How thick is a three-sided coin?

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Information How thick is a three-sided coin?

Title	:	How thick is a three-sided coin?
Lasting	:	14.53
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Views	:	968 rb

I've seen a video on this problem pop in my recommended recently and you know what? This has been in my head for long enough, I'll just leave this here, the video that has put it in my head in the first place: I've calculated years ago that the coin ratio bcould/b be Sin(90-(15π/2))/Sin(15π/2) so 2293059806 or 0436098525brIf you can make that coin, feel free to try, but don't forget to test for variables that can influence the result such as spin or the type of throw, since this is not a perfectly symmetrical shape
Comment from : [T]aKer - Speedrun

3:44 but it DOES matter!brgyroscopic effects! a three-sided coin spinning around its axis of symmetry iresists/i tumbling forward or backward when it lands and gets a forward or backward torque from hitting one edge on the ground before the other
Comment from : TeslaInvestah

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
Comment from : idiot321321321 idiot

Nice idea - but that way, you'll only get the approximation, right? "The coin with the side X-wide, landed the closest to 1/3" You won't get the mathematically precise answer, ie WHY X?brTo calculate X, derive it from some laws, not approximate it - you would need solving some very complex system of physical equations, i guessbrbrAlternative idea: to appoximate faster and with less expense - use physical modelling in some 3D-engine, like PhysX
Comment from : adarsops

I'm still waiting for the answer -
Comment from : xgozulx

Would these numbers change depending on the material of the surface and coins, ie would more bouncy coins have a different outcome than non-bouncy ones, and if so, how mucch
Comment from : Piers Hanson

He didn't mean a dice He meant a die Dice is plural
Comment from : Kalridian

I think that 'Radius/Thickness' ratio should be 2
Comment from : Ved Kotadiya

The RAND Corporation toyed with this in the late 1940s Source: Prisoner's Dilemma: John Von Neumann, Game Theory, and the Puzzle of the Bomb by William Poundstone, Anchor Books, New York, 1992
Comment from : craigcarroll2000

There's a math answer, not just an experimental one But it's to do with the way the edge interacts with the facets too
Comment from : Kaiser Basileus

I… I don’t think you needed a statistician to tell you that it should be somewhere in between
Comment from : Jack Dog

Surely a 3 triangular prism with the triangle faces pinched to points would make a better d3?
Comment from : Jack Dog

I laughed when I watched the video and I laughed even more when I read the comments below - all the standard jokes about engineers, mathematicians and statisticians are included, plus a few new ones And we still don't know the answer - not only that, but there doesn't seem to be a way to get it, other than by trial and error OK, let's watch episode 2
Comment from : Chris England

Okay, so it's been 45 years What's the answer?
Comment from : Michael

if anyone throws me precise measurements and a formula in the repluies im gonna use a lathe and make some coins to flip
Comment from : LewggsvanSNAD

did an answer develop?
Comment from : Yooless

I wonder if you need more corners if every surface is slightly concave, it may enable the dice to be more "decided"brbrdesided?^^
Comment from : Yooless

It's a die Plural: dice
Comment from : Stuart White

I’ve flipped a coin and it landed on its edge brbrWhat were the odd of that happening please Matt
Comment from : darren collings

Two mathematicians walk into a pub with a cup full of three sided dice
Comment from : LEGO head 27

Did the three sided coin problem die a death? The think math link is down at least
Comment from : JohnyDL

If this has proved one thing, it's that "thick coins" are not suitable as dice Every coin that lands on its edge has a probability p > 0 to roll against something (wall, uneven surface, other coins) and eventually fall over But the probability for a coin that had landed on heads or tails to end up standing upright is much smaller because it stops moving almost immediately This has to be compensated by increasing the probability of falling onto the edge in the first place But there is no way to do this for all possible surfaces and environments, so every "thick coin" will be unbalanced ("unfair") in all but one very specific environment
Comment from : Fritz 46

Interesting how in the thin test, even ignoring the side flips, it seems like you have a statistically significant bias towards flipping Os vs Xs Your two values are 3-4 standard deviations away (assuming a poisson error just to get an order of magnitude)
Comment from : thisisnotdom

The last time we saw his head not shaved
Comment from : Mutoh

He means “die” (0:47) “Dice” is the pluralbrbrDIE (n)The singular of "dice" We seldom hear the word, because there is a prohibitory proverb, "Never say die" At long intervals, however, some one says: "The die is cast," which is not true, for it is cut The word is found in an immortal couplet by that eminent poet and domestic economist, Senator Depew:br A cube of cheese no larger than a diebr May bait the trap to catch a nibbling miebr— Ambrose Bierce, “The Devil’s Dictionary”
Comment from : Michael Lorton

you put a dice with only 3 numbers and each number is in 2 sides
Comment from : Charles Boys

I would point out that the surface you are throwing a coin influences how likely it is to land on it's 3rd side Since it can be seen as an asymmetric dice, how fair it is is dependent on the surface it landsbrbrTo put it rigorously, you have to specify that the coin needs to be thrown in a perfectly flat plane
Comment from : Alkis05

Me in 2018: "Oh, how interesting! I'm really looking forward to the next part"brAlso me, 4 years later: still waiting 😭
Comment from : David S

Ignoring the difference due to the rolling and flipping the “coins”, where I’msure physics and moments of inertia come into play, I’m curious to know how the three sided coin would be different from a 3 coloured die where 4 “edge” sides have the same colour?
Comment from : Sayu Sekhar

Me trying to chose my Pokémon Legends starter with a three sided coin
Comment from : Duskplains

Just roll a D6 and take a third from that 12-34-56
Comment from : WAMTAT

Did you get the data yet
Comment from : nordykid

I hope you realize that the number of dice in each throw will also affect the probability of landing on edge Dice that land on edge will be counted as landing on their face if they happen to roll into another die, tip over, and stop
Comment from : Fourth Root

So, has this been solved?
Comment from : Henrix98

You rolled the coins like dice, rather than flipping them like coins
Comment from : Nerd Made

The relatively large difference in the counts for the circular ends itself looks weird How these counts are going to come out will depend on the tossing process and the loading into the cup When throwing a die whose faces are all symmetrical, it will, pardon the pun or not, shake out You probably do not have a random tossing process Jen could calculate how likely that is Fix the tossing process before making any more empirical tests
Comment from : Seeking The Love That God Means

To have a equal probability the surface area for landing needs to be equal brbrCentrifugal force also skews the outcome The coins that land on 3 having centrifugal force are less resistant to land on 1 or 2 rolling off the table
Comment from : Matt Something

Obviously, you should have used 1/3 the vorea of the sphircle instead
Comment from : cfsterpka

8:03 What is that bit of notation? They keep talking around it
Comment from : alissa mower clough

Maybe I'm being a bit too simplistic for this, but maybe you should just try a thickness of 5D? That way the heads, tails, and edge each have an equal surface area in the side-on view you used
Comment from : dmdz

isn't there a flaw in this experiment with some coins interacting with others effecting the outcome of the "spin"shouldn't they have been flipped individually, its fairly random,,but there is still interference,,
Comment from : whoarethebrainpigs

my confusion lies in the orientation of the ' flip' in one rotation , the heads and tails have ( if you discount the improbable edge landing for now ) 180 degrees each of the rotation upon which to win the toss i imagine that the angle at which it lands need not deviate from the mid point by much to topple it one way or the other widening the base of the mid point ie by widening the edge would alter the angle at which it topples due to an altered symetry of weight distribution there is conveivably a potential for it to land on it's new ' edge ' where heads or tails meet the disputed edge the heads and tails being flat plains and the edge ( the original edge not the new edge ;) ) being a 360 degree flat plain ( yes i know its a circular section of a cylinder but the bit that rests on the table after the flip is flat ) so to not alter the angle of collapse the width of the edge i think would need to equal the width of the heads and tails herein lies a problem because at this point you have created an unequal share of the potential results with 2 quarters going to the edge and one each to the heads and tails in the other rotation whereby heads and tails remain in situe and the edge rotates, the edge has 100 of the potential to win the toss in conclusion i would say that you are missing some variables ;)
Comment from : Mark Darkling

Is it considered a coin or a die?
Comment from : Jared

The throws of ten "coins" had too much interaction between each other, with the probability of a coin being knocked onto 1 side by another, causing lower numbers than ideal
Comment from : Matt

Did someone calculate the p-value for the second case?
Comment from : Francisco Bruno Dias

What about The value of the angularmomentum, when the coin edge hitting The ground, makes a diffrence whether The coin flips over The edge or notbrbrThe the propabilityfunction affects how often The coin flips over its edgebrbrHence The probability of each side landing on the ground depends on how the angular Momentum is distrebuted after rolling the dice
Comment from : SN

R for the circle should be 1 and the hight of the cylinder should be 1/2
Comment from : Alomir K

Great video, but these people may as well have been talking Martian for all I know! :D
Comment from : RichardM8422

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 videobrAlso we did testing to verify that its true, its not just theoretical
Comment from : Miguel Nasarre Budiño

From a physics perspective, shouldn't you just make sure the center of mass is on equal distance from every surface, seems the simple solution to me And calculatable relatively easy
Comment from : Abihef

I like this
Comment from : Frances Bernard

Pi
Comment from : 2LPfan

Sorry to be the bearer of bad news but I think this won't work because you'd need to standardize the flipping method, whereas a normal die(dice) or coin doesn't have any such restriction as long as the flip is sufficiently chaotic along the right dimensions Perhaps Persi Diaconis could help
Comment from : Clayton Rabenda

wen update
Comment from : Luis López

Since 410 was too fra on the edge:brWhich number would be acceptable from a thousand throws?
Comment from : Morten Nørgaard

The answer is 2 Because it depends on when the center of gravity of the coin passes on the vertical of the edge To give a probability of 1/3 the angle has to be 60°, this makes a 1 unit high cylinder for 2 units diameter
Comment from : Septentrion

I suspect the result depends on whether the landing is an elastic or inelastic onebrIf you have a completely inelastic landing (no bounce at all, the coin hits the table, stops, and settles to a stable equilibrium) I think the solid-angle calculation (using the center of gravity of the coin as the center of the sphere) would be the correct onebrbrOn the other hand if the landing is completely elastic, the coin would never stop bouncing The bounce height would vary though at each bounce as a cocked landing would transfer energy between spin and linear motion/heightbrbrThe intermediate case, where the coin bounces but loses some height each time, is the most realistic, but I think it must involve some consideration of the height of the center of gravity on each bounce If the coin lands on edge, it only needs a little residual energy to raise the center of gravity enough to topple onto one of the faces On the other hand, if the coin is nearly settled on its face, it takes a lot more energy for it to be able to pop up again onto its edge My mind is heading over the statistical mechanics where the probability of being in any particular state depends on the negative exponential of the state's energybrbrThis is all assuming that the toss generates a completely random orientation, which may not be the case It also ignores aerodynamic considerations
Comment from : Kevin Martin

Centre of mass is lower if it lands on a face so a tumbling coin is less likely to bounce from face to edge than vice versa However, centripetal force will keep it upright if it’s tumbling lengthwise (like a wheel staying upright when it’s moving)brbrIt’s not just picking a random point and concluding the areas have to be thirds because the coin is tumbling when it landsbrbr(I’m too lazy to switch these 2 paragraphs)
Comment from : White Shadow

Still nothing new?? Come on Matt…
Comment from : lasser

It seems as though the required thickness would vary depending on how you actually do the flipping
Comment from : Dan Wylie-Sears

Ok this is a good mathematically approach, but it's not only the point it's landing on The size of the face also matters The tape will land on the round side way more often than standing up That's because is can easily rest on the round sides but is unstable standing up When you throw it it will have angular momentum which is enough to push it from standing to sitting but not from sitting back up to standing
Comment from : Thinamo

"That's how engineers are born" Nearly correct, but you forgot to mention the ritual in which you publicly acknowledge that e and pi are both the same as 3
Comment from : Marcus Willbrandt

Well the issue with Mr Hunts approach is that his calculations are only true for a rotation around the non cylindrical axis of the cylinder For a rotation on the cylindrical axis the probability of the - side is higher ( not a 100 because of rotational wobbles ) which nicely aligns with the experimental findings So we would have to calculate the probability of each rotational axis happening in the throw of a dice
Comment from : TheKoenigsParkour

The trite submarine certainly decide because llama pharmacodynamically shiver apud a abounding beach scarce, puzzling chinese
Comment from : Joanne Howard

stupid coin story: So a buddy of mine was playing fire emblem three houses, and he couldn't decide which one to play, so we flipped a coin and it landed on its edge, it was a nickel though So he ended up playing as neither and picked the only house he didn't want
Comment from : Noah Zoldak

dropping a coin from a cup is not flipping a coinbrbrflipping a coin makes it flip around an axis along the diameter of the circle doing so causes the rim surface to be a flat plane so, to make this flat plane equal in chance to be landed on as 1 of the other 2 sides, it'd have to be as thick as the radius of the circle, because there are 2 of the previously mentioned flat planes to be landed on 2r is equal to 1d, clearly where it gets trickier is considering angular momentum, as landing on the rim is equally likely, but it is more likely to tip over and fall onto another side if it lands on a rim edge this is where the problem gets complicated, and i'm not sure how to overcome this problem besides compensated weight distribution
Comment from : Nathan Himmerich

sqrt(3) Holds in a situation where all angular momentum is lost at the moment of contact with the table I think a better lower bound would be found, if the angular momentum of the coin is taken into account Take the range of most common angular momentums of the coin spins while tossing it Find the probability to have each of the momentums(probably a bell curve), find the angle of landing such that the momentum can bring the coin upright, find the probability on landing such an angle, and calculate angle1prob * angle1momentumprob + angle2prob*angle2momentumprob + + angleNprob * angleNmomentumProb This could give a better lover bound, as it takes some of the physics into account
Comment from : marko9900

@Stand-up Maths 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
Comment from : J Smith

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
Comment from : J Smith

Pedant here "Die" is the singular of "dice"
Comment from : RWB

My guess is a cylinder with an equal height and radius or maybe a slightly smaller height than radius
Comment from : Your Greatest Ally

I think the math is probability of LANDING on a side, but not of staying on that side brbrIf we assume we have a coin on its edge, a sufficient amount of energy will tip it and cause it to expend its gravitational potential energy brbrMeanwhile if we have a coin that is already on a face, it takes a significant amount of energy to overcome the gravitational "hill" required to become upright again brbrThis skews the probabilities, but the actually number is going to depend on friction, and kinetic energy as well as area
Comment from : Andrew F

The variance between the heads and tails only without regard to the third side suggests there is something wrong with the experimental methodology They should go back to the drawing board on this one
Comment from : Andrew Ferguson

What if you work out the angle the coin should hit the ground in to land on one of it's sides? What do we get if we set that angle to 180°/3 and solve for the thickness
Comment from : ThisIsNotMyRealName

Wow, I just learned about this stuff in school
Comment from : James Marker

Still no result video?
Comment from : Alexander Assies

I believe a 'fair' die must consist of equal surface areas symmetrically distributed around a single point or a single line (as in a wheel) Therefore, a coin with any thickness can never truly be a 'fair' D2 die A 'fair' D3 die might be made from a D12 with 4 of the faces marked as either A,B or C
Comment from : Cran Cowan