How

Kuh! Spoopid "experts"! Fake News! Make Mathematics Great Again I say!

Three logicians walk into a bar. The bartender asks, "do you all want the special?". After a pause, the first logician says, "I don't know". Then the second logician says, "I don't know". Then the third logician says, "Yes".

@@AlZimmermann Ooh I like this one.

Al Zimmermann That's brilliant.org!

Fudmottin i mean, you are why there is never a solid answer to this question Mean anomaly. :p

LOL! I just couldn't resist this one. Now if you want relevant maths, how do you check a credit card number is valid? Hint. They match a checksum. This ties into the cryptography of things.

You don't have enough trials. You need many more. Once the wrapper breaks I suspect you will revert to the mean.

Some websites trust the checksum too much and you can checkout with mr jhon doe's 5555 5555 5555 5557.

The checksum is really meant to just catch typos. You check the number again on the back end before completing the order. It also reduced load on the rear end since the checksum is super fast to check.

Still waiting.

@@webrusheriii8934lol that’s crazy u just replied 15hrs ago

I think spinning coins might be the reason ...The sqrt(3) answer might work well for non-spinning coins. But if the coin spins and tilts, I think it rolls on a curved path like on a circle due to precession/spinning top physics. And then you get this centrifugal force effect, which works against the coin tilting and creates a bias towards falling on the side. So one possibly needs to decrease the area of the side and make the coin slimmer to raise the probabilty for it to fall on the two faces. Which would result in a required ratio bigger than sqrt(3) .... Just a speculation 🤔

that was possibly the nerdiest exchange I've ever seen I loved it

My favourite triangle is : 45° , 60° , 75° .

vihart

My favourite triangle has sides 5, 7, and 8 and has one 60 deg angle.

my fav triangle is the 3, 4, 5 right angle pythagorasly correct triangle

I like my triangles even, there is a mathematical beauty to them.

Wow that's mean (pun intended).

Man, I don't get any of these jokes

@Keynsie I have an A-level in math and still didn't get it.

I think jen calculated the average/mean of the 2 products

It works generically as a stats joke, but also specifically because Matt's coin was too thin, Hugh's coin was too thick, and Jen, the statistician says yeah, you got it someplace in the middle. Yes... the joke is about statisticians calculating the mean of a distribution and the common misinterpretation that it applies to a real world situation. *sigh*

Literally the most useless bit of input too ffs

It's a binomial and a normal distribution, everyone knows how they work.

Not everyone.

As an engineer, I'm sorry, as a male engineer, I'm real sorry. Lol.

Sometimes it's just to confirm that one is thinking the along the same lines.

I did a paper on this, but it's in Serbian. When exams are over, I will translate it and send to Matt.

@@DanijelAleksicMath Could you link the paper anyway?

@@4M4D3USM0Z4RT I posted a new comment with a link.

@@DanijelAleksicMath could you post the link again, I cant find your comment. thanks.

The square root of 2 plus 1 : 1

relatable, retweet

Noah Knox came to the comments just to see if anyone else had thought this

I smell some integrals here. I'm going to do some quick calculations

That's what I thought too.

so should the D/t ratio be 2 then?

Using this method I got 1/sqrt((2*sqrt(3)-3)/3) which is approx. 2.54 which is within the range indicated by their experiment

Jake It’s the calculus we all learned sneaking into our brain! On one side, P(“middle”) is near zero, on another side, P(“middle” is nearly 1, so one would hope that if it’s a continuous function whose inputs are widths and outputs are probabilities of edges, somewhere in the middle the IVT says we can get 1/3rd!

@@AlisterCountel Then it pulls a sneaky and there it's not continuous for some reason

@@gabemerritt3139 I don’t really have the statistical background to justify it, but the elongation of the width is a continuous function and calculating the probability of landing on each side should just be a matter of shape and density (which vary continuously with width). So there’s reason to assume that the P(width) is continuous

@@AlisterCountel I was mostly joking, it's likely a function of how you roll, the side edge to side ratio, center of mass, among other smaller factors. Each should be continuous so the resulting function should be continuous. I'd be surprised if density played a factor however.

geocarey interestingly the diameter method seems to be closer to the correct ratio than the sphere method

I like this idea a lot! I am not an expert in physics/Newtonian motion, so I went with another approach... why not just try to equalize surface area? On a die, all faces have the same surface area and the same chance of being rolled, so it should work for this as well, right?

Branching off, the rate at which mechanical energy leaves the system should be factored in. If the 'coin' has fully inelastic collisions and/or is being flipped in a viscous syrup opposed to air, it will lose nearly all its energy upon first contact with the ground and thus would not be able to flop to a new position from the one it landed on. In that case the sphere model might be more accurate. However, if it's the case that the coin is in a near vacuum and loses very little mechanical energy per contact, perhaps each face of the coin could better be modeled as a several potential wells in spherical coordinates, potential wells that correspond to the mechanical energy required to move the coin to a new position. Multiple potential wells seems to be a problem encountered in quantum mechanics and thus may already have a solution, but I can't make any sense of the math.

It is not true that a 6 sided dice will require the same energy to change a number. It will require the same amount of energy to change from 1 to a 2,3,4, or 5, but a different energy to change to a 6 since it needs enough energy to change sides twice.

I think what they meant was the energy to change from a 1 to not a 1 is the same is from 2 to not a 2 and so on

Is this not solved by the spherical model, though?

No. The spherical model says nothing about the initial conditions of the flip. Using the picture drawn on the chalkboard in the video, it is being assumed that coin is being flipped so that Y axis (going perpendicular through the chalkboard) is the axis of rotation. Now imagine what happens if you were to flip the coin along the X axis. You would effectively be spinning the coin along its "banded" edge, and it would be the ONLY side of the coin to ever make contact with the ground. You would have a 100% chance of getting an edge and 0% chance of landing on either of the disc shaped faces. The problem is, you can't ever perfectly flip a coin along the Y axis. The more "sloppy" your flip is, the more it deviates towards the X axis, increasing the odds of an edge landing. The issue has nothing to do with the model, the issue is that the model is predicated on an inaccurate assumption. Case in point, if the spherical model solved the problem, then the emprical evidence would have matched the data, but it didn't.

I think there's probably a way of finding functions in terms of angle, integrating and solving for the ratio between height and width.

Davy Jones, interesting; my own thoughts while watching the video were similar. The problem is almost trivial if you drop the "coin" straight down with 0 angular speed from an initially randomised orientation. (Although the "randomised orientation" itself is a non-trivial thing and needs to be considered carefully! For example, for computer simulation of a toss with *uniform* distribution you need to use spherical coordinates.) It gets ridiculously complex once you introduce flipping, because stuff like angular velocity, friction, moment of inertia start to matter. So not only you have to define the direction of the flipping, you also need to make assumptions about all those factors. What's the probability distribution for the velocity? etc. Since I'm not a mathematician (or a physicist), I'd love to see someone tackle the problem with a proper, rigorous approach, instead of this half-assed stuff.

_given random conditions_ dbaston530, please define "random conditions". That's the whole point I'm having difficulty with. If we want to only approach a solution experimentally, that's easy enough, and with enough repetitions (many more than in the video) we can have an arbitrarily high degree of certainty. We just have to be very strict about the experiment conditions - eg. have _x_ dice in a box of dimensions a*b*c, shake it by hand (which imposes a limit to stuff like momentum), then drop them from height _y_ . Those results would be useful and good-enough if you wanted to market those dice as a product, *but* also impossible to generalize. If you blindly grab a regular die and roll it only on the z-axis (looking at it from the side, I assume), 1 and 6 are excluded as much as 2 and 5, or 3 and 4 are excluded. It's still fair. But you cannot blindly grab a cylindrical die.

Biggest cliffhanger on KZfaq!!

@@Reptonious there's the enigma video of code bullet

How are you verified?

@@Zuiker1 he’s not it’s just an emoji

Take the diagonal cross-section. Make it so that the coin is balanced at a 45 degree angle, so that the coin is just as likely to go either direction.

As another chemical engineer student I can confirm that.

Finally the answer I need.

Tom Nicholson Yes, that‘s the reason why you can‘t just make the three faces have equal area.

This was my question exactly. I kept waiting for it to addressed. It seems the coins they tried have a higher center of gravity on edge than when the coin is sitting flat. and that would lead to less edges than predicted by their area calculation. But the theory wasn't strongly supported by the data. My guess would have been any area type calculation would have led to less sides than the area theories predicted and that wasn't the case. However, if the coins were flipped similar to the way a coin is normally flipped there might be a lower edge percentage than with the cup technique they used. Also materials might come in to play here. You might get one result with mostly elastic collisions and a different result with mostly inelastic collisions.

Forget it man, its a Parker problem now.

Parker square.

19 months.... Still waiting

22 months now? I couldn't possibly lose more sleep over something than I did with this

@@rewrose2838 add a day to the counter

I have to agree there would be lots more physics to consider. If the flip is non-random, then you have angular momentum/ gyroscopic effects to consider (is it flipped like a coin head over tails about a horizontal axis, about a vertical axis like a coin spinning on a table, or rolling like a wheel?) Then, even if you drop it on a perfectly flat and level gymnasium floor (no walls/ obstacles), what's your coefficient of friction between your chosen material and the floor? What if the edges are rounded with wear or manufactured for safety? It seems this is a different question than what cylinder has the same stability (lowest mechanical potential energy?) whether resting on face or edge (would that not be height equal to diameter, or is that complicated since more mass in the middle when on its side instead of evenly distributed when on a face?)

I think that if the coins were made out of Tungsten, there would be no edge wear. But I think that lego plastic should be sufficient based on very little edge wear happening each time due to the low weight of the piece relative to the strength of the piece.

You are just being pedantic since not even dice manufacturers have that level of scrutiny

Yes. I think there are quite a few other factors that could potentially play into things too like drop height, elasticity of material, type of surface.

Wouldn't flipping them introduce significantly more rotational momentum, making it more likely to fall over onto a face?

@Simon Read correct, it's just {ans}

I think that's where the root three ratio comes from. If the diameter is root three times the thickness, then when rotating the coin 180 degrees over its edge from the heads down position to the heads up position, for the first 60 degrees the centre of gravity will be over the heads face, for the second 60 degrees the c.o.g. will be over the side of the cylinder, and for the third 60 degrees the c.o.g. will be over the tails face.

TRUE, SOMETHIGN LIKE LEAD IF IT LANDED ON ITS SIDE, WOULD NOT BOUNCE CAUSING IT TO STAY ON ITS EDGE, SOMETHIGN LIGHT WOULD BEHAVE DIFFERANTLY

damn capsloc, too lazy to edit XD sorry im not yelling at all

Slight material variances at different parts of coin can cause interesting magnetization and, depending on surface, adjust the trajectory of coin too in a very complex way. Also, the stability of the landing surface (e.g. is it a concrete road or a table or maybe even a floating plank) also matters a lot. Yea, there are indeed many factors.

I suggest to make a physics simulation of this - using a standard computer physics engine (e.g. Havoc). Let the computer generate coins with steadily increasing thickness while it virtually flips (care must be taken here so it is as random as possible) each generated coins some million times and records the results. Once you have a coin that is a near 1/3 chance of landing on either of its three sides - you got the "correct" coin. Then produce a physical coin of identical dimensions and do real life tests to verify simulation results.

I feel like this is a problem that has no perfect solution. A standard 6-, 12-, or 20-sided dice is going to behave in a very predicable way because the way it "lands" on a particular number is the same for every number, plus you have very symmetrical moments of inertia. This disc coin is likely to have a very different moment of inertia for one axis than for an axis that is perpendicular to that axis. Thus the "randomness" of a roll is immediately called into question. I would also suspect that the "bounciness" of the dice and table materials are going to play a role (pun?) in how it behaves. A pure physics solution accounting for center of mass and trigonometry and angular momentum and point of impact, factoring all of those things in, would still only be ideal for a perfectly non-bouncy surface. Because a bounce changes the axis of rotation, and because the moments of inertia for each axis are so different, and because you've got a mix of right angles and curved surfaces, I feel that the bounce changes the ideal radius to length ratio. And I haven't yet touched on another point of concern: the nature of the rolling environment. The scientists wanted to avoid having the dice roll off the table, so they made a wall to contain them. The problem is, if a dice is rolling on its edge on a limitless surface, it will eventually come to rest on its edge due to friction. But if it hits a wall, there is a chance it will topple over. The wall therefore creates a bias against edge results that isn't present when there is no wall. I don't fault them for not coming up with an answer because, as I'm sure you'd agree, there is none. But I do wish they had been more thorough in the discussion of just how complicated this question is.

When I was a kid some sixty years ago, I flipped a nickel into the air and it landed on its edge and rolled up against the base of a floor lamp; it never did fall over. I suspect this is more likely with a nickel than with other coins owing ti its relative thickness. What is the likelihood given thickness as a function od diameter?

I also flipped a penny a thousand times and was surprised I got exactly 500 heads.

Exactly. Their solution only works if the coins "freeze" in place as soon as they touch the table. A real coin will fall over one way or the other, changing the probabilities.

yshwgth that is what's trying to figure out. At what ratio is the probability of it landing on the edge in a stable equilibrium equal to the probability of it landing on a specific face or the edge so as to collapse onto said face.

Yes, I feel like they should be thinking about the three angles (or two since it's a cylinder) that it forms with the ground at the landing and make intervals of angles for each of the three coin results. All of this while assuming that the coins wont bounce, of course.

Yeah, and I think the moment of inertia is going to be a factor too. There's a LOT going on in this problem

i wonder if the ratio will depend on the exact dynamics of the throws, like the initial momentum and elasticity of the materials involved.

"In reality, as soon as you have some horizontal velocity, the edge is far less likely due to the center of mass being higher up." --yeah, that's putting into precise terms what I was thinking all along :) When it lands on the sides it does not stand up, but when it lands on the edge it can fall down. So this has to be calculated in, making it less a problem of geometry and more one of mechanics -- about centre of mass and velocity, just as you said. That sounds like a fun task for a computer simulation of newtonian mechanics, eh?

Even if it has no horizontal velocity, there will be a similar effect from it bouncing on the edges. If it lands with either face being closer to down, it is most likely going to fall onto that face, but if it lands with the side close to down, it will be pushed into a rotation, and there is a high chance that the bounce from the table impact will give it enough time to spin so that one of the faces is down. This is not to mention rotational motion it may have before impact, even falling perfectly vertically. In the basic principle, the first idea of calculating the edge ratio in 3D space is more correct - while all axial rotations are technically equivalent, there are still more of those equivalent states for the side-landing case, when considering the steradians represented by an edge landing versus a face landing. However all the other points mentioned here make the edge far less likely, due to being so high up causing it to balance poorly, which is why the thicker coin resulted in being closer to 1/3 ratios despite the principle of calculating using the 2D profile being incorrect. Fundamentally dice work by not only having equal geometry, but also all sides being the same energy state - they all have the same vertical centre of mass, resulting in them all being equally desirable to land in, whereas the three-sided coin must contend both with geometry and energy states, not just one of the two. This is why a d3 dice is (usually) just a triangular prism.

I guess one has to assume that there is equal velocity in rolling as well as spinning. Now, if the distribution should be like half of them rolls and the other half rolls (or it should be random if they spin or roll), or if all coins should have some roll and some spin idk

No, we're looking for an answer that we can experimentally validate, so assuming that will do no good. It will only result in a more complicated version of the 2*sqrt(2) and the sqrt(3) models. Another hurdle would be the bounciness of the material, giving these "coins" more chances to hit the ground with a flat side after bouncing, and thus more chance to stabilize on this flat side (let's note that the effect of landing on the slice and rolling will also give birth to some chances to hit a border and come down flat) So even if we find a ratio that is experimentally verified, there is a great chance that it would be only true for the material used and that a more bouncy or less bouncy material would required another diameter/width ratio. Thus, the answer to the problem is depending on many more variables that just the ratio, the primary ones being: -elasticity of the coin material -friction parameter of the coin material upon the table material - the topology of the table (infinite? borders to bounce on?) - sharpness of the coin borders? tl;dr: No obvious and analytical solution if we want something that actually works in a default D&D session

In the real world, the edge was the most likely result with the thicker sample. Did you even watch the video?

typically Numberphile

"Pythagorus"? Never heard of that. But that sounds suspiciously like a math guy named Pythagoras.

Or calculating any distance ever

Nope, @@noahdoss1967. Not for things that require pi, such as circles and sines, etc., and relatively small measurements that just take simple unit conversions, etc.

To get the wrong answer? 😀

yeah I had the same philosophy. It is good that you also ended up on one third because it is about the ratios between the flat side and the edge. But that is linear anyway so your approach is kind of the same.

Well, to be fair; A common way to solve math problems is to just start running numbers and then looking at the data after-the-fact to see if you can figure out the why in reverse. So this is totally the first half of a mathematician's answer to the question. Unfortunately, they didn't finish (yet?)

Avatar pic checks out

The Parker square meme is a Parker square of a meme

I think this problem depends too heavily on the the physics of the throw and the surface it bounces on to get a solution that always works. It may be fair in one set of parameters, but change anything, and it wouldn't be fair.

The only way to get equal probabilities is to rely on symmetry, which you can't use here. There is almost certainly no unique solution, it will depend on all sorts of physical parameters. That's also why the "5-sided die" is a really bad idea.

Maybe it will become the million dollar problem of the 2020s

@@bwayagnes2452 more like a 100 thousand dollar problem

Actually, if using sphere, try edge length of 3.39694097682 to 1. For circle, try 1.59868289539 to 1. I think the circle one seems more right, but I haven't yet been at the pub long enough to get the courage (or 3d printer) to try it. If anybody tries it, tell me so I can cry at your ability to afford a 3d printer.

@@jiralishu Might give this a go tomorrow. Not sure a 3d printer exists which is 16bit precise tho.

@@jiralishu Why?

Yep. Languages are weird

Matt believe that even single piece of dice should be called dice, even it's technically a die. I think "dice" means "a dice", but "die" means "the die", that's some meaningful explanation for this language intuition; it makes sense for me

a dice sounds very strange to me

So you could say "I've thrown six dice and five of them are all 2, but look at this particular 4 die! It's totally some Parker square die."

now im just confused

Well...if you think about what he did with his car, it makes sense...

Marrying a witch as an engineer is your ticket to joblessness

He likes muggles so bad that he finished Hogwarts and went to Cambridge so he could learn how they manage to handle nowadays situation ( without using magic, of course )

@@johanncmelo i feel accomplished for understanding harry potter jokes, i've only gotten into it over the past 2 months

Ron Weasley's dad is Father Brown.

Not really. A d3 (at least, all the ones i've ever seen) has 3 equal sides of shapes and size and angles and whatnot. The Parker-D3 has different side shapes, which is an entirely different problem, and much more difficult that cutting a cylinder into a 3 sided shape, and rounding the edges.

take a D6 and have all numbers have 2 sides.

Canine Kitten The *proper* D3. About time.

meanwhile DnD players invented a perfect D3 decades ago by rolling a D6 and either taking mod 3 (so (1,4) (2,5) (3,6) give the same result) or dividing by 2 rounding up (so (1,2) (3,4) (5,6) give the same result) or flipping the dice over if it shows 4+ (so (1,6) (2,5) (3,4) give the same result) or the good old catch-all method to make any size dice smaller than an available dice: if you roll a number higher than what would be available on the dice you want, reroll. repeat until you roll a number available on the dice you want. (for example if you have a D4 but want a D3, anytime you roll a 4 on the D4, just reroll until you roll a 1-3)

I own a D3. Its just a Prism of a regular triangle and pyramids added to the triangular faces.

So are you upset, or happy?

I understand what they are talking about however when it comes to doing that stuff myself, I find it very difficult for some reason. I tend to easily get confused in a way unlike with any other type of maths problem...

dude noice!

my reaction is just plug the number into the graphical calculator instead of calculating it

Uuuuuuuh... None of that made any sense.

Lol. Practical application.

That makes alot of sense, considering a perfect 3 sided coin would have to be essentially unstable. Just as likely to fall on a side as an edge.

Oscar Mulin yes please

Oscar Mulin so If the height of the coin is the diameter/sqrt(2) you should have a pretty even 3 sided dice.

How did you randomise the nature of your flips.

According to the results of the video the ratio should be between 2*sqrt(2) and sqrt(3) but sqrt(2) is way beyond that range. I think there's something wrong with your simulation.

Technically not, since there is an infinitely small amount of time during which the coin has landed either on heads, tails, or the side. You should say: "...will destroy your timeline _after_ any coin lands on tails or heads..." and then claim that "every non-destructive flip would result in a coin on its side."

If the universe is governed by hard determinism, that isn't an experiment I would want to perform.

If the universe is governed by hard determinism, you don't really have any choice in whether to perform that experiment, do you?

True, but then you have no choice about whether you would want to, either.

+David Conrad That depends on your definition of ‘choice’. From a subjective perspective, that is, one where you don't have a view of the entire universe past, present, and future, you could say you have a choice even under hard determinism. The unknown variables could be considered, in a sense, your degrees of freedom.

This! I was excited to watch this video because I never learned about how the momentum of spinning irregular objects works on impact. I thought I'd learn about it here. :(

There's a video where they did this on a computer simulation and elasticity, mass and initial momentum all changed how thick the coin needs to be.

I can't wait until the new IDO (Initial DiceCoin Offering)

I thought this said dogecoin and you were rich.

so many more factors than sphere and circle

Orin Johnso

2 sqrt2 is about 2.83 and sqrt3 is about1.73. Wouldn’t it be hilarious if the physically tested answer turned out to pretty much be dead on 2.0?

Orin Johnson this was the first thing i thought of after reading the video title. Good on you. Glad to see your work and that the result seems to be between the high/low prob of the sphere and circle methods.

Equal areas for each "side" is what I originally thought this would be, so why did they rule this probability out?

I thought this too, but the ratios they used are in terms of the diameter, not the radius. In terms of the radius √3/2 < r/h < √2 or 0.866 < r/h < 1.414

Brute force/trial and error is the only option. Problem is too chaotic.

Yes, his non parker-head.

HAHAHA

Well... Not so much

I think it's hilarious his bald spot looks like a gigantic part in his hair

Or it was recorded later and his hair grows really fast.

Iacta alea est

The pedant in me is _dicing_ to explain the difference to you!

shocka, julius caesar spoke english?

Please do, I think it's interesting to learn that.

singular dice is just awkward dont use it please

Honestly, that would be the biggest news for me in the past 22 months

@@rewrose2838 Someone made a video where he found the answer was silver ratio. (1 + sqrt(2))

why? i was thinking, 'why don't they just use a cube die, then mark 2 sides with an x, mark another 2 sides with an o and leave the other 2 sides blank? then there are 3 outcomes with a cube die. (shrug)

rictus grin At least somebody has my back

The singular for dice is also dice, according to the dictionary. That said, both are acceptable, although I don't often hear people using "die" irl

If so many people have used dice incorrectly to mean a single die, then the dictionary will accept this as the new normal word. It also gets seen as pedantic to correct someone from 'a dice' to 'a die'. It's still rude to call someone wrong for using the originally correct definition as Matt did here. If the word dice is allowed as singular now, then there are just two acceptable words. Officially the most annoying thing I've seen Matt do in any video.

Lol.

Nicholas Kam My dictionary (Oxford English) lists "dice" as plural of "die", noting that "die" is less common. But I wouldn't correct anyone either way, even after I have checked both English and American dictionaries :-)

it is bigger than sqrt(3) and smaller than sqrt(8)

Isn’t the center of gravity the same as Hughs circumference division method? The center of gravity should lie exactly on the diagonals of his rectangle.

Yes, but its only about a 7% difference from 2√2. Given just how unlikely 2√2 was(p = 7e-48), I doubt that √7 is the answer, it could be, but I wouldn't bet the farm on it.

That model doesn't take into account the rotational energy the coin has when it hits.

Evyatar Baranga The answer is 1/sqrt((2*sqrt(3)-3)/3) approx. D=2.54T Coincidentally that means that a 3 sided coin would be 1 inch in diameter and 1cm thick (off by a few microns, but I think that’s close enough)

Well, whilst he is wrong, it _is_ *A* die, singular, not plural, still worth hanging in there. We all make mistakes?(?)

"Poe's law is an adage of Internet culture stating that, without a clear indicator of the author's intent, it is impossible to create a parody of extreme views so obviously exaggerated that it cannot be mistaken by some readers or viewers as a sincere expression of the parodied views."

In other words, I didn't unsubscribe at all. Matt does great things. The "forever" was meant to be so extreme as to obviously be sarcasm. Poe's Law wins.

Sad to say, that humour is not at all obvious sometimes, here in the wild and woolly world of YT commentary. Just waiting for someone to leap in here blaming it all on Hilary. It happens.

For sure. I've actually given up entirely on trying to have conversations on KZfaq. Just find the correct subreddit for the video and discuss there. Still get the usual internet laws, but chance of meaningful conversation is considerably more likely than on KZfaq.

Yes, it would be like that if there were only X and O, but there were 3 possibilities, the -- possibility either being higher or lower would basically randomly replace either the X or 0 result. So unless they were all equal 33%, which they were not in either case, the X to O ratio itself is insignificant because since they have the same surface we know as a fact it's equal chance for one of them to land.

I don't know. I think that X and O should always be equal on a cylinder, regardless of the side probability. On a coin, X and O (heads and tails) are equally likely, so it stands to reason that as the side or edge becomes more likely, the probability of landing on X or O should decrease equally, meaning if there was a 10% of landing on the side, there should be a 45% each of landing on X or O. The ratio between them shouldn't change.

methodology problem, manufacturing problems. There is definitely more than enough data here to say its likely the coins are biased between the x and o faces as well as biased for and against the edge between the two thicknesses. Possibly thicker density on one side from being printed. better off being cast from a mold.

Yea, that's how it would work in theory, but it's all 1 statistical result, and as Graeme Evans stated it's also biased because of manufactoring issues, and in fact you could argue the artificial "barrier" they set up to collect the data could have influenced the outcomes quite a bit too.

More possible outcomes means more variance for the occurrences of each outcome, and therefore larger standard deviations even between the two equal sides... They do approach each other, but it should be not anywhere near as quickly as they would with tests done on flat coins. Though even then, I think they worked out the standard deviation to be around 15 (sqrt of 222.2), as difference of 80 is still pretty unlikely if the sides really had equal probability, so I'm guessing minor machining errors.

Statistics is just mathematicians take on physics.

I'm pretty sure that posit only works if the sides are the same shape. If t = r, then D = 2t. But it was determined that the statistical answer lies somewhere between D = root(2)t and root(3)t. The answer of 2t is already out of the bounds of the results.

Marc Hutley yeah you are right. Then it must violate my posit, which means that the cylinder must be a bit shorter than what I have stated.

No, the answer lies between sqrt(3) = 1.73 and 2sqrt(2) = 2.83, so a ratio of 1:2 is definitely possible.

Ah you're right, i missed the 2.

A chi^2 goodness of fit test returned a p-value of 0.005, implying that there was a 1 in 200 probability of seeing these results given that the two opposite sides of the dice are equally likely. I would assume that they would have thought about this before posting the video, but there doesn't seem to be any reasonable explanation. tl;dr: This is a **very** significant difference.

I guess football is more important these days...

I feel betrayed.

Yep, that's why the models are both off. They said straight off at the start of the video they didn't think it was that simple. Really just looking at the sphere example you can see how that's the case pretty easily, near the edges of the 'band' part, your 'die' if it is as wide as 't' would be clearly be leaning far enough to one side when it lands that it would tend to flop over instead of ending up on the 'edge'. and that's not even taking into account the rotational dynamics of the thing. Basing calculations on the position of the center of gravity rather than surface area would be more accurate.

exactly

I bet even the mechanical properties of the "coin" and the landing surface are factors....

Somehow I think you need to take into account the likelihood of landing on the edge and falling over onto a face, vs landing on a face and falling over onto the edge. I don't think geometric equality is what you should be striving for. It seems like the thin (sphere) coin is way more likely to land on the edge and fall over onto a face than land on a face and fall over onto the edge.

And how is 1.33 ratio explaned? They had sqrt(3)=1.73 and 2*sqrt(2)=2.83 and fair ratio must be between these(if they tossed it right). Anyway there are many parameters in your simulation such as: center of mass, average angular velocity while landing, restitution of the bounce, friction coefficient, etc.

That’s interesting! Did you try varying any of the physical parameters such as coefficient of restitution, friction between the cylinder and the surface, or gravity? Did it make any difference, if so?

I'm sure I did but I did this almost 10 years ago and I've lost the source code.

It's possible that by throwing the coins, it makes it more likely to land on the edge since once they start rolling, they stay on the edge. In my simulation, all the coins had angular velocity but not linear velocity.

@radixvinni You’re confusing the radius and the diameter. If h = 1.33r that’s 2.666 of the diameter, which *is* between √3 and 2√2.

(I know you commented years ago, but...) I think he was differentiating between a die as in a gaming die (dice) and a molding/manufacturing die.

What's molding/manufacturing die?

@@heimdall1973 it's probably more general than this, but I remember old timey coins being made by taking a disk and hammering a thing onto them which shapes the face of the coin. that thing is called a die. But then I searched and out "Die (manufacturing)" is a wikipedia article, so probably has a much better answer than I gave. I'm probably wrong, anyway, that thing I described is probably called a stamp.

@@silverharloe Thanks

@@heimdall1973 en.wikipedia.org/wiki/Die_(manufacturing) For example, most coins are made using dies to press the image into their faces.

well a hypersphere would have some issues :D

+

I was wondering the same thing

A hypersphere would be possible to calculate, but more complex than I want to get into right now. A 1D "sphere", though, is easy: its surface only has two points, so there can only be two sides.

sqrt(7)

what?

siggiarabi what are you confused about?

Im in the us i can flip coins

Lol, I think he means he wants to visit them and run the trials using their equipment. The way it's worded is funny, though xD

If ypu are on the us, you already got the rolls :)