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

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  • เผยแพร่เมื่อ 20 ธ.ค. 2021
  • This extra footage follows the main 10-minute video at: • Bertrand's Paradox (wi... (watch this first)
    More links & stuff in full description below ↓↓↓
    3blue1brown video on the shadow a cube: • A tale of two problem ...
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    Grant Sanderson on The Numberphile Podcast: • The Hope Diamond (with...
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ความคิดเห็น • 2.3K

  • @toughnerd
    @toughnerd 2 ปีที่แล้ว +1110

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

    • @jamaljohnson2414
      @jamaljohnson2414 2 ปีที่แล้ว +36

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

    • @gerrybaker7055
      @gerrybaker7055 2 ปีที่แล้ว +42

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

    • @jamaljohnson2414
      @jamaljohnson2414 2 ปีที่แล้ว +22

      @@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
      @minikretz1 2 ปีที่แล้ว +23

      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
      @snickers10m 2 ปีที่แล้ว +20

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

  • @jajohnek
    @jajohnek 2 ปีที่แล้ว +2541

    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
      @linuskerr 2 ปีที่แล้ว +227

      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
      @alkylperchlorate388 2 ปีที่แล้ว +104

      “That’s a great question” -grant

    • @IceMetalPunk
      @IceMetalPunk 2 ปีที่แล้ว +105

      @@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
      @heartles_xyz 2 ปีที่แล้ว +24

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

    • @damamoot2291
      @damamoot2291 ปีที่แล้ว +4

      **AN* answer

  • @iamnorwegian
    @iamnorwegian 2 ปีที่แล้ว +508

    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
      @ThePondermatic 2 ปีที่แล้ว +52

      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
      @sender1496 2 ปีที่แล้ว +20

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

  • @martijn3151
    @martijn3151 2 ปีที่แล้ว +574

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

    • @sujals7108
      @sujals7108 2 ปีที่แล้ว +31

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

    • @SilverLining1
      @SilverLining1 2 ปีที่แล้ว +59

      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
      @Tony-cm8lg ปีที่แล้ว +14

      @@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
      @elidrissii ปีที่แล้ว +2

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

    • @jamesonskalinski6910
      @jamesonskalinski6910 หลายเดือนก่อน

      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.

  • @Rubrickety
    @Rubrickety 2 ปีที่แล้ว +2809

    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
      @TheGreatAtario 2 ปีที่แล้ว +51

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

    • @stephenbeck7222
      @stephenbeck7222 2 ปีที่แล้ว +58

      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
      @starpawsy 2 ปีที่แล้ว +27

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

    • @pawelkorzeniewski4897
      @pawelkorzeniewski4897 2 ปีที่แล้ว +8

      He he. Tele prompter goes brrrr.
      Seriously though impressive

    • @DynestiGTI
      @DynestiGTI 2 ปีที่แล้ว +11

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

  • @dojelnotmyrealname4018
    @dojelnotmyrealname4018 2 ปีที่แล้ว +1056

    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
      @nahblue 2 ปีที่แล้ว +12

      Grant speaks on that topic at 15:15 i think

    • @samuela-aegisdottir
      @samuela-aegisdottir 2 ปีที่แล้ว +87

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

    • @ismailshtewi8560
      @ismailshtewi8560 2 ปีที่แล้ว +12

      @@samuela-aegisdottir 7x6 😉

    • @666Tomato666
      @666Tomato666 2 ปีที่แล้ว +3

      @@samuela-aegisdottir The Ultimate Question!

    • @mjj29
      @mjj29 2 ปีที่แล้ว +11

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

  • @Salchipapafied
    @Salchipapafied 2 ปีที่แล้ว +834

    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
      @andreybashkin9030 2 ปีที่แล้ว +6

      Watch his interview to Lex Fridman, then.

    • @Daiwie44
      @Daiwie44 2 ปีที่แล้ว +5

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

    • @paulomartins1008
      @paulomartins1008 2 ปีที่แล้ว +8

      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
      @miscellany_media 2 ปีที่แล้ว +8

      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
      @Fred-tz7hs ปีที่แล้ว +1

      it's still edited

  • @JaapvanderVelde
    @JaapvanderVelde 2 ปีที่แล้ว +104

    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.

  • @tyranneous
    @tyranneous 2 ปีที่แล้ว +1494

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

    • @lb5299
      @lb5299 2 ปีที่แล้ว +62

      This guy is brilliant, this was a perfect analogy

    • @Wheat_Thinn
      @Wheat_Thinn 2 ปีที่แล้ว +34

      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
      @Wheat_Thinn 2 ปีที่แล้ว +7

      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
      @nathanepstein4079 2 ปีที่แล้ว +4

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

    • @Wheat_Thinn
      @Wheat_Thinn 2 ปีที่แล้ว +9

      @@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

  • @willhastings731
    @willhastings731 2 ปีที่แล้ว +1685

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

    • @finnwestergren8670
      @finnwestergren8670 2 ปีที่แล้ว +71

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

    • @phmfob
      @phmfob 2 ปีที่แล้ว +152

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

    • @peterhaugen887
      @peterhaugen887 2 ปีที่แล้ว +9

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

    • @xenonchikmaxxx
      @xenonchikmaxxx 2 ปีที่แล้ว +27

      yes, Its always pleasure to spectate really smart people arguing

    • @aapjeaaron
      @aapjeaaron 2 ปีที่แล้ว +10

      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.

  • @jimsmedley234
    @jimsmedley234 2 ปีที่แล้ว +246

    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
      @Idorise 2 ปีที่แล้ว +9

      "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
      @LeeClemmer 2 ปีที่แล้ว +21

      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
      @aenetanthony 2 ปีที่แล้ว +16

      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
      @Idorise 2 ปีที่แล้ว

      You have a Book of Lazlo Meró on this subject.

    • @mibber121
      @mibber121 2 ปีที่แล้ว +10

      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

  • @realchrisward
    @realchrisward 2 ปีที่แล้ว +151

    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
      @6612770 11 หลายเดือนก่อน +2

      Please explain...

    • @highviewbarbell
      @highviewbarbell 10 หลายเดือนก่อน +14

      ​@@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
      @jukmifggugghposer 6 หลายเดือนก่อน +3

      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
      @FuzzyLogic0 3 หลายเดือนก่อน +1

      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.

  • @jvcmarc
    @jvcmarc 2 ปีที่แล้ว +385

    "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
      @PetWolverine 2 ปีที่แล้ว +50

      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
      @madisondampier3389 2 ปีที่แล้ว +20

      @@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
      @johanlopez655 2 ปีที่แล้ว +10

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

    • @JohnMeacham
      @JohnMeacham 2 ปีที่แล้ว +7

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

    • @SoleaGalilei
      @SoleaGalilei 2 ปีที่แล้ว +4

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

  • @Henrix1998
    @Henrix1998 2 ปีที่แล้ว +375

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

  • @dominicveconi4301
    @dominicveconi4301 2 ปีที่แล้ว +86

    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
      @bonezman5 2 ปีที่แล้ว +12

      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
      @LK90512 2 ปีที่แล้ว +1

      ftw

  • @MrDiscoTube
    @MrDiscoTube 2 ปีที่แล้ว +127

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

  • @macronencer
    @macronencer 2 ปีที่แล้ว +664

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

    • @skilz8098
      @skilz8098 2 ปีที่แล้ว +8

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

    • @otonanoC
      @otonanoC 2 ปีที่แล้ว

      And his saucy voice too.

  • @krish4288
    @krish4288 2 ปีที่แล้ว +106

    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
      @alynames7171 2 ปีที่แล้ว +5

      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
      @KT-dj4iy 2 ปีที่แล้ว +3

      @@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
      @ExplosiveBrohoof 2 ปีที่แล้ว +3

      Mathematics is nothing more than the hunt for interesting tautologies.

  • @daniellambert6207
    @daniellambert6207 2 ปีที่แล้ว +40

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

  • @oshuao414
    @oshuao414 2 ปีที่แล้ว +55

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

  • @BowlOfRed
    @BowlOfRed 2 ปีที่แล้ว +663

    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
      @vell0cet517 2 ปีที่แล้ว +9

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

    • @kevinsmith9385
      @kevinsmith9385 2 ปีที่แล้ว +8

      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
      @sam5992 2 ปีที่แล้ว +3

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

  • @rainzhao2000
    @rainzhao2000 2 ปีที่แล้ว +128

    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.

  • @JP.Q
    @JP.Q 2 ปีที่แล้ว +234

    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
      @Sarimae23 2 ปีที่แล้ว +3

      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
      @jameswood8021 2 ปีที่แล้ว +1

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

    • @thecommexokid
      @thecommexokid 2 ปีที่แล้ว +28

      @@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
      @victorquesada7530 2 ปีที่แล้ว +2

      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
      @watchm4ker 2 ปีที่แล้ว +17

      @@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
    @malcolmw513 2 ปีที่แล้ว +15

    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.

  • @zachrodan7543
    @zachrodan7543 2 ปีที่แล้ว +237

    "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
      @madisondampier3389 2 ปีที่แล้ว +14

      It fills you with determination!

    •  2 ปีที่แล้ว +1

      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
      @JohnMeacham 2 ปีที่แล้ว

      I feel like there should be a word for this.

    • @zachrodan7543
      @zachrodan7543 2 ปีที่แล้ว +11

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

    • @Merthalophor
      @Merthalophor 2 ปีที่แล้ว

      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.

  • @yashrawat9409
    @yashrawat9409 2 ปีที่แล้ว +30

    Following it up as soon as I finished the first video

    • @supu8599
      @supu8599 2 ปีที่แล้ว +2

      Me too 😀

  • @joemonster55
    @joemonster55 9 หลายเดือนก่อน +2

    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.

  • @verdigris9742
    @verdigris9742 2 ปีที่แล้ว +5

    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!

  • @columbus8myhw
    @columbus8myhw 2 ปีที่แล้ว +22

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

  • @proxyprox
    @proxyprox 2 ปีที่แล้ว +10

    21:09 This is the real proof by intimidation

  • @ericbright1742
    @ericbright1742 2 ปีที่แล้ว +14

    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
      @TristanCunhasprofile 2 ปีที่แล้ว +2

      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
      @dominicveconi4301 2 ปีที่แล้ว +2

      @@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.)

  • @bleikeze
    @bleikeze 2 ปีที่แล้ว +5

    "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]?"

  • @frankjohnson123
    @frankjohnson123 2 ปีที่แล้ว +54

    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_
      @_ilsegugio_ 2 ปีที่แล้ว +4

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

  • @asamenechbayissa553
    @asamenechbayissa553 2 ปีที่แล้ว +49

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

    • @digitig
      @digitig 2 ปีที่แล้ว +1

      All of mathematics is an extension of logic.

    • @marsulgumapu2010
      @marsulgumapu2010 2 ปีที่แล้ว

      Thus was born the quantum computer

    • @digitig
      @digitig 2 ปีที่แล้ว

      @@marsulgumapu2010 Or not. Until someone observes it.

    • @alxjones
      @alxjones 2 ปีที่แล้ว

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

  • @Joshman0131
    @Joshman0131 2 ปีที่แล้ว +8

    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

  • @Yupppi
    @Yupppi ปีที่แล้ว +8

    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
      @soymilkassassin ปีที่แล้ว +4

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

  • @Lemon_Inspector
    @Lemon_Inspector 2 ปีที่แล้ว +9

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

  • @zygoloid
    @zygoloid 2 ปีที่แล้ว +126

    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
      @sheridanvespo1269 2 ปีที่แล้ว +13

      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
      @matthewhubka6350 2 ปีที่แล้ว +1

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

    • @columbus8myhw
      @columbus8myhw 2 ปีที่แล้ว +28

      @@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
      @ancientswordrage 2 ปีที่แล้ว +3

      I wonder what the symmetry there is

    • @addy9961
      @addy9961 2 ปีที่แล้ว +1

      @@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

  • @borisgrozev2289
    @borisgrozev2289 2 ปีที่แล้ว +22

    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.

  • @aajpeter
    @aajpeter 2 ปีที่แล้ว +6

    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.

  • @linkspring1287
    @linkspring1287 2 ปีที่แล้ว +33

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

    • @andrewharrison8436
      @andrewharrison8436 2 ปีที่แล้ว +7

      Actually Grant has an extensive colllection of D4 dice.

    • @QuantumConundrum
      @QuantumConundrum 2 ปีที่แล้ว

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

  • @momerathe
    @momerathe 2 ปีที่แล้ว +207

    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
      @huckthatdish 2 ปีที่แล้ว +29

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

    • @adelarscheidt
      @adelarscheidt 2 ปีที่แล้ว +6

      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
      @finn8518 2 ปีที่แล้ว +19

      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
      @lonestarr1490 2 ปีที่แล้ว +6

      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
      @evanrosenlieb8819 2 ปีที่แล้ว +6

      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.

  • @zacfubar1
    @zacfubar1 2 ปีที่แล้ว +7

    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
      @Evan490BC 2 ปีที่แล้ว

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

    • @garychou9621
      @garychou9621 6 หลายเดือนก่อน

      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.

  • @Klarpimier
    @Klarpimier 9 หลายเดือนก่อน +1

    “IM GOING TO BURN YOUR HOUSE DOWN”
    “That’s a good question”

  • @jacemandt
    @jacemandt 2 ปีที่แล้ว +39

    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
      @LucenProject 2 ปีที่แล้ว +1

      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
      @timothybexon6171 2 ปีที่แล้ว

      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.

    • @DavidGuild
      @DavidGuild 2 ปีที่แล้ว +1

      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
      @bendotc 2 ปีที่แล้ว

      @@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.

  • @rustymustard7798
    @rustymustard7798 2 ปีที่แล้ว +9

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

    • @Lemon_Inspector
      @Lemon_Inspector 2 ปีที่แล้ว +2

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

    • @tim40gabby25
      @tim40gabby25 2 ปีที่แล้ว

      @@Lemon_Inspector So "." is unambiguous?

    • @Lemon_Inspector
      @Lemon_Inspector 2 ปีที่แล้ว

      @@tim40gabby25 Yes.

    • @tim40gabby25
      @tim40gabby25 2 ปีที่แล้ว

      @@Lemon_Inspector Interesting...

    • @Lemon_Inspector
      @Lemon_Inspector 2 ปีที่แล้ว

      @@tim40gabby25 No.

  • @jukmifggugghposer
    @jukmifggugghposer 6 หลายเดือนก่อน +3

    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.

  • @eranbernstein7972
    @eranbernstein7972 2 ปีที่แล้ว +15

    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 :)

  • @jounik
    @jounik 2 ปีที่แล้ว +55

    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
      @iurigrang 2 ปีที่แล้ว +4

      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
      @pxlated2366 2 ปีที่แล้ว

      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
      @buttonasas 2 ปีที่แล้ว +1

      "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
      @jounik 2 ปีที่แล้ว

      @@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
      @buttonasas 2 ปีที่แล้ว

      @@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.

  • @stevescott2819
    @stevescott2819 2 ปีที่แล้ว +103

    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
      @fruitshuit 2 ปีที่แล้ว +2

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

    • @Aleksandr011
      @Aleksandr011 2 ปีที่แล้ว +1

      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
      @randomnobody660 2 ปีที่แล้ว

      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
      @fwiffo 2 ปีที่แล้ว +5

      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
      @Aleksandr011 2 ปีที่แล้ว +4

      @@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.

  • @wiseSYW
    @wiseSYW 2 ปีที่แล้ว +13

    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
      @dougaltolan3017 2 ปีที่แล้ว +3

      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
      @jezennalakomne 2 ปีที่แล้ว

      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
      @artsmith1347 2 ปีที่แล้ว

      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.

  • @russellthorburn9297
    @russellthorburn9297 ปีที่แล้ว +1

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

  • @protocol6
    @protocol6 2 ปีที่แล้ว +80

    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
      @hens0w 2 ปีที่แล้ว

      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
      @protocol6 2 ปีที่แล้ว +10

      @@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
      @lachlancooke 2 ปีที่แล้ว +6

      quantum woo 😂

    • @recompile
      @recompile 2 ปีที่แล้ว +3

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

    • @jessejordache1869
      @jessejordache1869 2 ปีที่แล้ว

      @@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).

  • @gleedads
    @gleedads 2 ปีที่แล้ว +19

    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
      @ricobarth 2 ปีที่แล้ว

      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.

  • @qo92
    @qo92 2 หลายเดือนก่อน

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

  • @kaidenschmidt157
    @kaidenschmidt157 2 ปีที่แล้ว +2

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

  • @ten.seconds
    @ten.seconds 2 ปีที่แล้ว +25

    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
      @okuno54 2 ปีที่แล้ว +2

      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
      @JohnMeacham 2 ปีที่แล้ว +1

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

    • @glenmatthes8839
      @glenmatthes8839 2 ปีที่แล้ว +3

      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
      @LeonardEisen 2 ปีที่แล้ว +1

      Yup, makes sense

    • @jamesflames6987
      @jamesflames6987 2 ปีที่แล้ว

      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.

  • @kasuha
    @kasuha 2 ปีที่แล้ว +51

    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.

    • @VAFFANFEDE18
      @VAFFANFEDE18 2 ปีที่แล้ว

      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
      @blacklistnr1 2 ปีที่แล้ว +6

      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
      @Varksterable 2 ปีที่แล้ว +2

      @@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
      @frechjo 2 ปีที่แล้ว

      @@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
      @blacklistnr1 2 ปีที่แล้ว +1

      @@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".

  • @Vacuon
    @Vacuon 2 ปีที่แล้ว +2

    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.

  • @Kenneally151
    @Kenneally151 2 ปีที่แล้ว +1

    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

  • @fangjiunnewe3634
    @fangjiunnewe3634 2 ปีที่แล้ว +17

    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
      @nestoreleuteriopaivabendo5415 2 ปีที่แล้ว

      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.

  • @vitorbortolin6810
    @vitorbortolin6810 2 ปีที่แล้ว +71

    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
      @cparks1000000 2 ปีที่แล้ว +23

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

    • @marsulgumapu2010
      @marsulgumapu2010 2 ปีที่แล้ว +8

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

    • @ricobarth
      @ricobarth 2 ปีที่แล้ว

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

  • @rianantony
    @rianantony ปีที่แล้ว +8

    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.

  • @bencheevers6693
    @bencheevers6693 2 ปีที่แล้ว

    When grant said ascribing your numbering set is choosing it all clicked for me, I loved Brady for saying that though because it felt so right and I was right there with him

  • @Nikception
    @Nikception 2 ปีที่แล้ว +7

    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
      @andrewharrison8436 2 ปีที่แล้ว +1

      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).

  • @timseguine2
    @timseguine2 2 ปีที่แล้ว +21

    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.

  • @brycehuff
    @brycehuff 2 หลายเดือนก่อน

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

  • @umbertorodrigez8213
    @umbertorodrigez8213 2 ปีที่แล้ว +1

    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.

  • @TheBasikShow
    @TheBasikShow 2 ปีที่แล้ว +25

    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
      @ricobarth 2 ปีที่แล้ว +3

      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
      @bendotc 2 ปีที่แล้ว +5

      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
      @joeeeee8738 2 ปีที่แล้ว

      @@bendotc great explanation

    • @rafaelmarkos4489
      @rafaelmarkos4489 2 ปีที่แล้ว +1

      @@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.

  • @jorgechavesfilho
    @jorgechavesfilho 2 ปีที่แล้ว +142

    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
      @franciscofernandez8183 2 ปีที่แล้ว +51

      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
      @Bovineprogrammer 2 ปีที่แล้ว +23

      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
      @mzg147 2 ปีที่แล้ว +7

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

    • @beartankoperator7950
      @beartankoperator7950 2 ปีที่แล้ว +1

      fantastic question

    • @Aaron-th7xx
      @Aaron-th7xx 2 ปีที่แล้ว +9

      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.

  • @panchajanya91
    @panchajanya91 2 ปีที่แล้ว

    I always enjoy listening to Grant explaining something, anything and going deeper and deeper.

  • @gusshultz5342
    @gusshultz5342 2 ปีที่แล้ว

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

  • @cheshire1
    @cheshire1 2 ปีที่แล้ว +114

    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
      @beetehotraroy3468 2 ปีที่แล้ว +18

      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.

    • @user-ks6qr9qw7h
      @user-ks6qr9qw7h 2 ปีที่แล้ว

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

    • @travismyers3396
      @travismyers3396 2 ปีที่แล้ว +3

      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
      @sidsixseven 2 ปีที่แล้ว +3

      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
      @strehlow 2 ปีที่แล้ว +2

      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.

  • @wktodd
    @wktodd 2 ปีที่แล้ว +6

    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.

  • @SoleaGalilei
    @SoleaGalilei 2 ปีที่แล้ว

    This was wonderful. Brady asks the best questions and Grant gives the most fascinating and cogent answers.

  • @charlieandfedericabeattie3432
    @charlieandfedericabeattie3432 ปีที่แล้ว

    I’ve watched this pair of videos three times. Love them.

  • @Tumbolisu
    @Tumbolisu 2 ปีที่แล้ว +40

    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
      @passerby4507 2 ปีที่แล้ว +8

      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
      @W0X42A 2 ปีที่แล้ว +12

      "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
      @Tumbolisu 2 ปีที่แล้ว +2

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

    • @jpdemer5
      @jpdemer5 2 ปีที่แล้ว +7

      @@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
      @passerby4507 2 ปีที่แล้ว +2

      @@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?

  • @thomaswalters7117
    @thomaswalters7117 2 ปีที่แล้ว +7

    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.

  • @s.avtarsingh7590
    @s.avtarsingh7590 ปีที่แล้ว

    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
      @s.avtarsingh7590 ปีที่แล้ว

      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

  • @reluginbuhl
    @reluginbuhl 2 ปีที่แล้ว

    And Grant's voice is so wonderful to listen to! It just sounds so relaxing, warm and friendly :)

  • @dhonors999
    @dhonors999 2 ปีที่แล้ว +7

    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 ;)

  • @reaganshonk
    @reaganshonk 2 ปีที่แล้ว +30

    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.

    • @VAFFANFEDE18
      @VAFFANFEDE18 2 ปีที่แล้ว +3

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

    • @HilbertXVI
      @HilbertXVI 2 ปีที่แล้ว +3

      @@VAFFANFEDE18 They're equivalent

    • @98danielray
      @98danielray 2 ปีที่แล้ว

      every net has a convergent subnet

  • @limbridk
    @limbridk 2 ปีที่แล้ว +2

    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

  • @eaglegosuperskarmor
    @eaglegosuperskarmor 2 ปีที่แล้ว +2

    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.

  • @ntucker2377
    @ntucker2377 2 ปีที่แล้ว +4

    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.

  • @IllidanS4
    @IllidanS4 2 ปีที่แล้ว +8

    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.

  • @xenontesla122
    @xenontesla122 2 ปีที่แล้ว +1

    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%.

  • @yeet3673
    @yeet3673 2 ปีที่แล้ว +1

    This was one of the best Numberphile vids.. and there have been a loooot of great ones, so that's saying something

  • @ClydeCoulter
    @ClydeCoulter 2 ปีที่แล้ว +6

    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
      @jacksonbrim7359 2 ปีที่แล้ว

      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
      @martinhall1188 2 ปีที่แล้ว

      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
      @tb45g 2 ปีที่แล้ว

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

  • @mickschilder3633
    @mickschilder3633 2 ปีที่แล้ว +18

    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
      @cubing7276 2 ปีที่แล้ว

      I don’t get it with the tangent thing

    • @mickschilder3633
      @mickschilder3633 2 ปีที่แล้ว

      @@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

  • @salsa123freak567
    @salsa123freak567 2 ปีที่แล้ว

    Unbelievably great discussion with on the spot answers that are awe inspiring. Great, great respect for your intelectual prowess, Grant!

  • @richardpike8748
    @richardpike8748 2 ปีที่แล้ว +1

    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?'

  • @kylebowles9820
    @kylebowles9820 2 ปีที่แล้ว +18

    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

    • @VAFFANFEDE18
      @VAFFANFEDE18 2 ปีที่แล้ว

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

    • @jakobr_
      @jakobr_ 2 ปีที่แล้ว +7

      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
      @cadekachelmeier7251 2 ปีที่แล้ว +3

      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
      @donaldasayers 2 ปีที่แล้ว

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

    • @mjeffery
      @mjeffery 2 ปีที่แล้ว

      @@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.

  • @KoHaGames_
    @KoHaGames_ 2 ปีที่แล้ว +10

    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
      @colinree6131 2 ปีที่แล้ว

      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
      @stargazer7644 2 ปีที่แล้ว

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

    • @KoHaGames_
      @KoHaGames_ 2 ปีที่แล้ว

      @@stargazer7644 how do you mean?

    • @BryantHarrisonCville
      @BryantHarrisonCville 2 ปีที่แล้ว

      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
      @stargazer7644 2 ปีที่แล้ว +1

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

  • @Mythobeast
    @Mythobeast 2 ปีที่แล้ว +1

    This is really a question of "what problem are you trying to solve?"
    - If you're looking for the distance tunneled through the earth between two random destinations, then the first answer is correct.
    - If you're looking for the shortest straight line that passes through a destination inside the earth, the second is correct.
    - If you're trying to figure out the path length of neutrinos passing through the planet, then the third one is accurate.
    This is where many engineers fail -- they don't drill down to figure out the problem that's being solved.

  • @IwoIwanov
    @IwoIwanov 2 ปีที่แล้ว

    Great addition to the main video right here. Thanks. Much appreciated.

  • @jurajorsulic6032
    @jurajorsulic6032 2 ปีที่แล้ว +18

    There is something else very unsettling about this not mentioned: choosing a random chord sounds like a contrived problem, but it is actually equivalent to choosing a point in the circle, as the second method shows.
    Depending on the method chosen, we can get different densities of random points in a circle.

    • @griffine6111
      @griffine6111 2 ปีที่แล้ว +2

      I believe choosing a random point on a disk is still well defined and perfectly clear - note how only the second method of choosing chords is equivalent to choosing a random point. The other methods give different answers for average chord length and therefore a different average point distance from the center. Choosing a random point in the disk is invariant under translation and I believe unambiguous

    • @jurajorsulic6032
      @jurajorsulic6032 2 ปีที่แล้ว +1

      @@griffine6111 true, it is obvious that the random points on a disk should have translation invariance, and therefore be chosen using the appropriate method.
      However, if you’re not careful, you can end up naively using a seemingly logical and fair method of picking random points, and end up with what corresponds to one of the other two chord picking methods.

  • @WarHero56
    @WarHero56 2 ปีที่แล้ว +21

    1/2 by far makes the most sense here. I gotta agree with Jaynes. Method 1 and 2 of generating chords are not proper randomness, they are clearly biasing the chords.

    • @jerryhu9005
      @jerryhu9005 2 ปีที่แล้ว +3

      "I hereby induct thee into the Church of Bayes"

  • @MostlyIC
    @MostlyIC 2 ปีที่แล้ว +1

    an improperly defined question (no definition of a random chord) leads to chaos, no surprise there and nothing to do with statistics, but what I like about this is how it leads to an understanding of Bayesian verses Frequentist points of view, something I've had a hard time remembering since I didn't learn it in school.

    • @IsaacKuo
      @IsaacKuo 2 ปีที่แล้ว +1

      Here's my take on Frequentist vs Bayesian analysis:
      Frequentist analysis will give you an imprecise but correct answer (error bars with uncertainty level according to the limited number of samples).
      Bayesian analysis will give you a precise but wrong answer (a precise number which could be anything based on your priors, but will almost certainly be wrong).

    • @nickergodos1554
      @nickergodos1554 2 ปีที่แล้ว +1

      Yes, Bertrand's paradox is all about the philosophical interpretation of probability. It's strictly speaking not a math problem, or even a stat problem. It's a problem strictly within the philosophy of mathematics.

  • @ForsakenDAemon
    @ForsakenDAemon 2 ปีที่แล้ว

    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