How thick is a three-sided coin?

[@jamdecks1]

A mathematician, an engineer and a statistician walk into a lecture hall. They conclude that they got it wrong.

[@Megasterik]

How

[@poneill65]

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

[@AlZimmermann]

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".

[@tissuepaper9962]

@@AlZimmermann Ooh I like this one.

[@NYsummertimeCHI]

Al Zimmermann That's brilliant.org!

[@Fudmottin]

I threw forty quarters into the air and every last one of them landed on the edge!!! Perhaps I should have taken them out of the roll.

[@amahlaka]

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

[@Fudmottin]

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.

[@Markle2k]

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

[@satibel]

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

[@Fudmottin]

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.

[@davsch22]

Me in 2018: "Oh, how interesting! I'm really looking forward to the next part." Also me, 4 years later: still waiting 😭

[@webrusheriii8934]

Still waiting.

[@LordMarshyMan]

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

[@rsinoradzki897]

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 🤔

[@SaloCh]

​@@rsinoradzki897 yeah you probably need a phisicist instead of a mathematician lol

[@SeanStephensen]

@@SaloChas the video concludes, don’t we just need a Monte Carlo?

[@noahknox8245]

I would think the center of gravity would be more of a factor than the surface area. The 3 sides need to have the exact same distance of CG from the surface of the table. That way there wouldn’t be any torques causing it to be unstable. If you think why a coin wouldn’t land on the edge, it’s not as much the surface area but when the coin is on its edge it’s CG is directly over a very narrow edge and as soon as the CG goes beyond the support it falls.

[@doctorphrog]

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

[@Henrix1998]

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

[@sebastianjost]

That's what I thought too.

[@earthbjornnahkaimurrao9542]

so should the D/t ratio be 2 then?

[@CrashTestCoder]

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

[@jamesfilosa6277]

4:49: "It's one of my favourite triangles. It's very triangle. Big fan."

[@vitakyo982]

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

[@jercki72]

vihart

[@gordonstallings2518]

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

[@qwertyuiop-cy5en]

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

[@cavemann_]

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

[@chuuuu1131]

4:50 You know they are mathematicians when they say "It's a great triangle"

[@burn1none]

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

[@samotten9874]

I like to imagine that Matt and Hugh realised that the mechanics involved were frustratingly complicated and decided to use stats to come up with some sort of approximation. They then realised that stats is really boring and tedious and so just outsourced all the work.

[@despacitotv7906]

relatable, retweet

[@KingofJ95]

I like when the Engineer showed off that thing he knew about statistics to the Statistician. Very engineer thing to do.

[@HoStallin]

Literally the most useless bit of input too ffs

[@jons2cool1]

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

[@liveportalfeedenglishchane5928]

Not everyone.

[@percyvile]

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

[@TimLF]

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

[@tparadox88]

Any update yet? KZbin decided I should see this video again.

[@DanijelAleksicMath]

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

[@4M4D3USM0Z4RT]

@@DanijelAleksicMath Could you link the paper anyway?

[@DanijelAleksicMath]

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

[@akirakato1293]

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

[@EGMusic12]

The square root of 2 plus 1 : 1

[@geocarey]

I don't think the perimeter ratio is an effective way of approaching the problem. How about looking at the energy required to topple the 'coin' from on state to another? A 6 sided dice will clearly require the same energy to change a number. To topple the 'object' the centre of mass has got to be raised and moved over the point of contact with the ground until it is able to flop to a new position. In a more dynamic approach I suspect moment of inertia, conservation of angular momentum and all sorts will need to be factored in.

[@Septimus_ii]

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

[@rohitramnath5401]

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?

[@gx1501996]

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.

[@matthewhartley3147]

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.

[@Thiodexal]

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

[@jek__]

I love how you start this, it's just like a mathematicians brain. "I saw a thing that had certain factors" "then I saw another thing that had opposite factors" "what is the thing that has these factors in balance?". Looking directly at space you can't see, but know must be there

[@AlisterCountel]

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!

[@gabemerritt3139]

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

[@AlisterCountel]

@@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

[@gabemerritt3139]

@@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.

[@jordanharkness]

Matt, Hugh and Jen go hunting for ducks. A duck flies into view... Matt shoots a meter too low. Hugh shoots a meter too high. Jen jumps up and yells, "yay! We got him!"

[@shredboy9163]

Wow that's mean (pun intended).

[@salj.5459]

Man, I don't get any of these jokes

[@Orincaby]

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

[@eadweard2214]

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

[@jordanharkness]

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*

[@stomponpie]

Does the fact that the edge of the cylinder has a high centre of mass matter? Even if there is an equal chance of each of the sides any small amount of momentum would be more likely to cause the cylinder to fall towards the side with a lower centre of mass.

[@lenn939]

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

[@davefoc]

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.

[@pfeilspitze]

So we're at 1.7 < r < 2.8, roughly? Did anyone try just 2?

[@iycgtptyarvg]

Mr Parker, it has been 9 months now. Why haven't we heard anything yet? Hasn't anyone supplied data to you? Seeing this is by far the most interesting video I watched from you I'm really bummed out about you leaving us hanging. At least let us know if you abandoned this project.

[@bilalbaig8586]

Forget it man, its a Parker problem now.

[@jiralishu]

Parker square.

[@squeakybunny2776]

19 months.... Still waiting

[@rewrose2838]

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

[@Zaderos]

@@rewrose2838 add a day to the counter

[@vivaldirules]

This is a fairly complicated physics problem, it seems to me. Things like edge sharpness and what motions of “flipping” are permitted are probably important as may be material density, even. Still, I was pleased to see this including the nice p-value calculation. Thank you.

[@any1alive]

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

[@any1alive]

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

[@AdriansNetlis]

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.

[@jarls5890]

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.

[@tom_something]

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.

[@lc7269]

I think that part of the problem is with the coin’s centre of mass: even if the coin lands on its corner and has a larger surface area on its side, due to its centre of mass, it gains a torque turns to a face instead. I think the way one should approach this is to take the centre of mass into account along side the coin’s surface area/ perimeter. Im also interested in whether a hollow coin will have an effect on this

[@ijabbott63]

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.

[@rjdruhan]

I think there's a problem in the way you "rolled" the thick ones. i.e. you rolled them like dice out of a cup instead of flipping them like coins. Using that method you should expect to get significantly more "edge" lands because you are no longer ignoring the rest of your 3D surface like you did on the chalkboard.

[@chaos.corner]

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.

[@jeffc5974]

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

[@davypi2]

Back in the early 90s, there was a rather prolonged discussion about this problem on sci.math. The biggest issue, which is not discussed at all in this video, is that you have to define how you flip the coin. The hypothetical solutions being offered assumes that the axis of rotation goes perfectly through the center of the rounded edge. However, if the rotational axis goes through the disk edges, you are effectively "rolling" the coin along the round edge rather than flipping it and you are much more likely to get an edge than a side. When you throw the coins out of the mug as you did in your trial runs, you are getting a different axis of rotation every time, meaning each flip has a different probability of getting an edge; you are flipping the coin at different angles between these two extremes. In essence, there is no solution to the problem because you have to define a flip, and there is no empirical way to prove your answer unless you can get a machine to produce flips with a consistent axis of rotation. As a side note, we had a similar problem come up in one of my 500 level stats classes where we hypothetically flipped a die with uneven faces. We came up with the same issue - you get different probabilities for the outcome depending on how the flip is defined.

[@TheREALMcChimp]

Is this not solved by the spherical model, though?

[@davypi2]

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.

[@TheREALMcChimp]

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

[@jasondoe2596]

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.

[@jasondoe2596]

_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.

[@uruiamnot]

*Major Methodology Errors* As others have pointed out, the coins should not be dumped out, en masse, onto a table from originally having been placed into the holding containers. If each die were to have been labeled systematically (always X first) and placed into the jar the same way, then there will be bias in the container. I think you could get a BINGO tumbler that might fit the bill in which everything is tumbled and individual coins emerge. The other problem is the 3D printing of a piece of plastic or nylon, etc. Your "coins" might suffer quickly from edge wear and distortion, unlike metallic coins. Once an edge gets worn from a sharp (die-cut) to a dull (worn) shape, your results will increasingly skew away from the edge toss. The 2-sided coin does not have a critical need for a sharp edge. But a *3-sided coin has to have a critically shaped, distortion-free edge.* If a sufficiently brilliant coin-flipper understands the ramifications of edge wear, he can use it to his advantage by flipping his own set of coins that he knows to be flawed. I doubt a 3-sided coin can be mass-produced that is fair and immune to edge wear. And for a parting shot, I am going to discuss the landing zone. A sufficiently bumpy surface will have undulations that will tend to make rolling coins flip or flop to a flat side.

[@bluemalamute]

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?)

[@evannibbe9375]

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.

[@pedrolmlkzk]

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

[@GaneshNayak]

Disappointed that no update on this.. even if the experiments have failed, could have updated what happened

[@Reptonious]

Biggest cliffhanger on KZbin!!

[@favorite89103]

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

[@Zuiker1]

How are you verified?

[@detonater7441]

@@Zuiker1 he’s not it’s just an emoji

[@jsmith5443]

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.

[@parasietje]

Modeling & simulation engineer here! Math is nice, but the real world is unforgiving. The math assumes a coin flips randomly in 3D space, then stops and moves very slowly down onto the table. 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. So you will have different ratio's for different throw velocities. If I have time, I will work out a 2D simulation example.

[@olegwahl8031]

"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?

[@iskierka8399]

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.

[@PAI93]

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

[@MystDaLow]

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

[@Bridgeboy777]

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

[@vpumeyyv]

Iv'e done my research about this a few years ago, and Iv'e reached a conclusion that 3 sided dice like this are case specific. The friction of the table, the velocity and the angle of the throw may change the results whenever the faces of the dice aren't identical and symmetrical. Basically, the only 'perfect dice' are Platonic Solids and Catalan Solids.

[@williamnathanael412]

Finally the answer I need.

[@djwillcaine]

I thought both of these approaches seemed odd so I did some trig based on the angle at which the centre of gravity would cause it to fall on its edge 1/3 of the time (60° of range at top and bottom) but I also got sqrt(3). Obviously the momentum of the coin spinning means that there is a bias towards it falling flat.

[@hobrin4242]

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.

[@celestial_crash0]

0:47 ? Die is singular for Dice

[@MrSansbjerg]

Yep. Languages are weird

[@lagduck2209]

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

[@celestial_crash0]

a dice sounds very strange to me

[@lagduck2209]

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."

[@celestial_crash0]

now im just confused

[@brianfox340]

Once a close to correct ratio is found, I would love to see that ratio scaled down to even smaller and scaled up to larger (maybe the tape roll size) to see if it holds true for different sizes/masses. Clearly it theoretically should, but does it?

[@akshat9282]

March 2019 and still no solution. iS tHis tHe nEw fErmAt's LaSt ThEorEm?

[@bwayagnesarchives]

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

[@akshat9282]

@@bwayagnesarchives more like a 100 thousand dollar problem

[@jiralishu]

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.

[@1992jamo]

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

[@AlecBrady]

@@jiralishu Why?

[@inscept8485]

The dice manufacturer "GameScience" makes 5-sided dice based on the same idea. They are triangular prisms with supposedly exactly the thickness to give a ratio of "face" to "side" of 2/3. They claim to have arrived at that ratio by extensive experimentation. However-and I think that is something that you, too, should keep in mind-in practice I have found that it makes a big difference on what kind of surface (e.g. hard and slippery vs carpeted) the dice is thrown.

[@roderik1990]

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.

[@Kalumbatsch]

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.

[@markupnone733]

I think the solution should work out to be a cylinder that is square in cross section, where the thickness is equal to the diameter. The problem seems to not be a surface area problem but more like a center of mass problem. The center of mass is dead center in the middle of the cylinder. In cross section it would be the exact center of the square. If the coin is balanced on its edge, the arc that the center of mass will follow is identical if it is falling onto the flat side or the rounded side. While this is true of any rectangle on its edge, to be fair the coin must flip with equal probability off of each side, and this is only possible if the edges are equal in length. One can imagine a cycloid-like curve that would be traced by the center of gravity as the coin is rolled from face to face over its edges. If the face lengths are unequal, the curve would have two distinct points, one higher than the other. When the center of mass is higher the resting coin position is less stable, thus less fair. Well, that is my 12.533 cents (US); 13.355 cents (UK). :-)

[@hektik2074]

One time when I was a kid I flipped a nickel in the air and it landed on its side, and after maybe 3 or 4 seconds of rolling it tipped over. Little did I know that would be the greatest accomplishment of my life

[@rudykrutar3319]

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?

[@rudykrutar3319]

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

[@charliesmith1358]

All about the invention of the D3

[@kairotox]

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.

[@DekarNL]

take a D6 and have all numbers have 2 sides.

[@AmaroqStarwind]

Canine Kitten The *proper* D3. About time.

[@robinlindgren6429]

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)

[@TheCassiusTain]

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

[@GavinWilsonsPage]

Doesn't the momentum and location of centre of gravity affect which edge it will eventually fall on, since you don't have a regular shape?

[@yshwgth]

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.

[@qwertycal1707]

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.

[@soxfansince97]

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

[@Dongufo15]

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

[@Moonbrony]

Wouldn't air resistance also play a part? If the disc is being flipped, it will encounter less resistance spinning along the cylindrical axis and hence be more likely to land on an edge.

[@miguelnasarrebudino9005]

Recently i did a project that tried to calculate the probability of an irregular polihedron to fall in each side when you roll it. And one of the most interesting conclusions was that the size of the object matters, in fact it matters a lot. If you flip a bigger coin with the same proportions the probability of it landing on its side increases. I don't know how interesting this might be and the model i made was kind of a mess but it's still an interesting thing to add to this video. Also we did testing to verify that its true, its not just theoretical.

[@soulsilversnorlax1336]

I've recently started investing in Dicecoin trading.

[@boriscat1999]

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

[@freedomcaller]

I thought this said dogecoin and you were rich.

[@MattyFez]

This is a Parker Triangle of a solution

[@TheBoundFenrir]

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?)

[@jivejunior8753]

Avatar pic checks out

[@vojtechjanku2534]

The Parker square meme is a Parker square of a meme

[@eliasmg9144]

As a chemical engineer student, yes, this is how we get born, but my kind usually gets born from a very massive beaker

[@marcuswillbrandt5901]

As another chemical engineer student I can confirm that.

[@oscarmulin114]

Using Bullet physics engine and a basic evolutionary algorithm, and this is what algorithm converged to after 3-4 tries: D (diameter) = sqrt(2) * h (height) It's not within the bounds in the video, but it's the answer that had an error percentage of 1%, meaning every side had almost exactly %33 chance of landing. I can make a video explaining how I've found it if anyone's interested. The collisions were slightly elastic with coins bouncing off a bit and often changing sides upon falling.

[@sujaankumar30]

Oscar Mulin yes please

[@mirageinthedesert5448]

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

[@ianhawthorn1527]

How did you randomise the nature of your flips.

[@Sibula]

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.

[@dmdragonfly]

My friend fell out of the lighting gantry at a U2 concert. HE landed on The Edge.

[@davidr2421]

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

[@Metaporphsycosis]

This is the ultimate answer to the question ‘When would you ever use Pythagorus in later life’

[@ITS_MEEE333M]

typically Numberphile

[@HelloKittyFanMan.]

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

[@noahdoss1967]

Or calculating any distance ever

[@HelloKittyFanMan.]

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.

[@SuprousOxide]

To get the wrong answer? 😀

[@inchinaxp8663]

After my intro to stats class I'm very chuffed to say that I actually know what they're talking about

[@RC_Engineering]

So are you upset, or happy?

[@silkwesir1444]

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...

[@SpaghettiEnterprises]

dude noice!

[@oldcowbb]

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

[@MD-pg1fh]

Uuuuuuuh... None of that made any sense.

[@Busalonium]

I'm wondering if this can really work in practice. I'd figure angular velocity would be an issue. If it lands on the thin side then it won't take too much angular velocity to flip it over to a thicker side, if it lands on the thick side with the same angular velocity it probably won't flip. So, the harder you toss the coin, the less likely it is to come up on its side. That being said, maybe the effect is negligible.

[@tristanridley1601]

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. :(

[@JessHull]

I didn't know Ron Weasely's dad was an engineer...

[@theodoremurdock9984]

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

[@TheSam1902]

Marrying a witch as an engineer is your ticket to joblessness

[@johannbcmelo]

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 )

[@adrianatgaming8640]

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

[@clintoncoker6]

Ron Weasley's dad is Father Brown.

[@Dunkle0steus]

The Xs and Os seemed to be pretty far from each other in each case. Could that information be used to determine anything? Each side of the coin should be just as likely as the other, so over 1000 tests, you would expect to see those two numbers approach each other. If you took their distance to be an indication of some sort of error introduced, how would that affect the results? If you used the Xs and Os to measure uncertainty in the data. I took one probability class in university so I have no idea what I'm talking about. I'm imagining a target shooting range (guns, bows, doesn't matter). I imagine a person goes to shoot and the average distance from the target of each shot is 50 cm. You could use that data to come up with a value representing how accurate that person is. However if an expert shooter also tried at the same range and had an average distance of 20 cm, you would begin to suspect something was wrong: perhaps this range has some factor that makes shooting very difficult. By comparing the deviance in the expert's shooting at this range to their expected shooting (which should be highly accurate) you may be able to come up with a value representing uncertainty in shooting introduced by the range and get an adjusted value for the first shooter, taking that uncertainty into account. With the flipping, you expect X and O to be very close to each other, not too many standard deviations apart. If they are vastly different, you can use that probability in a similar way.

[@Skrymaster]

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.

[@Dunkle0steus]

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.

[@ge2719]

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.

[@Skrymaster]

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.

[@TheEvolNemesis]

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.

[@keiyakins]

The mental image of two guys at the pub, having had just enough alcohol to get an enjoyable fuzz but not enough to actually be drunk, just deciding to throw some dice for a few hours is far funnier than it should be.

[@quintessences]

When the video first came out all the statistics flew over my head, but now after taking a statistics course it’s satisfying to fully understand it

[@auroran0]

Going off what's probably oversimplified equal areas per side. a = 2πrh +πrr +πrr πr(2h + r +r) 2h = r -> r/h=2 which puts it between the √3 and 2√2 tests from the video.

[@jeromewink557]

Orin Johnso

[@jeromewink557]

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?

[@jamesmahoney8678]

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.

[@kevwang0712]

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

[@WispOTW]

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

[@DudeWhoSaysDeez]

I wanna roll the coins! I will do loads of flips But i live in the US

[@siggiarabi]

what?

[@DudeWhoSaysDeez]

siggiarabi what are you confused about?

[@Zalied]

Im in the us i can flip coins

[@Xrelent]

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

[@decs3491]

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

[@lidarman2]

This video popped up on my google feed again. Matt, where are you on this? I 3d printed some coins and found the thickness to get really close to 1/3 is super sensitive at the thickness between the two values you discussed and how you toss it. I think the probability curve vs thickness of it landing on edge is a steep sigmoid function between a zero thickness edge and an infinite thick edge.

[@gabemerritt3139]

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.

[@Confuseddave]

"...he means a dice." Ooooh you troll you

[@user-yw9mw9hv8o]

pls update on the state of this problem

[@rewrose2838]

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

[@lyrimetacurl0]

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

[@CalvinHikes]

This is great! Like a real show, with the practical application and bringing in experts.

[@JasonMitchellofcompsci]

Lol. Practical application.

[@RaunienTheFirst]

"The normal distribution" "Noice"

[@TriplePalindromous]

0:46 Oh yes, "The dice is cast", as Julius Caesar said.

[@ShinyRayquazza]

Iacta alea est

[@Nilguiri]

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

[@JimSteinbrecher]

shocka, julius caesar spoke english?

[@TriplePalindromous]

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

[@MikkoHaavisto1]

singular dice is just awkward dont use it please

[@adizmal]

This is probably my favorite standupmaths video. The two reasons are as follows... First, the introduction of the other participants in the video is very well paced and slick... nice production in that sense. Second, I like watching other nerds talk about innocuous stuff like the topic of this video in a relaxed yet stimulating manner, three people from different fields just having a chat about some nonsense - the three of you have a great rapport, very casual yet concise. Thanks for your videos, Matt.

[@LupeFenrir]

Appart from working out the chance of the 3 faced coin landing on each face you also would need to work out the chance of it landing on an edge and then toppling over to a face. That is, if you want to do it with math(s). Brute force trial and error is off course an entirely valdid option.

[@pizzahut3001]

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

[@dud621]

Holy shit why is no one talking about the flip at 5:50 that hyped me tf up

[@MorningDusk7734]

I've always enjoyed thinking of a three-sided dice as sort of an un-curved banana shape. Like a triangular prism with a gradual fillet down to a unified point at either end.

[@waltermcdonough]

what happens when mathematicians get too crazy at the pub, they start flipping special cylinders

[@adj6006]

If you create a device which will destroy your timeline whenever any coin lands on tails or heads, then the odds are 100% that every coin flip will end with it landing on its side

[@CJBurkey]

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."

[@badlydrawnturtle8484]

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

[@daicon2k6]

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

[@jeepien]

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

[@badlydrawnturtle8484]

+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.

[@silverchairsg]

I really like how the board has simple maths without calculus. I've been watching too many calculus and advanced maths video and it's a breath of fresh air to see such (relatively) simple maths on the board.

[@PromptCriticalJello]

the starting condition will drastically effect the outcome.... if the coin is flipping vs if the coin is rolling.

[@kailomonkey]

so many more factors than sphere and circle

[@peppybocan]

Right of the bat response: for a cube we have equal probability for each side 1/6, because the centre of mass is in equal distance from each side of a cube. So we need a cylinder such that each side is in equal distance from the centre of mass. Not sure if this is in any way, shape or form correct. :D

[@peppybocan]

Statistics is just mathematicians take on physics.

[@EtoileLion]

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.

[@peppybocan]

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.

[@jellevm]

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

[@EtoileLion]

Ah you're right, i missed the 2.

[@ftfracingnz]

There are additional factors involved here over and above the simple shape. But starting with the shape, when the coin lands on the edge, the centre of mass is higher than when it lands on a face. This means it's less stable and more likely to bounce from edge to face than from face to edge. Which brings me to the material used and the surface you are throwing the coins on - if these materials produce more bounce effect then the above cones more into play and the coin has to be thicker to compensate. The reason this does not matter for any other shapes of dice is that they are perfectly symmetrical. My bet is no matter how long you spend trying to work this out, without specifying the materials, you will not get a consistent answer to how thick a three sided coin needs to be.

[@smugless191]

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.

[@cseguin]

"Die" is singular for "dice" - so, he did mean "die".

[@isaacbriefer193]

rictus grin At least somebody has my back

[@iPhatDeluxe]

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

[@tristanridley1601]

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.

[@cseguin]

Lol.

[@RupertReynolds1962]

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 :-)

[@asailijhijr]

Matt, you've found a possible solution using a 2D sphere and a possible solution using a 3D sphere, now what's the ratio given by the same geometric formula on a 4D sphere?

[@Roescoe]

well a hypersphere would have some issues :D

[@vampyricon7026]

+

[@index7787]

I was wondering the same thing

[@Tzizenorec]

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.

[@ronraisch510]

sqrt(7)

[@rectorsquid]

Where is the link to the outcome of all of the experiments? I want to know the final best size to make!

[@tomeubank3625]

IMHO, no such three-sided coin can yield statistically consistent results due to the interaction of the momentum of the toss and the coificient of friction of the table surface. For fair tosses, it would be much simpler to use a cube with 1, 2, and 3 dots on opposite faces, but that shape would not qualify as coinage. A solid that could be suitable for fair-tossing, tri-surface coinage would look something like an American football with an equilateral triangle as the cross section at every point along its length -- providing, of course, the ends and edges were rounded a bit so they wouldn't quickly wear holes through ones pockets.

[@tiagotiagot]

If we're not restricting ourselves to just "coins", why not just get the intersection of a cylinder and a triangular prism, with the top angle of the triangle used to tune the probabilities of the elliptical faces versus the, I'm not sure what to call that shape, the resulting third side.

[@graphite2786]

TiagoTiago Or just get a regular six sided die and pair up the sides - 1+6 side (a) 2+5 side (b) 3+4 side (c) Statistically it's 3 sided.

[@matrixstuff3512]

I think that friction is exactly why you will get consistent results. Momentum may be a little more if a problem to hid under the rug

[@stigcc]

Yes, the american football can be used to create any dice. Just draw a circle around the poles and section them. Then flatten the sides so they allign

[@joeshoesmith]

I can't fathom that there would be no solution. Take your three-sided coin, and say it's weighted toward landing flat. Ok, we need a thicker coin. So make it considerably thicker, and it will naturally land on the side - the side is essentially the flat now, after all. By intermediate value theorem, it's pretty fair to say there's something between that has consistent thirds. This doesn't *feel* like it's a mathematical proof, but it's more like common sense anyway.

[@FrankLassowski]

Did I miss something? What's up with the 3-sided dice? Any progress?

[@VocalMabiMaple]

3 days to pi day! Didn't know where else to put this and this was your most recent video. Super excited about what you'll do this year.

[@evyatarbaranga5624]

could it be sqrt(7)? I used center of gravity, and that the chances of falling to each side are equal to determine that the oblique line is 2*sqrt(2), and from there the diameter is sqrt(7).

[@evyatarbaranga5624]

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

[@stefanozurich]

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.

[@ddxexex]

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.

[@mushroomcube623]

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

[@CrashTestCoder]

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)

[@RealLuckless]

I'm saddened by the sloppy distinction between dice and coins in this. You've ignored a key factor in that coins are FLIPPED. You need a design were the thickness of the coin is such that it has an equal chance of catching on edge and halting its motion around the flip axis. From there it could safely roll without affecting outcome chance. If we assume a coin flip mechanic, not a dice throw, then the "band area" may safely be larger than either of the face areas, as we become more worried about rotational cross section. So we need a shape that is balanced such that stopping on either side of the edge then equals the odds of stopping on a given face. (Because in a flip the edge is in play twice for each time a face is.)

[@RealLuckless]

Dice are tossed, but in my part of the world we flip coins. Held over the thumb and forefinger, and thrown into the air with a flick of the thumb... This is a fundamentally different mechanic as compared to the motion produced by throwing them with a dice cup.

[@hiveinsider9122]

A well thrown di(c)e has sufficient spin to make the difference insignificant. Or that's what I think anyway, plenty of people barely spin the di(c)e at all.

[@RealLuckless]

But a "randomly thrown coin" will have its spin on a random axis - A thick coin rotating like a spinning wheel is a different mechanic than one flipping between faces.

[@hiveinsider9122]

Anything when tossed or flipped etc will have a spin on a random axis, there is no other way to do that, the axis might change according to how you do it, but there is not real human way to keep it consistent all the time.

[@MrSpookersMcGeeThe1st]

I agree. But HOO BOY, all of that flipping would be tedious as hell!

[@daviidayala4987]

Hey! I believe I've got an idea that you can consider for the problem. So, you guys took the throwing of the dice as if it was that of a random variable going over from points on sphere and on points on a circunference, and that is a really neat aproach if you don't consider angular momentum. What I mean is that, in tye case of a dice, for example, its geometrically simetric, each face with respect to each face. So, any alleatory contribution of a angular momentum cancels out, making no preference to any face of the dice. In this order of ideas I believe that your model would aprouch results more neatly if in the throwing of the dice you do not invert angular momentum, like literarily just droping the 3-sided-coin from the hand without letting it roll or so (or preparing them randomly in a cup and holding it for that when you flip the cup, you led them just fall down taking away your hand holding it. But it would be grat to consider the effect of angular momentum as random variables too and taking the nonsimetry of the coin into account. Hopely this was for your interest! Love your work and books!

[@silverharloe]

"he means a dice" - uh, "die" is the singular of "dice" . Or is that different across the pond?

[@Roccondil]

(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.

[@heimdall1973]

What's molding/manufacturing die?

[@silverharloe]

@@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.

[@heimdall1973]

@@silverharloe Thanks

[@Roccondil]

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

[@RaymondJerome]

dimensions. a face is a 2 dimensional surface and is stable. an edge is a one dimensional line and is unstable. also an edge landed on not rolling form is potentially unstable, the center of gravity might be above the midpoint. i say a cylinder where height is same length as diameter. let me thing, is that between your upper and lower bounds? just came out of a pub so will not think to hard about that

[@sanjeen2503]

There're a lot of physical factors apart from working in the backstage governing the motion of the 'coin die', out of which the most significant ones for 'landing side' would be the centre of mass height, friction, how the coin is thrown and maybe the flatness of two circular faces (a slightly outside bulged one would make it roll over easily, compared to an inside one)

[@darcyhayes6664]

I think its impossible. The speed that the coin is turning would change the probability of it landing on the cylinder edge. if it wasn't turning at all then the ratio 1:sqrt(3) would be right. but if it's turning faster it will more likely land on the faces because the center of gravity is lower and would take more momentum to tip. So the person throwing the coin could alter the chance of it landing on the side by changing the rotating speed and the coin would be unfair no matter the ratio of sides.

[@SSJProgramming]

I think this is the correct answer. The initial angular velocity alters the probability to land on the edge vs side. This forms a relation between the thickness "H" they are looking for, and the initial angular velocity "V". So really, they are looking for the relation rather than a singular value.

[@guillaumepages9335]

I agree that the answer depends on the physical conditions of the problem. And the speed at which the dice looses its energy is critical ( this depends on the hardness of the dice material and the ground)

[@-eea32]

This seems logical

[@Cyrribrae]

This is probably true, but apply all of this same logic to a normal 6-sided die (or even a normal pyramid). These are affected by die material and surface being thrown on and angular velocity and etc. etc., but the result is still fair - tipping doesn't matter. We only think of "tipping" in this scenario because we conceptualize a normal coin when we think about it. Your own comment (unless I'm interpreting wrong, which is possible) proves that point. You are saying that if there were NO turning, then you would expect that coins that landed on their edge would just stay on their edge, right? And if there is turning (which there was), then you'd expect it to land on its faces more often. The data for the 1:root3 coins show exactly the opposite, don't they?

[@gmaasry]

Matt, love your videos. One question on this: isn't it a bit strange that the "thin" coins have 393 and 475 for X and O sides, respectively? That seems a statistically significant difference, far beyond what you'd expect if in fact the chance of landing on either flat edge were truly the same. Could this imply something about the coins themselves being flawed, if not the measuring technique?

[@benjamindarnell6296]

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.

[@nicholasboatright41]

Why are these the cleanest chalkboards I have ever seen?

[@johannesvahlkvist]

you're assuming that when the cylinder lands, all kinetic energy stops and your result is equal to whichever region the center of mass is directly above (+, O, -) which is wrong

[@TheEvolNemesis]

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.

[@kailomonkey]

exactly

[@mishoidenis]

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

[@ecsciguy79]

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.

[@Henrix1998]

So, has this been solved?

[@ADHDunce]

@Stand-up Maths I think you had the right solution, but the flips weren't fully random as the coins knocking into each other as they fall out of the tosser's cup/hand and knock into the table will create more outward momentum with which the coin *is more likely to stabilize into a rotation instead of continuous flipping* which is why you got more edges than expected!

[@Caramelhorse1]

Will there ever be a follow-up to this video or have I just missed it?

[@lordofcastamere9376]

A Parker Coin

[@gabrielkwiecinskiantunes8950]

Or is it a Parker Die? 😵

[@D34dlyTidsy]

Man I loved this so much. I did maths at uni and haven't done it in years and this just bought back great memories

[@connorwilliams3451]

It seems like the solution to this one would be waaaay easier than they are making it out to be. The coin just has to have a thickness of R^2/2R! The easiest way to make the probability of a coin landing on a face equal for both faces and the side is to have both circular faces and the "edge" have the same area. The faces are both circles so A=Pi * r^2. The edge, if unfurled, is just a rectangle with an area of B*H where the height H is equal to the circumference of the circles or Pi*2r. Both circular faces will of necessity have the same area so practically all you have to do is to make the area of the rectangle equal to the area of one of the faces. Doing this out: Area of circle = Area of rectangle, where height H = Circumference of circle, or Pi*2r Pi*r^2 = B * H Pi*r^2 = B * Pi*2r Divide both sides by Pi r^2 = B * 2r B= r^2/2r So if I decided the radius was 5cm, Pi*(5^2) = B * Pi*(2*5) 25Pi = B * 10Pi 25Pi/10Pi = B 25/10 = B, or (r^2/2r = B) B = 2.5cm I wonder if they will eventually try this solution because I'm pretty sure I am correct unless there is something wacky about the physics of a coin landing where equal areas does not actually make for equal probability...

[@superjugy]

Connor Williams that is kind of what they are doing, except they are mapping it to a sphere since just throwing a cylinder with the same area doesn't mean it will land on the face exactly. It will be rotated in all 3 axis (like a sphere) so they map the point it would hit to a face. In any case it is wrong too.

[@CrashTestCoder]

I found my answer by positioning the center of gravity directly over the edge of the cylinder and the face and used the law of sines to find the optimal dimensions for this. I came up with a cylinder that is 1 inch in diameter and 1cm thick It’s quite the coincidence that the two measurement systems line up in this application the way they do

[@Ajomoni]

Firstly, why don't you divide out the R in your formula? R^2/2R = R/2, given that R != 0 Secondly, I found it really nice that you and this fellow commenter came to the same conclusions in a different way: kzbin.info/www/bejne/Y6LUgX6BhLpgr7s&lc=UgzS9CjX4R6AXE5aTIN4AaABAg

[@garythomas4030]

Oh yep. Well they said that thickness is 1. So we just have to use your doohickey so it equals 1 and work back from there.

[@garythomas4030]

Ryan McHale I did an idiot thing and came up with 2.526 which is sort of 1 inch:1 cm.

[@desia.brimou]

you can tell the video was prerecorded because he still has hair.

[@kavigollamudi]

Yes, his non parker-head.

[@steliostoulis1875]

HAHAHA

[@JPsk8core]

Well... Not so much

[@Derkman96]

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

[@Malkovith2]

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

[@uPPyinmate]

This is great. It serves as evidence for my theory that everything is statistically equally probable based on the understanding that anything and everything is possible. Your ability to create the purpose of the sides of the die being the constant or infinite variant and the evaluation or concept at question being infinitely used to determine the purpose.

[@Schmogel92]

A typical coin flip makes the coin spin around an axis parallel to its flat surface. Your methodology does not take that into account.

[@fleecemaster]

That's why they have been rolling them like a dice

[@WillHirschUK]

Quite the opposite. Hugh's estimate is what you get if you consider rotation ONLY about that axis. If you rotate a coin at an angle to that axis then the points where it contacts the circle it traces out don't form a rectangle that is as wide as the coin is thick like the rectangle in Hugh's diagram.

[@lukes2219]

Schmogel92 but they are not trying to GE that they are trying to get somthing that will land evenly on each side

[@sander_bouwhuis]

Wouldn't it be fairer to just drop them from 1 meter high or so and then let them bounce?

[@brianmacker1288]

Flipping a coin is very different than spilling out of a cup. The angle of rotation matters a lot on these three sided coins. If spinning in the direction of the cylinder axis, with the axis near parallel with the table top, the coin will always land on its edge and then roll. A roller. There is no chance a roller will flip to another side, because its spin is in the direction of more edge. Spilling out of a cup tends to produce many more rollers. In fact a very long cup with 3 sided coins at the bottom when spilled should produce mostly rollers. The diameter of the cup also probably matters to imparting a roll. For each side . Height of center of mass, vs base width matters. Amount of original spin speed matters. Orientation of spin matters. How the spin is imparted mattets (if it skews the orientation of the spin angle to axis). None of these sphere area, circle circumference, or surface area calculations are complex enough. Because the edge and faces have different geometries they will have different odds of bouncing to the same or differemt face. They also have different heights for center of mass, etc. Plus spin angle effects a face differently than an edge. They cannot be ignored, which is what is happening with simple calculations.

[@jeremiahbaker985]

I'm still waiting for the result. What is the answer?

[@IceBrys]

I'm of the opinion that there is no solution on the grounds that depending on how you throw the coin/dice, it will have different probabilities of landing on the "edge". For instance, if you flip it as you would a coin, it will have a different probability than if you threw it perpendicular to the ground spinning as though it were a tire, in which case it would more likely land on its edge. Only if we agreed that the coin had to be thrown a certain way each time could we consistently get the same probabilities.

[@scottyscroggins1071]

That seems right to me as well... do you think even something like how far the coin falls would affect the probability?

[@bacovey]

Did the project to solve this come to a conclusion?I am curious.

[@memesfurmich5259]

But is your coin flipping realy random? By "sliding" them out of the bear mug you don't have a random orientation of the coins hitting the desk. I propose you do do it all again but flip the coins properly;)

[@Tfin]

Flipping induces its own bias. They need to shake a cube containing the 10 dice.

[@luminicadeesuuu281]

The mug is not a flat surface, it has curvature making certain things more likely than others. If instead of sliding over an X or O it rolls from the mug from the third side the coin has no rotation on 2 axis, making it almost guaranteed to land on the side. Now there are other factors into play as well that complicate things a lot, but I highly doubt a mug is a proper coin tossing tool and it can skew the results, although admittedly I did not watch the video showing their throwing method.

[@jarhead9853]

R3Testa - Well, since we are talking about flipping coins...

[@complexeddrummer]

The issue I have with your calculations are that they answer the question of: What is the thickness of a cylinder such that if thrown into the air, it will have equal probability of initially hitting the normal surface on any of its three faces. Rather, the result you want is such that when thrown, after hitting the normal surface, it will have equal probability of SETTLING on any of its three faces. You want to design the coin then around the idea that if the coin strikes the normal surface on one of its two edges ( where the two opposite faces meet the circumferential face ) it will be equally likely that it will tip towards one of the planar faces or the circumferential face.

[@twistedtachyon5877]

I'm a bit sad that the engineer didn't spot that. This seems to be a tricky dynamics-dependent physics problem that is probably highly dependent on how you actually throw the dice. Also disappointing is that I don't see a final answer response video here after all this time. We need to know!

[@augustus3024]

7:58 had she been silently waiting for her introduction the whole time?

[@rewrose2838]

Probably waiting in the kitchen , making some tea for the two

[@SwagnerCountsThings]

2 years later and I still think the coin derived from the circle is accurate. I just think that you were treating your "coins" as "dice".

[@DanielSchwen]

I think the way to go is to make a potential energy argument. The stored potential energy of a coin on its side needs to be equal to the potential energy of the coin on its edge. You basically need to integrate over horizontal slice area times height h above the table (times dh). Numerical integration (yes, I know), shows that the potential energies for the side and edge configuration are identical if the thickness is equal to the radius of the coin.

[@karlmuster263]

Here's the easy way to make a fair D3: Start with a fair D6 and paint opposite sides the same color. The 3-fold symmetry about a corner ensures that it's fair.

[@brcoutme]

Easy to avoid one side depending on the angular momentum basically anyone can intentionally 'roll' the die so it is improbable for one of the sides to appear. (Making 1/3 odds closer to 1/2 and 2/3 odd nearly guarantee ). Of course if instead you don't do opposite sides but rather adjacent ones then a reasonable throw would be mostly fair. I.E. on most standard dice rolling a 1,2; 3,6; or 4,5 as opposed to your proposition of rolling a 1,6; 2,5; or 3,4.

[@diggle149]

Couldn't you just run a large simulation for several different thicknesses and just let it run by itself?

[@the-goat]

I was thinking the same thing. Due to the centre of mass issue I also have a feeling that this may have many solutions depending on the materials involved, both for the cylinders and also the surface, whether we allow for real world defects, the velocity vector of the throw, the angular momentum, etc. For starters, a random initial rotation would be required to avoid the rolling side bias that comes if our throws tend to roll. Just the shape of the human hand or any other throwing vessel immediately introduces a non-random element that may bias the outcomes.

[@RKBock]

@@the-goat not really. by assuming a constant density over the whole body, the mass becomes irrelevant for the rotation, if we ignore air resistance. (it would be relevant for the calculation of the falling time, but that's not really an issue) in other words: the simulation would be rather easy to do, by calculating the inertia tensor for the body (since it's a cylinder, it's pretty easy), then generate random numbers to determine the force and point of impact (simulation of a flip) and then let it conclude where and how it lands. the initial height should be constant.

[@zexot]

Try a coin of thickness that is half of the radius

[@davidpacker7602]

@@RKBock wouldnt heavier objects have more angular momentum and thereforen be more likely to land on the head/tail sides then the edge. it seems likely

[@RKBock]

@@davidpacker7602 you mean momentum, not inertia. you can't have 'more' inertia. and... no.

[@hovikgasparyan9729]

This has to be the Parker square of standupmaths videos. When are we getting the flippin' results!!!

[@jhawk2402]

Guys, I just realized something. If you wake the weighted average of the results: (132(2√2) + 410(√3))/542 guess what you get. You get 1.999064965995824!!!!! This means that the correct thickness is probably exactly 2!!!

[@simon-pierrelussier2775]

Yes, I don't see how they analyze the problem so wrong that a result of a diameter much smaller or much greater than 1/3 of the sum of all sides would possibly be the answer... D = 2 for t = 1, it seems obvious to me.