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

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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 ...
3blue1brown: / @3blue1brown
Grant Sanderson on The Numberphile Podcast: • The Hope Diamond (with...
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20 дек 2021

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Комментарии : 2,3 тыс.   
@jajohnek
@jajohnek 2 года назад
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 года назад
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 года назад
“That’s a great question” -grant
@IceMetalPunk
@IceMetalPunk 2 года назад
@@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 года назад
@@linuskerr P != NP proved! by way of a Ruger 57
@damamoot2291
@damamoot2291 Год назад
**AN* answer
@toughnerd
@toughnerd 2 года назад
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 года назад
That literally had me laughing out loud. ...and I'm a trucker.
@gerrybaker7055
@gerrybaker7055 2 года назад
@@jamaljohnson2414 how does being a trucker change your ability/willingnrss to laugh though?
@jamaljohnson2414
@jamaljohnson2414 2 года назад
@@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 года назад
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 года назад
His complete deadpan seriousness at considering the problem had me in stitches. Easily my new all-time favorite numberphile moment.
@iamnorwegian
@iamnorwegian 2 года назад
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 года назад
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 года назад
Rewatching that part after reading this comment was the funniest thing ever
@martijn3151
@martijn3151 2 года назад
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 года назад
IKR, I was completely in awe of how quickly he could find the flaw and point it out so clearly
@SilverLining1
@SilverLining1 2 года назад
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 Год назад
@@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 Год назад
I love Grant tbh. I wish there were more people like him in the world.
@jamesonskalinski6910
@jamesonskalinski6910 7 дней назад
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 года назад
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 года назад
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 года назад
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 года назад
ANd no ums and arhs, either. Very unusual. Agree wtih you compltely.
@pawelkorzeniewski4897
@pawelkorzeniewski4897 2 года назад
He he. Tele prompter goes brrrr. Seriously though impressive
@DynestiGTI
@DynestiGTI 2 года назад
The more you know about the topic, the easier it is to talk about it.
@dojelnotmyrealname4018
@dojelnotmyrealname4018 2 года назад
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 года назад
Grant speaks on that topic at 15:15 i think
@samuela-aegisdottir
@samuela-aegisdottir 2 года назад
"What's the answer?" "42" "What's the question then?"
@ismailshtewi8560
@ismailshtewi8560 2 года назад
@@samuela-aegisdottir 7x6 😉
@666Tomato666
@666Tomato666 2 года назад
@@samuela-aegisdottir The Ultimate Question!
@mjj29
@mjj29 2 года назад
@@ismailshtewi8560 Canonically the question is 'what is 6 times 9'
@Salchipapafied
@Salchipapafied 2 года назад
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 года назад
Watch his interview to Lex Fridman, then.
@Daiwie44
@Daiwie44 2 года назад
I am just disappointed he doesn't actually have the eyes of his logo
@paulomartins1008
@paulomartins1008 2 года назад
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 Год назад
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 Год назад
it's still edited
@BowlOfRed
@BowlOfRed 2 года назад
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 года назад
I agree. The original video was awesome, and this one was even better.
@kevinsmith9385
@kevinsmith9385 2 года назад
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 года назад
omg yes, I was dying to hear whether or not he'd mention infinite chords.
@tyranneous
@tyranneous 2 года назад
Grant's on-the-fly analogy of the multi-sided die really got me there with this paradox. Absolutely nailed it!
@lb5299
@lb5299 2 года назад
This guy is brilliant, this was a perfect analogy
@Wheat_Thinn
@Wheat_Thinn 2 года назад
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 года назад
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 года назад
@@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 года назад
@@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 года назад
Actually got chills when Grant so quickly and eloquently explained the failings of Brady’s proposed method.
@finnwestergren8670
@finnwestergren8670 2 года назад
yeah it's inspiring. as someone who is so bad at making sense I really admire this.
@phmfob
@phmfob 2 года назад
Him explaining the random dice analogy felt like I was back in school ignorantly asking a silly question
@peterhaugen887
@peterhaugen887 2 года назад
@@phmfob you mean picking the low hanging fruit on your knowledge tree?
@xenonchikmaxxx
@xenonchikmaxxx 2 года назад
yes, Its always pleasure to spectate really smart people arguing
@aapjeaaron
@aapjeaaron 2 года назад
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.
@JaapvanderVelde
@JaapvanderVelde 2 года назад
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.
@realchrisward
@realchrisward 2 года назад
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 9 месяцев назад
Please explain...
@highviewbarbell
@highviewbarbell 9 месяцев назад
​@@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 5 месяцев назад
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 2 месяца назад
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 года назад
"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 года назад
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 года назад
@@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 года назад
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 года назад
I think a whole lot of people are going to be doing that after watching this video. I know I will.
@SoleaGalilei
@SoleaGalilei 2 года назад
I am literally lying in bed at night thinking about Bertrand's Paradox right now!
@Henrix1998
@Henrix1998 2 года назад
This should be part 2 on the main channel, too important to miss
@mahxylim7983
@mahxylim7983 2 года назад
true
@LeonardEisen
@LeonardEisen 2 года назад
100%
@jimsmedley234
@jimsmedley234 2 года назад
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 года назад
"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 года назад
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 года назад
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 года назад
You have a Book of Lazlo Meró on this subject.
@mibber121
@mibber121 2 года назад
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
@MrDiscoTube
@MrDiscoTube 2 года назад
I love the way Grant smiles throughout - he is having such fun, and so are we!
@macronencer
@macronencer 2 года назад
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 года назад
If it's not him then I'd say Ben Eater is another great communicator.
@otonanoC
@otonanoC 2 года назад
And his saucy voice too.
@krish4288
@krish4288 2 года назад
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 года назад
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 года назад
@@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 года назад
Mathematics is nothing more than the hunt for interesting tautologies.
@dominicveconi4301
@dominicveconi4301 2 года назад
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 года назад
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 года назад
ftw
@daniellambert6207
@daniellambert6207 2 года назад
I love how Brady pokes to inject emotion into the discussion. A true master of the craft. 15:03, 21:05
@rainzhao2000
@rainzhao2000 2 года назад
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.
@yashrawat9409
@yashrawat9409 2 года назад
Following it up as soon as I finished the first video
@supu8599
@supu8599 2 года назад
Me too 😀
@JP.Q
@JP.Q 2 года назад
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 года назад
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 года назад
If all the chords are vertical, wouldn’t that mean you have a preferred direction?
@thecommexokid
@thecommexokid 2 года назад
@@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 года назад
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 года назад
@@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 года назад
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 года назад
"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 года назад
It fills you with determination!
@MikkelHojbak
@MikkelHojbak 2 года назад
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 года назад
I feel like there should be a word for this.
@zachrodan7543
@zachrodan7543 2 года назад
@@JohnMeacham If it were up to me... "mathocistic pleasure?"
@Merthalophor
@Merthalophor 2 года назад
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.
@proxyprox
@proxyprox 2 года назад
21:09 This is the real proof by intimidation
@oshuao414
@oshuao414 2 года назад
Grant's reaction after Brady saying "draw every possible chord, each one is numbered... just wonderful
@verdigris9742
@verdigris9742 2 года назад
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 года назад
"Infinity comes with its own little first-aid kit." That's a great quote!
@asamenechbayissa553
@asamenechbayissa553 2 года назад
"Probability is the extension of logic" This line will stay with me
@digitig
@digitig 2 года назад
All of mathematics is an extension of logic.
@marsulgumapu2010
@marsulgumapu2010 2 года назад
Thus was born the quantum computer
@digitig
@digitig 2 года назад
@@marsulgumapu2010 Or not. Until someone observes it.
@alxjones
@alxjones 2 года назад
There's actually a formalization of this exact idea called fuzzy logic.
@Joshman0131
@Joshman0131 2 года назад
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
@ericbright1742
@ericbright1742 2 года назад
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 года назад
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 года назад
@@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.)
@frankjohnson123
@frankjohnson123 2 года назад
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 года назад
also it's impressive how quick he can reasonably answer Brady's questions being thrown at him
@Lemon_Inspector
@Lemon_Inspector 2 года назад
Alternative title: The 7 Stages of Jaynes "I don't like Jaynes. His smug aura mocks me." ... "So in conclusion, I agree."
@Yupppi
@Yupppi Год назад
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 Год назад
You'll probably like Grant's videos on his channel...
@borisgrozev2289
@borisgrozev2289 2 года назад
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.
@rustymustard7798
@rustymustard7798 2 года назад
"Choose a random chord." Define choose, random, and chord.
@Lemon_Inspector
@Lemon_Inspector 2 года назад
Define "define", "and", "a", and ","
@tim40gabby25
@tim40gabby25 2 года назад
@@Lemon_Inspector So "." is unambiguous?
@Lemon_Inspector
@Lemon_Inspector 2 года назад
@@tim40gabby25 Yes.
@tim40gabby25
@tim40gabby25 2 года назад
@@Lemon_Inspector Interesting...
@Lemon_Inspector
@Lemon_Inspector 2 года назад
@@tim40gabby25 No.
@zygoloid
@zygoloid 2 года назад
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 года назад
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 года назад
@@sheridanvespo1269 I presume it’s the same as what he gives
@columbus8myhw
@columbus8myhw 2 года назад
@@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 года назад
I wonder what the symmetry there is
@addy9961
@addy9961 2 года назад
@@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
@kaidenschmidt157
@kaidenschmidt157 2 года назад
Brady: “I’m gonna burn your house down” Grant: “ok fair question”
@jukmifggugghposer
@jukmifggugghposer 5 месяцев назад
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.
@jounik
@jounik 2 года назад
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 года назад
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 года назад
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 года назад
"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 года назад
@@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 года назад
@@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.
@linkspring1287
@linkspring1287 2 года назад
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 года назад
Actually Grant has an extensive colllection of D4 dice.
@QuantumConundrum
@QuantumConundrum 2 года назад
@@andrewharrison8436 that is a genius level trap for a street performer right there.
@joemonster55
@joemonster55 8 месяцев назад
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.
@aajpeter
@aajpeter 2 года назад
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.
@momerathe
@momerathe 2 года назад
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 года назад
We have a transitive symmetry in the points on a sphere case. That method is not invariant over that transitive symmetry.
@adelarscheidt
@adelarscheidt 2 года назад
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 года назад
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 года назад
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 года назад
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.
@stevescott2819
@stevescott2819 2 года назад
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 года назад
Thank you, phrased like that it does sound much more natural!
@Aleksandr011
@Aleksandr011 2 года назад
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 года назад
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 года назад
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 года назад
@@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.
@zacfubar1
@zacfubar1 2 года назад
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 года назад
I love the Bayesian approach (I did my doctorate on it) but by itself doesn't solve Bertant's "paradox".
@garychou9621
@garychou9621 4 месяца назад
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.
@wiseSYW
@wiseSYW 2 года назад
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 года назад
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 года назад
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 Год назад
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.
@gleedads
@gleedads 2 года назад
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 года назад
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.
@protocol6
@protocol6 2 года назад
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 года назад
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 года назад
@@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 года назад
quantum woo 😂
@recompile
@recompile 2 года назад
Never listen to anyone who uses the term 'woo'. They're easily suckered by con-men like James Randi.
@jessejordache1869
@jessejordache1869 2 года назад
@@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).
@eranbernstein7972
@eranbernstein7972 2 года назад
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 :)
@russellthorburn9297
@russellthorburn9297 Год назад
I was somewhat lost until 17:32 and the analogy with the dice. Well done Grant.
@jacemandt
@jacemandt 2 года назад
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 года назад
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 года назад
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 года назад
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 года назад
@@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.
@fangjiunnewe3634
@fangjiunnewe3634 2 года назад
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 года назад
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.
@charlieandfedericabeattie3432
I’ve watched this pair of videos three times. Love them.
@panchajanya91
@panchajanya91 2 года назад
I always enjoy listening to Grant explaining something, anything and going deeper and deeper.
@Nikception
@Nikception 2 года назад
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 года назад
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).
@ten.seconds
@ten.seconds 2 года назад
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 года назад
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 года назад
Yeah, the 2 choices exactly line up with the 2 symmetries. Makes sense. Cool.
@glenmatthes8839
@glenmatthes8839 2 года назад
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 года назад
Yup, makes sense
@jamesflames6987
@jamesflames6987 2 года назад
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.
@qo92
@qo92 27 дней назад
Brady asking the best questions, Grant giving the best answers. Top notch!
@AlphahawkA25
@AlphahawkA25 7 месяцев назад
“IM GOING TO BURN YOUR HOUSE DOWN” “That’s a good question”
@kasuha
@kasuha 2 года назад
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 года назад
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 года назад
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 года назад
@@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 года назад
@@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 года назад
@@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".
@TheBasikShow
@TheBasikShow 2 года назад
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 года назад
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 года назад
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 года назад
@@bendotc great explanation
@rafaelmarkos4489
@rafaelmarkos4489 Год назад
@@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.
@Mythobeast
@Mythobeast 2 года назад
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 года назад
Great addition to the main video right here. Thanks. Much appreciated.
@vitorbortolin6810
@vitorbortolin6810 2 года назад
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 года назад
It's not unsolved. It's undefined. You need to define WHAT random means. You have to define the distribution.
@marsulgumapu2010
@marsulgumapu2010 2 года назад
I'd say there's a 50/50 chance it will be solved.
@ricobarth
@ricobarth 2 года назад
@@cparks1000000 The problem is how do define it, and that problem remains unsolved.
@timseguine2
@timseguine2 2 года назад
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.
@yeet3673
@yeet3673 2 года назад
This was one of the best Numberphile vids.. and there have been a loooot of great ones, so that's saying something
@limbridk
@limbridk 2 года назад
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
@reaganshonk
@reaganshonk 2 года назад
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 года назад
But remember that we are working in R^2 and so compact means closed and bounded
@HilbertXVI
@HilbertXVI 2 года назад
@@VAFFANFEDE18 They're equivalent
@98danielray
@98danielray 2 года назад
every net has a convergent subnet
@rianantony
@rianantony Год назад
This is an old discussion, but I'd like to add that if you picked a set of paralel cords, you can say pretty confidently that half of them are longer than the lengh of the side of the triangle. And because a set of paralel cords could be made with any orientation, it would make sense that any one of them would be representative of the whole. Edit: a day later I return to add. If random means that you have a flat distribution of every possible lengh of cord, than that means that you could pick one of each possible lengh of cord and that set of cords would be representative of the whole. One of each possible lengh of cord, if arranged, parallel to each other, in order can be shaped into a circle, first one half in ascending order until the diameter and then the rest in descending order. Now that we have arranged a representative set of every cord in order we could draw the triangle with one side parallel to them and see that exactly half of them are longer than the sides (with the same trigonometry in the video). I'm not saying anything new (idk if this has already been said in the videos and I forgor💀). But this is my take either way, if I had to pick an answer with a gun to my had it'd be this one.
@RajaAnbazhagan
@RajaAnbazhagan Год назад
Yup.
@salsa123freak567
@salsa123freak567 2 года назад
Unbelievably great discussion with on the spot answers that are awe inspiring. Great, great respect for your intelectual prowess, Grant!
@gusshultz5342
@gusshultz5342 2 года назад
I'm not even a maths guy, but your enthusiasm and clarity makes this a joy to watch. Subscribed!
@ntucker2377
@ntucker2377 2 года назад
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.
@thomaswalters7117
@thomaswalters7117 2 года назад
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.
@reluginbuhl
@reluginbuhl 2 года назад
And Grant's voice is so wonderful to listen to! It just sounds so relaxing, warm and friendly :)
@SoleaGalilei
@SoleaGalilei 2 года назад
This was wonderful. Brady asks the best questions and Grant gives the most fascinating and cogent answers.
@cheshire1
@cheshire1 2 года назад
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 года назад
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 года назад
Well said. I was a bit confused about that part and your interpretation helped a lot.
@travismyers3396
@travismyers3396 2 года назад
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 года назад
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 года назад
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.
@dhonors999
@dhonors999 2 года назад
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 ;)
@matthewwadhwa2582
@matthewwadhwa2582 2 года назад
Really enjoyed this vid. Thank you for creating and sharing!
@rinner2801
@rinner2801 2 года назад
Excellent. I'm also well into the 1/2 camp since I've become so fascinated with symmetry in Mathematics.
@Tumbolisu
@Tumbolisu 2 года назад
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 года назад
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 года назад
"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 года назад
The idea that "lines outside the circle don't exist" just feels artificial to me. Why shouldn't they exist?
@jpdemer5
@jpdemer5 2 года назад
@@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 года назад
@@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?
@robindebreuil
@robindebreuil 2 года назад
Great videos :). From the perspective of programming, it seems to me this is one of those problems that seems fully defined but actually isn't when you go to code it. When you say chord, there are two points, and it is important to know you are talking about the probability of the second point once the first is fixed, or two random points. The 1/3 probability is akin to asking what the average distance is to all points of the circumference from a fixed point (rotating the inner triangle is the equivalent of fixing the point). The 1/2 probability is akin to asking what the average height (bottom to top of a vertical line) of the circle is, which will be the same in any orientation. I feel the 1/4 answer is the same thing, but asking about the average height of half a circle. In this way it reminds me of the Boy or Girl Probability Paradox from that Zach Star video. Once you 'fix' one of the two children the probabilities change.
@PTNLemay
@PTNLemay Месяц назад
I love these kind of discussions. Digging into the philosophy oh math
@Ojisan642
@Ojisan642 2 года назад
Brady you ask such great questions that make your guests think really differently about these problems and also make me laugh
@wktodd
@wktodd 2 года назад
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.
@jorgechavesfilho
@jorgechavesfilho 2 года назад
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 года назад
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 года назад
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 года назад
This is true. There is a paper showing that, but I cannot find it now :(
@beartankoperator7950
@beartankoperator7950 2 года назад
fantastic question
@Aaron-th7xx
@Aaron-th7xx 2 года назад
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.
@peterwolf8092
@peterwolf8092 2 года назад
Chills! Literally chills!
@ForsakenDAemon
@ForsakenDAemon 2 года назад
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
@IllidanS4
@IllidanS4 2 года назад
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.
@ClydeCoulter
@ClydeCoulter 2 года назад
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 года назад
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 года назад
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 года назад
Isn't that technically just the same as the third method?
@Thrashenizer
@Thrashenizer 2 года назад
I can think of very few people who injected as much peace of mind into the _soul-tormenting_ shadow of mathematical uncertainty! Thanks to you both, Brady & Grant, truly ...
@heaslyben
@heaslyben 2 года назад
These were very interesting and satisfying and clear! Thank you!
@mickschilder3633
@mickschilder3633 2 года назад
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 года назад
I don’t get it with the tangent thing
@mickschilder3633
@mickschilder3633 2 года назад
@@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
@WarHero56
@WarHero56 2 года назад
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 Год назад
"I hereby induct thee into the Church of Bayes"
@aaronherman1056
@aaronherman1056 3 месяца назад
Seeing the two circles next to each other really made it clear to me why the radial point solution satisfied the constraint of being preserved in translation, because I could imagine an infinite grid and see how uniform distribution along the radius was analogous with uniform distribution along one of the axes of the grid.
@thelocalsage
@thelocalsage 2 года назад
the connection of enumerating uncountable sets with selecting a probability distribution was insane, like a nucleation point where the whole rest of the content could crystallize. grant is a brilliant communicator
@djspacewhale
@djspacewhale 2 года назад
for me, the ambiguity here is that we didn't name the distribution to start out - there's many different measures we could give to the space of chords through the circle, and once we name one the question loses its ambiguity. also: lol the probability of choosing a given chord is 50/50 bc you either choose it or you don't
@possessedchair8144
@possessedchair8144 2 года назад
That’s the classical solution, aka the question isn’t well-posed. The alternate viewpoint insists that the question is well-posed and that there is this unique concept of uniformity in picking a chord at random.
@columbus8myhw
@columbus8myhw 2 года назад
@@possessedchair8144 That feels very "I don't know what you care about" vs "I know what _I_ care about". Jaynes doesn't seem to consider that other people might care about different things.
@G12GilbertProduction
@G12GilbertProduction 2 года назад
This is looks like a emotional fallacy. Positive seemings it's not in subjective meaning of topic really objective, pessimistic seemings it's not in objective meaning of topic really subjective what there are etc.
@jerryhu9005
@jerryhu9005 Год назад
That sounds like saying probability of hitting the bullseye is 50/50. You either hit it or you don't
@kylebowles9820
@kylebowles9820 2 года назад
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 года назад
So the 1/" method is the best one? You are chosing the first and second component independetly
@jakobr_
@jakobr_ 2 года назад
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 года назад
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 года назад
3 degrees of freedom for the line, position of a point on the line x,y and orientation.
@mjeffery
@mjeffery 2 года назад
@@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.
@shpensive
@shpensive 2 года назад
Great episode!
@Vacuon
@Vacuon 2 года назад
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.
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