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

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

@@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.

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

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

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."

“That’s a great question” -grant

@@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."

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

**AN* answer

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.

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

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

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!

@@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.

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

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.

Grant speaks on that topic at 15:15 i think

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

@@samuela-aegisdottir 7x6 😉

@@samuela-aegisdottir The Ultimate Question!

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

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

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.

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

He he. Tele prompter goes brrrr. Seriously though impressive

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

Watch his interview to Lex Fridman, then.

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

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.

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.

it's still edited

This guy is brilliant, this was a perfect analogy

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.

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

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

@@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

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

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

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

yes, Its always pleasure to spectate really smart people arguing

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.

"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".

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.

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.

You have a Book of Lazlo Meró on this subject.

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

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.

@@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!

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

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

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

true

100%

Please explain...

​@@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

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.

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.

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 ;)

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

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

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!

@@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.

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.

@@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? 🤓

Mathematics is nothing more than the hunt for interesting tautologies.

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

And his saucy voice too.

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.

ftw

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

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.

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

It fills you with determination!

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

I feel like there should be a word for this.

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

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.

Me too 😀

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?

@@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.)

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

All of mathematics is an extension of logic.

Thus was born the quantum computer

@@marsulgumapu2010 Or not. Until someone observes it.

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

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.

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

@@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.)

I wonder what the symmetry there is

@@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

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

Actually Grant has an extensive colllection of D4 dice.

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

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

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

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 :)

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.

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.

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

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

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?

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.

@@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.

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

@@Lemon_Inspector So "." is unambiguous?

@@tim40gabby25 Yes.

@@Lemon_Inspector Interesting...

@@tim40gabby25 No.

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.

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.

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.

@@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.

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.

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.

"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).

@@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.

@@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.

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)

@@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.

quantum woo 😂

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

@@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).

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.

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?

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

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.

Yup, makes sense

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.

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).

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.

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

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."

@@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.)

@@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?

@@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".

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)

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

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.

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

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

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

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.

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.

@@bendotc great explanation

@@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.

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

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.

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.

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.

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

fantastic question

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.

Yup.

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.

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

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.

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.

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.

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

"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.

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

@@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.

@@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?

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

@@VAFFANFEDE18 They're equivalent

every net has a convergent subnet

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.

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 :)].

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

The issue and question seems to end up being: where are you evenly distributing the chords? Your method gives a uniform distribution across the area of the circle, but if you look at your rotationally-locked example it seems the endpoints of the chords aren't evenly distributed on the perimeter of the circle. So which of these is more important to keep uniform?

@@shadefangkweep in that arrangement, every point on the circle is an endpoint of a single chord. Its perfectly even.

@@PopeGoliath That doesn't actually follow; we are dealing with infinities, so the density can vary. Like how every number between 0 and 1 is exactly one value, but has a different probability in a gaussian distribution.