative to the circle. But that's not the case here.

[@colinree6131]

It's clearly 1/2 if you construct the chord using polar coordinates. Choose an angle for the radius, then a position along that radius, both with uniform distribution. Rotationally and radially independant, right? The issue is, when you look at that result with points instead of lines, you clearly see severe biasing towards the middle. The radius chosen affects the distance between increments of angle. Personally I'm attached to the 1/3 answer, because it removes that second dimension and any influencing that can occur because of it. If we think about the circle not as a 2d disk but as a 1d circumference, then we just need two random numbers from [0, 2*pi] and voila. The point is, you get two different answers if you see chords as every set of possible lines that contact a disk, versus every set of possible points on a ring. There's no reason to say one's more valid than the other.

[@stargazer7644]

Except that.....you CAN tell the odds if you do it correctly.

[@KoHaGames_]

@@stargazer7644 how do you mean?

[@BryantHarrisonCville]

This is super clear. What I don't get is why the 1st method doesn't produce the same answer. - choose your two points on the circumference - draw the line - rotate the line so it is parallel with the horizon - repeat Why is this method not evenly distributed like the 3rd method. I can't figure it. Furthermore, if you're moving the circle across a plane full of random lines. Wherever you stop the circle, you can visualize that circle as a bunch of paired points touching the circumference. How is that more random and more distributed than picking random points on the circumference to begin with?

[@stargazer7644]

@@BryantHarrisonCville In the first case the lines have a higher probability to be shorter, so it skews the result.

[@xenontesla122]

There's actually a *fourth* way to interpret a "random chord". For a circle of diameter d, generate a line with random length between 0 and d. Place one end at a random angle along the circle and tilt it until it's a chord. Because we already defined the length as being part of a uniform distribution, the chance of it being larger than the triangle length is just 1 minus the triangle length over d. So a chance of 13.4%.

[@thelocalsage]

the connection of enumerating uncountable sets with selecting a probability distribution was insane, like a nucleation point where the whole rest of the content could crystallize. grant is a brilliant communicator

[@rianantony]

This is an old discussion, but I'd like to add that if you picked a set of paralel cords, you can say pretty confidently that half of them are longer than the lengh of the side of the triangle. And because a set of paralel cords could be made with any orientation, it would make sense that any one of them would be representative of the whole. Edit: a day later I return to add. If random means that you have a flat distribution of every possible lengh of cord, than that means that you could pick one of each possible lengh of cord and that set of cords would be representative of the whole. One of each possible lengh of cord, if arranged, parallel to each other, in order can be shaped into a circle, first one half in ascending order until the diameter and then the rest in descending order. Now that we have arranged a representative set of every cord in order we could draw the triangle with one side parallel to them and see that exactly half of them are longer than the sides (with the same trigonometry in the video). I'm not saying anything new (idk if this has already been said in the videos and I forgor💀). But this is my take either way, if I had to pick an answer with a gun to my had it'd be this one.

[@RajaAnbazhagan]

Yup.

[@bleikeze]

"What is the probability that a random person comes up with an answer to this question that is less than k for each k in [0, 1]?"

[@ntucker2377]

I think a great illustration/analogy of the difference between the three methods of choosing chords is to take the problem of choosing a random point on a horizontal line, and deciding that it seems fair to do so by standing somewhere near that line and pointing at it with a laser pointer held at a randomly chosen angle. This warps the answer space to the method chosen: it will seem evenly distributed, but the answers will be clustered more tightly along the place where the laser beam is perpendicular to the wall. I think it would be interesting to illustrate the different chord choosing methods in terms of the others, for example, plot the density of the chord midpoints when choosing by randomly choosing the two points on the edge, and compare this to the randomly-chosen chord midpoints. If you can come up with a way to intuitively compare the distributions of the A B and C methods can more concretely illustrate the arbitrariness of those methods. The plot of the chord midpoints chosen using the Jaynes method, for example, will obviously be much denser around the center of the circle. This doesn't really answer which is more correct, of course, but it does point out the difference in methods that "intuitively" seem equivalent but really aren't.

[@qo92]

Brady asking the best questions, Grant giving the best answers. Top notch!

[@eaglegosuperskarmor]

I also feel that working with circles can cause issues. If you try to pick a random point using polar coordinates, you'll be biasing the point towards the centre. If you were to draw the lines of all of your angles that you can sample from, you'll notice that those lines get further apart as you move towards the edges, which is why polar form integrals have the correction factor of multiplying by the radius. It might feel more intuitive to use polar form whenever you're working with circles, but it also subtly distorts the world you're working in, and you might not notice it.

[@reaganshonk]

I saw the word "compact" and my brain immediately went: "Every open cover has a finite subcover!" I'm both disappointed and proud of myself.

[@FManga18]

But remember that we are working in R^2 and so compact means closed and bounded

[@HilbertXVI]

@@FManga18 They're equivalent

[@98danielray]

every net has a convergent subnet

[@cristianstoica4544]

I think although two random points generate a cord, the opposite is not true: a random cord does not generate two random endpoints. The endpoints positions are correlated to each other.

[@dominicveconi4301]

What's a "random cord" though?

[@centripetal25]

@@dominicveconi4301 a chord generated by plotting 2 random coordinates

[@dominicveconi4301]

@@centripetal25 That seems like a weird definition of “random cord” if we’re asserting that “a random cord does not generate two random endpoints”

[@centripetal25]

@@dominicveconi4301 they dont generate random endpoints on a circle when plotted using its circumference. but i would like to think? when you generate 2 random points on a grid which the circle exists in, it will have symmetry. The bias of the paradox comes from trying to generate chords specifically with features of the circle itself. This is a odd limitation. This is what Jaynes is getting at

[@dominicveconi4301]

@@centripetal25 Ah, so by "2 random coordinates" you meant two coordinates in a Cartesian grid, not two points on a circle. My mistake. Is it an odd limitation though? Lots of perfectly natural and physically relevant probability distributions aren't invariant under the symmetries of the underlying space (normal and exponential, for example). And lots of probability distributions are defined only on weird fractal objects (see "strange attractors"), so geometric and topological structure definitely can get involved in defining probability measures. To me (and I think to Grant), demanding translational invariance is a choice, no more or less valid than other choices for defining "random cords".

[@IllidanS4]

A metaquestion now: Considering that these three methods could be essentially defined by the distribution of the angle difference between the starting point and the end point, does every resulting probability correspond to a possible distribution? If so, how many of those distributions actually correspond to constructible chords? But considering that you could essentially pick any distribution, I liked the "balanced" answer of 0,5, like an average over all the infinite possible distributions.

[@YousufAhmad0]

This should definitely have been on the main channel as a part two to the first part. I couldn't imagine any numberphile being OK with being left hanging the way part one does! Personally, after watching this, I don't see how there's any ambiguity left to argue over: in the absence of any qualifying context, "random" is universally accepted to imply uniformly random, and the uniform distribution is symmetric by definition, so any uniformly random distribution of chords on a circle must account for all possible symmetries in 2D space, namely both translational and rotational, so the only correct answer must be the one we get from Jayne's radial method, i.e., 1/2

[@gusshultz5342]

I'm not even a maths guy, but your enthusiasm and clarity makes this a joy to watch. Subscribed!

[@newsfromthefrunk]

What does the distribution of chord lengths look like in the three scenarios?

[@ThePetersNation]

I thought the same thing. Seems like that would help explain that they are just randomly picking lengths from different distributions.

[@kazedcat]

The distribution is the probability. On the first method 1/3 of the chord length is less than S on the second 1/4 is less than S and on the third method 1/2 is less than S they are all linearly distributed the only difference is the slope of the distribution.

[@dhonors999]

For the method of averaging the answers of different methods: If we pick a random method for generating chords, it will generate an answer between 0 and 1. Assuming a uniform distribution, the average answer will be 1/2 ;)

[@robindebreuil]

Great videos :). From the perspective of programming, it seems to me this is one of those problems that seems fully defined but actually isn't when you go to code it. When you say chord, there are two points, and it is important to know you are talking about the probability of the second point once the first is fixed, or two random points. The 1/3 probability is akin to asking what the average distance is to all points of the circumference from a fixed point (rotating the inner triangle is the equivalent of fixing the point). The 1/2 probability is akin to asking what the average height (bottom to top of a vertical line) of the circle is, which will be the same in any orientation. I feel the 1/4 answer is the same thing, but asking about the average height of half a circle. In this way it reminds me of the Boy or Girl Probability Paradox from that Zach Star video. Once you 'fix' one of the two children the probabilities change.

[@s.avtarsingh7590]

Since the very first video I watched back then in 2016 or 2K17, don't remember exactly, but the only thing kept striking me was, "Why did this man has named his channel to "3Blue1Brown" ?" And today I got the answer, which is really soothing and I'm solaced after that long period of time ! You named the channel, AFTER YOUR RIGHT EYE !!!

[@s.avtarsingh7590]

I really didnt know that you mentioned the reason for your channel, UNTIL I watched this video till end ! Genuinely I noticed your eye color at exactly 13:00 minute

[@cyb3r._.]

here’s a different way to think about the “1/2 method”: imagine you have a circle and you also have all of the infinitely many chords running vertically in the circle, then you also have all of those chords rotated at every fraction of a degree from 0 to 360; there’s a 1/2 chance before we rotate them, and rotating them doesn’t change anything so it’s still 1/2 also here are two more ways of choosing a chord: 1. you start with one point on the circumference of the circle, draw that triangle again with one vertex on that point, then you draw a second point anywhere within that circle. the probability then turns out to be about 0.61 (exact answer of (3 sqrt(3) + 2 pi)/6 pi) 2. you choose two random points within the circle, connect them, then extend that segment until it reaches the circumference of the circle (i suspect this method will come out to be 1/2, but i have no idea how to compute it)

[@vibaj16]

For your second way of choosing a chord, idk how to solve it either, but I wrote a program to check it (the way 3b1b did for the other methods) and it seems like the probability is 3/4.

[@mickschilder3633]

I think i prefer the two dots on the circle with symmetry of moving the dots over the symmetry of translating the circle, because it is less degenerate. With this i mean, when translating the circle is your symmetry and you apply it to a cord, you can not guarantee the image if that cord will still be a cord. So the symmetry is not even well defined. With the symmetry of rotating the second point, the only edge case you get is the two points being the same. Hiwever, not only does this edge case have probability 0, but you can also solve it by saying a tangent line of the circle is the cord defined by two identical points. TLDR: the solution of introducing symmetry by translating the circle does not define an action in cords, whilst the symmetry of rotation the second point does, therefore i’d venture to argue the latter is more canonical.

[@cubing7276]

I don’t get it with the tangent thing

[@mickschilder3633]

@@cubing7276 so if the two points agree, the cord you take is the tangent line to the crircle at that point. So the length of the cord would be 0

[@chriskondak1867]

Really liked Grant's perspective on the question: "What does this paradox make you feel"! Great conversation. I do think that the 2/3 1/3 method does more intuitively model the possible set of chords. If you realize only a small set of possible lines cross the center, but every chord passes the edge, you would expect the distribution to be progressively "more covered" at the edges than at the center! At the risk of repeating someone else' idea, I have a fourth suggestion: what if you chose a random point on the circle, and a random angle from 0 to 180deg? Almost 2/3 of the chords will be less than 60deg or greater than 120deg. Zero degrees is obviously not a chord, and both 60 and 120 deg are equal to the triangle chord, so the edge cases make it slightly less than 2/3. I've been thinking about this one all day!

[@Zxv975]

This scenario is the same as the very first case. The two omitted scenarios you listed have measure 0 so excluding them doesn't make it slightly less. You have probability 0 of landing on exactly 0° or 180°.

[@kevinbruce5981]

I have to agree, ultimately the definition of a cord is only a line with both ends on the circle. Thus this 1/3 method seems to be the right one to me the other methods use secondary properties that are as they are due to the interplay of definitions. Thus if we are talking an even distribution it should be of the primary definition we should not expect even distribution when looking at secondary properties.

[@zacfubar1]

@@kevinbruce5981 Except that it is clearly not a uniform distribution when plotted empirically. And, as mentioned, the term "pick a random X" is almost always taken to imply a uniformly random "X". The issue is that the meaning of a "uniformly random chord" is ambiguous, but I think that when viewed empirically, anyone would be hard-pressed to justify that the 1/3 method is the one that arrives at uniform randomness.

[@kevinbruce5981]

@@zacfubar1 seems uniform to me

[@zacfubar1]

@@kevinbruce5981 Empirically, there is a gaping hole in the middle of the circle though, which is clearly not spatially uniform. That's the whole point of the video though: you need that transitive property to arrive at spatial uniformity. There is not as pleasing of a physical meaning to the end points of the chord being uniform on the edge of the circle, because the edge of the circle isn't the only thing that exists.

[@timseguine2]

Even in Jaynes' view, the answer can be affected by the priors. That's the entire point of Bayesian analysis, which he seemed to miss. The "unambiguous" answer of 1/2 only arises from Jaynes' interpretation of what the priors are. The translation invariance property is a reasonable prior belief but it is by no means baked into the problem.

[@richardpike8748]

Grant is actually so good at coming up with counterexamples and explanations to Brady's questions off-the-cuff. So good. Especially with his analogy to "I have this selection of random dice, what's the probability I roll a 5?" (Brady frustrated at 16:49, Grant's analogy at 17:12, Brady frustrated again at 17:41 lol). That really made it click with me. It's like, 'oh, we don't actually have a well-defined starting point here. What lines are we dealing with? Just like what dice are we dealing with?'

[@brycehuff]

I love this explanation! Reminds me of the Hitchhiker’s Guide “42” How you define the parameters of the question impacts the answer.

[@kylebowles9820]

I always think of information degrees of freedom, the circle and the chord has 2 degrees of freedom, so all generated lines must uniformly explore the OUTPUT space on the circle in position and angle, otherwise you only sample a subset of the problem space, or a non-uniform sampling density

[@FManga18]

So the 1/" method is the best one? You are chosing the first and second component independetly

[@jakobr_]

You’re right, but, the choice of what variables you assign those degrees of freedom to is (at first) arbitrary. Imagine trying to make a uniform distribution of points across 2D space. You have to decide whether to use polar or rectangular coordinates. In polar coordinates, you choose one distance (r) and one angle (theta) to give your degrees of freedom to. In rectangular coordinates, you assign those degrees of freedom to two distance variables (x, y). If you make a uniform distribution using polar coordinates, and take a bird’s eye view of the plane, you’ll notice a disproportionate clump of points around the origin and you might say “hey this isn’t right”. If you make a uniform distribution of points using rectangular coordinates, then look from inside the plane (at the origin), you’ll see a bias towards further distances, and you might again say “this isn’t right”. In the chord problem, if you define the “output” space by two variables corresponding to the positions of two points, you get the first answer 1/3, which is biased against chords that are close to the center. If you define the “output” space by seeing all the ways a single line can intersect a circle (by translating and rotating it, one position variable and one angle variable), you get the third answer, 1/2, which is biased against chords that intersect two nearby points of the circle. The answer is really not as clear-cut as you might think.

[@cadekachelmeier7251]

All 3 methods have 2 degrees of freedom. The first one can be generated by two random angles. The second one can be generated by an x/y coordinate pair and throwing out the ones outside the circle. The third one can be generated by an angle and a number between 0 and r.

[@donaldasayers]

3 degrees of freedom for the line, position of a point on the line x,y and orientation.

[@mjeffery]

@@donaldasayers That's only 2 degrees of freedom: Once you pick an orientation (first degree), you can find every line by sliding perpendicularly to it (second degree). Another way to look at it is that picking an X, Y, and orientation picks not only a line, but also a point on that line. The choice of that point is one extra degree of freedom, but since you didn't need that point, you can throw it away with the extra degree of freedom. So you used 3 - 1 = 2 effective degrees to pick the line.

[@wktodd]

There's an infinite number of lines greater than and an infinite number shorter, so obviously the probability is minus one twelfth of the sqaure root of an apple tree.

[@jonathanshott6571]

I'm no mathematician, but would this method give a definitive answer: Consider a unit circle centred on the origin of an X-Y coordinate system. Any possible chord on said circle can be translated - by rotation - into an equivalent chord that is parallel to the X-axis and intersects the Y-axis somewhere between Y=0 and Y=1. Therefore any possible chord on the unit circle can be defined by an appropriate angle of rotation and a corresponding intersection point on the Y-axis between Y=0 and Y=1. In the context of Bertrand's Paradox we are only interested in the length of any given chord. So we don't need to worry about the angle of rotation because it will not affect the length of the chord. That means any possible chord on the unit circle can now be completely defined by only one parameter, its intersection point on the Y-axis between Y=0 and Y=1. When an equilateral triangle is inscribed into a unit circle it can similarly be rotated so that one of its sides is parallel to the X-axis and intersects the Y-access between Y=0 and Y=1, just like our set of all possible chords. We can use trigonometry to deduce that the point at which said side of the equilateral triangle intersects the Y-axis is at Y=1/2. We have shown that all possible chords, once rotated so as to be parallel to the X-axis, can be completely defined by their intersection point on the Y-axis. If a chord intersects the Y-axis above Y=1/2 then we know it must be shorter than the side length of the equilateral triangle. Similarly, if a chord intersects the Y-axis below Y=1/2 then we know it must be longer than the side length of the equilateral triangle. Therefore P(l > s) = 1/2

[@rainzhao2000]

You have just rephrased the 3rd method presented in the previous video. Can you see how?

[@whydontiknowthat]

@@rainzhao2000 yeah but I only understood it until this guy explained it.

[@vitorhugotokenshiambrosio795]

The whole paradox thing is that there is no definition of what makes a chord random. Say there is a big cilyndrical silo in the middle of a firefight, for whatever reason, in which there is no way of knowing where the bullets are coming from (but they are moving perpendicular to the height of the cylinder). Because of all the grain inside the silo, the bullets that hit the silo will have enough energy to leave ONLY if they cross less than sqrt(3)*the radius of the silo. What is the proportion of bullets that will stop inside the silo? 1/2. That is the 3rd method. Now say that 2 people will sit around a circular table at a completely random place around it's edge to eat. If the two people can only pass plates to each other if they are less than sqrt(3)*the radius of the table away from each other, how likely are they to be able to do so? 2/3. That's the 1st method. If we defined the problem in those ways, it is not that hard to come up with answers, but the thing is that both questions can be fairly described as "say there is a random chord in a circle".

[@jonathanshott6571]

@@vitorhugotokenshiambrosio795 My thinking was that to calculate probabilities accurately we need to identify the set of all possible chords on the unit circle. We shouldn't make any assumptions about 'how' to select a chord at random. With a pack of playing cards, for example, we wouldn't make any assumptions about 'how' a card is selected at random from the pack. We know there are 52 different cards in the pack, so the probability of obtaining any one particular card is 1/52. The problem with chords on the unit circle is that there are infinitely many, so how do we count them all? Well, since we only need to know the probability P(l > s) we don't need to know how many chords there are in total. We just need a way to partition the space of all possible chords into those where l > s and those where l < s. This is why I based my reasoning around the third method shown in the videos, as pointed out by Rain. When considering the first two methods, it was not intuitively clear to me whether the processes involved where definitely covering the set of all possible chords on the unit circle. Maybe they were, but it wasn't obvious, I couldn't convince myself. But with the third method it 'did' seem clear to me that all possible chords we being considered. All chords are intersected at their mid-point at a 90° angle by the radius of the circle. That to me seems intuitively true for any possible chord - I feel sure that it is true for the set of all chords. Then, by the trigonometric arguments provided by Grant in the videos, we know that any given chord will only have length l > s if its intersection point with the radius is less than 1/2 from the centre of the circle. This is true irrespective of the angle of the radius that intersects the chord. This seems to me to give a way of partitioning the set of all possible chords into two distinct subsets: those that intersect a perpendicular radius at distance < 1/2 from the centre of the circle and those that do not. Whilst we still cannot count the total number of possible chords in each subset - because there are infinitely many - it seems to me that each subset must be exactly equivalent in size. This is because one subset will correspond to the real numbers between 0 and 1/2, and the other subset will correspond to the real numbers between 1/2 and 1. This line of reasoning would seem to reduce the problem of P(l > s) to the equivalent problem of 'what is the probability that a real number between 0 and 1, chosen at random, will be less than 1/2'.

[@legendgames128]

Indexing system for chords: Ordinals. Use countably infinite numbers, and extend to infinity, then when you run out of every possible natural ordinal, start again with .1 added to it, and repeat.

[@bayleev7494]

this video has always been amazing! a quick note on grant's points re. symmetries determining uniform distributions (in light of a measure theory course i'm taking): the technical concept he's referring to here is something called the haar measure. basically, you can prove for any compact lie group G that there is a unique "reasonable" measure that does not care about the action of G. in more detail, you can assign a positive number to each (reasonable) subset of G, in such a way that the number doesn't change when you take g*(everything) for some g in G. thus, if you have a symmetry on a set X which is transitive, you can transfer that measure over to X and use it to define a probability. if you're interested in learning more, definitely look into measure theory! it's one of those subjects which deconstructs your preconceptions about maths and makes you feel like you understand everything a bit more deeply.