## Coin Conundrum

Posted by rajesh on July 7, 2011 – 1:14 am

The chances of a tossed coin falling head or tails is almost 50-50 with an extremely small chance of it falling on its edge.

How will you make a coin so that it has a probability of 1/3 to fall heads, 1/3 to fall tails, and 1/3 on its edge?

July 7th, 2011 at 3:03 am

Just make it thick?

July 7th, 2011 at 4:13 am

Yes, Chono, but how thick?

I would say that, over the full 360 degree range of possible landing angles, the centre of gravity (CoG) would be such as to cause it to land on either side or its edge one third of the time in each case. So the angle at which the coin lands on its final bounce determines its final position.

This angle should favour the near side two to one over the coin’s edge. This is because if the landing angle favoured the far side on another toss then it, too, should favour that side two to one over the edge. The edge would then have a 1/2*(1/3 + 1/3) = 1/3 chance, as required.

So the CoG has to be at a 30 degree angle to the edge. This means that the coin’s edge has to be twice as thick as its diameter.

July 7th, 2011 at 4:18 am

if i assume that when the size of the surfaces are equal the chance of landing on it is equal i get the following:

pi*r^2=2*pi*r*t or r=2*d

so if you make the coin such that radius is 2 times the thickness then it will fall 1/3 heads, 1/3 tails and 1/3 on the edge.

i am really unsure if my assumption is reasonable though.

July 7th, 2011 at 4:55 am

Further to my post # 2 above . . .

The CoG in the middle of the coin should subtend an angle of 120 degrees to each side and 60 degrees to each of the two edges to give each a 1/3 chance.

The coin’s side diameter then has to be twice its thickness.

That’s two different answers I’ve given, thus doubling my chances of winning the prize.

July 7th, 2011 at 5:01 am

but how thick do your answers have to be to triple your chances

July 7th, 2011 at 11:29 am

just keep the coin on its edge on ground and rotate it.

July 7th, 2011 at 5:20 pm

By making it a cylinder

July 7th, 2011 at 6:33 pm

@Tata, A coin is already a cylinder. No matter how thin the non-circleesque side is, if that is a word, it is still a cylinder.

July 7th, 2011 at 9:51 pm

Here are some possible solutions:

1. You would have to make the coin a 3-sided, 3-dimensional, regular figure, which I have no idea how to explain.

2. You could have a nonagon, with the heads emblem taking up three sides, the tails emblem taking up three sides, and three blank sides.

3. You could make the coin thicker until the edge takes up as much space as the faces do.

July 7th, 2011 at 10:14 pm

Take the coin and divice the diameter by 3 drawing lines on the coin to mark the divisions. Then using a vice or other tool bend the edge of the coin in till it reaches a right angle; do the same with the other side.

In this configuration there are four sides that the coin can possibly land on. Landing on the edges would be opposite either heads or tails; if the edges are opposite tails then the two other sides represent the outcome heads.

Even though the figure has four possible sides on which to land, there are still only three possible outcomes.

weather or not the probability is 1/3 for each outcome would require more than just a few miniutes of pillow thought before bed, but I hope someone can take my idea and show that I am either on the right track or just a hypnogogic pipe dream.

Enjoy,

Kent

July 8th, 2011 at 10:35 am

If we’re able to change the shape of the coin in this scenario. I would simply make a pyamid shape, with the bottom beveled so that it has to land on one of the three sides.

July 8th, 2011 at 10:41 am

Nope, the pyramid shape woulden’t work.

July 8th, 2011 at 10:43 am

How bout a ball with 3 labled flat spots on it.

July 8th, 2011 at 10:47 am

Or the shape of a three pronged jack. You spin it, and the prong that sticks up after it stops is the one in three that won.

July 8th, 2011 at 12:34 pm

H David. I think the only thing you are meant to do is to change the thickness of the coin. Wiz’s post 4 looks to me to be the right idea, but I haven’t checked his geometry.

July 8th, 2011 at 1:09 pm

i am really in doubt cause i am not gonna to mess with the center of mass and gravity but simply i would suggest that make the thickness of the edge half the diameter of the coin. Well by doing this we made the probability distribution as

sum of probability of coming head and tail=probability of coming the edge.

So, this time change the thickness of the edge to one-third the diameter. Now, the probability of coming edge=probability of coming head=probability of coming tail.

July 8th, 2011 at 1:16 pm

A three-sided die IS possible.

Think of three spheres sitting on a plane, intersecting at 120 degrees from each other.

The common space for these three spheres is a 3-sided die.

Another way to view it is to consider a triangle lying on a plane (probably in first class) … think of point centered on the triangle but away from the plane (+z axis). the lines from the point to each corner of the triangle is curved convex. Now add a point the same distance on the -z axis.

Each side of the die is a curved rocker I guess.

July 8th, 2011 at 1:19 pm

Another 3-sided die…

Consider a three-sided pyramid (which also has a bottom).

However, the bottom is rounded so that the bottom is never able to be on the bottom.

clear as mud?

Sorry if the words I use to describe what is in my head fails to do so.

July 8th, 2011 at 1:46 pm

If we simply want a three sided die, then an infinitely long Toblerone bar would do the trick. But the question makes it clear that we are looking at a coin like shape, not any shape that comes to mind.

I can say with certainty that a coin of the right thickness will do the job. A short cylinder such as a coin obviously is too thin, whereas a long cylinder (like a broom stick) is obviously too long. It is obvious that somewhere in between lies the correct thickness/length.

I think choosing the area of the rim being the same as the area of the usual faces is not precisely the correct trick, but I’m sure that it is in the right neighbourhood.

July 8th, 2011 at 2:07 pm

gotcha.

If we consider the cylinder on edge, with the exact opposite edge directly above it … draw a plane straight up.

If the two volumes are equal, the item has a 50/50 chance of falling on the face or the cylinder side.

I think this might actually provide an advantage to the cylinder side, since it has a 50/50 shot for each face.

If the cylinder, balanced on edge has a 2/3rds chance of falling on the face, then it has 1/3rd chance of falling on the cylinder side. Same for the other edge. This means a 1/3 chance for each (head, tails, side).

Need to think about how the length of the side affects the 2/3rds chance for it to fall on the face versus the edge.

July 8th, 2011 at 2:18 pm

Okay .. I see that Wiz and Jan explained this.

Not sure I can visualize it just yet.

I think I need to consider my thick coin on edge with the side at a 45 degree angle.

If the side is the half the length of the diameter, I think we have what Wiz and Jan described.

If I draw a plane from the point of the coin on the floor, straight up, segmenting the coin in two… the section on the downward facing face of the coin should have twice the volume as the other section.

However, I can’t remember my calculus enough to measure this.

Good puzzle. Thanks!

July 8th, 2011 at 4:40 pm

Hi cazayoux. The problem with the last half of your post 20 is that for the situation you describe, the coin will fall deterministically, not probabilistically. The problem seems very difficult to me – which guarantees that there is a very slick solution I haven’t the faintest idea how to solve it. If the coin was exactly balanced on one edge then I believe that it’s equally likely to fall in either direction, and the smallest deviation from the balanced state will guarantee which direction the coin will fall.

Of course it is possible that the thickness being half the radius is the right answer (I’m definitely not saying that it is wrong) – it just seems too naive to me. I can’t readily think of how to do the mechanics for a coin that is spinning and bashing onto a hard surface. I guess we assume that it is still being flipped normally, and so is only spinning in the head//tails sense.

July 8th, 2011 at 4:46 pm

I think the answer is much simpler – just take a normal die and “unify” its opposite sides (those who sum up to 7). You get what you want. Working with a cylinder is trickier and nowhere in the problem is mentioned that the coin shape should be preserved

July 8th, 2011 at 5:05 pm

Hi slavy. The words after the picture strongly suggest a conventional coin shape. For the die what dies “heads”, “tails” and “edge” mean?

But yes, of course the die shaped coin fits the bill perfectly. The symmetry of the die completely avoids the need to do sophisticated mechanics.

July 8th, 2011 at 5:25 pm

Another suitable shape could be described as a three sided rugby ball or a three sided banana (after being straightened out).

July 8th, 2011 at 6:37 pm

I now know what the answer is. The thickness is over half the radius (that suits my intuition), otherwise the coin is conventional.

However, I am dubious about the reasoning, despite it being made by a high order genius.

July 8th, 2011 at 7:20 pm

My dictionary defines a “coin” as a flat piece of metal, usually disc shaped. So I think that eliminates the more exotic shapes suggested above. Do any countries have any other shape than flat?

So I assume that we are looking for a cylinder, and only its proportions remain to be determined. I’m withdrawing my post # 2 above and going with my answer in post # 4, namely coin diameter twice its thickness.

Seeking experimental verification I took a small (full) paint tin of diameter 65 mm and height 50 mm and tossed it 50 times. I know it’s not a perfect cylinder, having rims etc, but I thought it was near enough. The results were: top 17, bottom 12, side 21. That seems to suggest that diameter double the thickness might be in the right ballpark. My dimensions are close to Chris’s in post 26 but the experimental results suggest otherwise. Maybe a larger sample of tosses is required.

Incidentally, the toss has to be end-over-end to get a fair result, as you would toss a coin. Tossing it with rim vertical means that it will nearly always land on its rim.

I realise that actually testing our theories in practice is contrary to the unworldly spirit of ToM, so this will probably get me charged with treason and thrown out of this group.

July 8th, 2011 at 7:36 pm

In order for the coin (cylinder) to land on any of its 3 surfaces with equal probability, the weight (=volume assuming uniform density of the material) below the centre of gravity must be equal – i.e. the cylinder is sliced down the centre lengthwise, or sliced through the centre crosswise. When considering a crosswise ’slice’ the volume of the semi-sphere below the centre line is pi*r^2*t/2 where t = thickness.

When considering a lengthwise slice, the volume below the CoG is (pi*r^2/2)/t

So, pi*r^2*t/2 = (pi*r^2/2)/t

minimizing the equation will give t=pi*r^2

or the thickness is equal to the numerical value of the surface area.

July 8th, 2011 at 7:39 pm

Correction to my post 28.

When considering a lengthwise slice, the volume below the CoG is (pi*r^2/2)*t

So, (pi*r^2)*(t/2) = (pi*r^2/2)/t

minimizing the equation will give t=pi*r^2

July 8th, 2011 at 7:41 pm

I’ll get it right in a minute :-

Correction to my post 28.

When considering a lengthwise slice, the volume below the CoG is (pi*r^2/2)*t

So, (pi*r^2)*(t/2) = (pi*r^2/2)*t

minimizing the equation will give t=pi*r^2

July 8th, 2011 at 7:52 pm

Hi Ray.

~~For a fairly symmetrical shape like the cylinder/coin~~For any shape at all, any plane passing through the geometrical centre will divide the shape into two equal volumes. So your initial calculations are pointless (sorry).Your pi*r^2*t/2 = (pi*r^2/2)/t is devoid of physical meaning, the RHS has the dimensions of volume, the LHS has the dimensions of length. As my dad likes to say, “you can’t equate apples with oranges”.

Solving your equation => t = 1. I have no idea how you minimized it.

July 8th, 2011 at 7:58 pm

Hi Ray. I see you posted a “correction”. Your “equation” is now

½ ∏ r² t = ½ ∏ r² t, i.e. half of the volume of the coin = half of the volume of the coin. So t can be anything at all, no matter what the value of r. How do you minimize it?

I agree the volume of half a coin is ½ ∏ r² t, no matter how you slice it, as long as that slice is through the centre.

July 8th, 2011 at 8:04 pm

Thanks for the feedback Chris. Not sure if you read my 2 updates, but correcting the RHS to read ((pi*r^2)/2)*t gives half the area*thickness and the LHS (pi*r^2)*(t/2) is area times half the thickness therefore both become apples, or both become oranges.

Hard to describe in mathematical terms (almost 50 years since I did this stuff in High school)

July 8th, 2011 at 8:09 pm

Hi Ray. Sorry, I am being mean to you. You’d better read my post 32, then the penny will drop (groan )

July 8th, 2011 at 8:18 pm

LOL – Groan, indeed.

July 8th, 2011 at 8:39 pm

I’ve now understood the calculation. Wiz’s (post 4) 120° is actually very close to the correct answer (which is a tad shy of 125°). But I think he’s got the angles reversed.

I suspect that Wiz and cazayoux are very close on the underlying physics.

Too stand a chance, don’t take into account spin, bouncing or other such real things. i.e. imagine a non-spinning coin dropping vertically downwards VERY slowly.

As I hoped, it is a slick answer (under the assumptions of the last paragraph).

July 9th, 2011 at 12:41 am

Cazayoux – try finding out about a Gomboc solid – it matches your 3 sided dice, but it is self righting!

http://en.wikipedia.org/wiki/Gömböc

Wizard of Oz – on the subject of coins – the Triganic Pu has its own very special problems. It exchange rate of eight Ningis to one Pu is simple enough, but since Ningi is a triangular rubber coin six thousand eight hundred miles along each side, no one has ever collected enough to own one Pu. Ningis are not negotiable currency, because Galactibanks refuse to deal in fiddling small change. From this basic premise it is very simple to prove that the Galactibanks are also the product of a deranged imagination. Hitchhikers Guide….

July 9th, 2011 at 1:00 am

My thoughts on this – and not being specific in any way…

1. Very Large Number/one of landing on its edge (and staying there…) The fact that the coin is tossed, infers it has momentum, so it would be a miracle to land on it’s side, as any slight off balance/movement would instantly make the coin fall on a face. Work backwards from here….

2. Use a coin with the smallest ratio of diameter to thickness.

So, let’s simplify things. Let’s take a coin as being a metal cylinder x mm in diameter, and y mm thick. With real-life coins, y is much smaller than x, but let’s take “coins” out of that real-life range. With a coin 2 mm in diameter and 20 mm long, you would expect it to land almost always on its edge. With x and y close, say both being 10 mm, you would expect a non-zero probability of landing on each face and on the edge, and you might also expect the probability to vary depending on how the coin is spun, how far it falls, and how much it bounces when hitting the floor.

3. Surely there is a very good chance of it landing on its side. but not staying on its side. Landing to me indicates that it is the first part to hit the surface after leaving your hand

July 9th, 2011 at 1:11 am

A bit more research on coin flipping shows that it is unfair….

http://www.codingthewheel.com/archives/the-coin-flip-a-fundamentally-unfair-proposition

Fascinating article!

July 9th, 2011 at 1:48 am

I’m still thinking of the coin on edge at a 45 degree angle with a vertical plane.

r = radius

If the coin is 2r thick, then plane bisects the cylinder, and there is a 50/50 chance of it falling on the face or the side.

Some have considered that the side is r, but then the plane does not divide the volume at 1/3 and 2/3s.

Solved integral [-1..1] of 1/2 * (1 – x^2) * dx.

The coin must be thicker than r.

I’m still trying to figure out the length.

July 9th, 2011 at 5:21 am

Hi. Cazayoux’s post (#20) first half is correct and is quite independent of the thickness of the coin. Ignore the second half.

If there is a slight deviation from that unstable equilibrium position, the coin will then fall in the direction of the deviation. If the coin was thin, then the balanced position would be when the coin is almost on its edge. If the coin was extremely thick, then in the balanced position, it would almost be on one of its faces. So the thickness of the coin determines the balance angle.

For the expected answer, the coin could be flipped about any axis – it isn’t necessarily spinning heads over tails, but could be spinning about the cylindrical axis too.

The coin could be in any orientation, and there is no bias to point in one direction or another. i.e. we are dealing with a uniform probability distribution with respect to orientation.

As another (lateral thinking type) clue is that you will need to be familiar with Archimedes’ method for calculating the surface area of a sphere in terms of its bounding cylinder.

July 9th, 2011 at 7:34 am

The surface area of a sphere is 4 x pi r squared. It’s volume is 4/3 pi x r cubed, which is 2/3 the volume of the cylinder of the same height and width…. I think.

I’m not going to progress further on this until Chris says my vague recollections of meeting with Mr Archie Medes of Syracuse, NY has been validated.

Although the standout figure is 2/3 – does this mean that the coin needs to have an edge that is 2/3 the diameter??? (Doubtless, not this simple!)

July 9th, 2011 at 7:38 am

I am now following Wizard of Oz into the pit of eternal damnation and gluing some 2p pieces together and trialling a small set of outcomes at various width/diameter ratios!

I was going to start with £1 coins, but realised I don’t get paid enough to waste even one of them!

July 9th, 2011 at 8:01 am

LOL. Hi Karl. Archimedes was able to relate both the surface area and the volume of a sphere to the enclosing cylinder. In each case you can even find the surface area and volume of a “disc” slice through the sphere.

I am quite sure that the theoretical 1/3 will not be verified by experiment. The theoretical value, although very nice, doesn’t take account of many realistic considerations.

I’ll calculate the probabiliy for a real coin, but give me half an hour or so, as I’ve got some work to do on my dad’s PC. I suspect that it will say that you are much more likely to flip an edge than you’ll actually find.

July 9th, 2011 at 9:10 am

For a British 2p coin (since 1992, would you believe!) the diameter is 25.9 mm (so the radius = 12.95 mm) and the thickness is 1.85 mm. According to the naive theory, the probability that a 2p coin will land on its edge is approx 7.61% i.e. way too high. From a certain point of view, this is correct. Sadly the theoretical formula doesn’t take dynamics into consideration. So although the coin might be said to literally land on its edge (in a sense), it most likely won’t stay there as it usually will have some kinetic energy and so topple over after landing.

An “advanced” consideration is that for a 2p coin, when it’s on it’s edge, the CoM is high, and the coin would “prefer” to lie flat so the CoM is low – that minimizes the potential energy of the coin (I wonder if that’s what Ray was thinking of!).

July 9th, 2011 at 10:11 am

By assuming, the more reasonable, regular head over tails flip, I calculate for a 2p coin a probability of approx 2.26% of it “landing” on its edge, Again this refers to the instant at which the coin first makes contact with the table. It assumes that all the KE is magically removed at that instant.

For head over tails flips, to get a probability of 1/3, we need t/r ≈ 0.577. Again this ignores dynamics.

This is also easier to calculate – no need to drag Archimedes out of his grave.

July 9th, 2011 at 3:39 pm

The diameter of the faces of the coin should equal the width of the edge

July 9th, 2011 at 4:18 pm

Hi Your Name. Can you justify your assertion, or is it simply a wild speculation?

July 9th, 2011 at 5:14 pm

Aaaarrrggghh. I’ve got to eat dirt. Wiz had the answer in his post (#2).

Bow and scrape. Wiz gave a better answer than the one that I’m sure Rajesh is looking for (and which I was also going with before post #46). But he made a simple calculational error. 30° doesn’t correspond to t = 2d, it corresponds to t = d / √3 ≈ 0.577 d ≈ 1.155 r.

It also means I’ve made a similar slip in my post (#46), I’m out by a factor of 2.

Explanation. If the coin is only executing a heads over tails flip, then when it lands (think of it nominally being edge down) if the just balanced angle is at ± θ from exactly upright / on edge, then we want 2θ = 180°/3 = 60°. So θ = 30° – just as Wiz said. But when the coin is at 30°, we have tan(30°) = t/d = 1 / √3 => t ≈ 0.577 d.

For the 2p coin, tanθ = 1.85/25.9 => θ ≈ 4.0856°, so the probability of landing on its edge is 2 * 4.0856° / 180° ≈ 4.54%. I hope I’ve finally got that right. I keep goofing because of diameter and radius mix ups (and general stupidity).

So, is anyone feeling up to dealing with the case of the coin spinning about any axis? NB If we were spinning the coin about the cylindrical axis, and the cylindrical axis was horizontal, the coin would definitely land on its edge. I like the official answer because of the neat geometrical arguments involved, and because that version of the problem was supposedly solved in about 20 seconds (mentally) after being heard, by at least one smart dude.

July 9th, 2011 at 6:44 pm

Okay ladies and gents … still stuck on my path with a depressing realization that calculus is just too far in my past.

I have enough to define my formula, but not to solve it.

I know it will involve a substitution such as x = sin(t), dx = cos(t) and possibly afterward even require a by-parts approach.

I’m hoping someone will find this formula fun to solve and help me to do it.

Intgral [0..1] (sqrt(1-x^2) – h)^2 dx – Intgral [sqrt(1-h^2)..1] (-sqrt(1-x^2) – h)^2 dx = (1/3)*pi*r^2*(r+h)

solve for h

This is essentially a plane intersecting a cylinder, where the plane is at a 45 degree angle from the side (and face), intersecting the edge of one face and on the other face is in between the center and edge. The length of the side is r + h. I’m trying to find h where the plane separates the volume of the cylinder 1/3 and 2/3rd. I’m using a integral estimating volumes by using triangles (bh/2) as the dx shape.

I realize this line of thinking for this puzzle has already been dismissed.

While I appreciate that, I’d still like to find h in this formula.

Any takers for helping with the formula?!

July 9th, 2011 at 8:38 pm

Hi cazayoux. If you take a sphere of radius r and wrap a cylinder of length 2r around it (so the cylinder exactly contains the sphere), then cut two planes parallel to the equatorial plane at a distance h1 and h2 from that plane, then the surface area of the “ring” on the sphere contained between the two planes is equal to the area of the ring ptojected onto the cylinder. So that area is, 2πr(h2-h1). For the whole sphere, h2 = r, h1 = -r, so the surface area is 2πr(2r) = 4πr².

A thin ring at the equator obviously has the same area as the ring on the cylinder. But one away from the equator at an angle θ (equator = 0) clearly has a smaller radius (r cosθ), but its width measured on the surface is 1/cosθ times its vertical width (height) – the two effects exactly cancel.

No calculus required. Archimedes worked that out. He also worked out a similar relationship for the volume of the sphere, but you have to bung a right circular cone into the considerations.

Extra big hint. If you paint say 10% of the surface of a sphere and drop it to the ground (in a uniformly randomly way), then the probability of it landing on the painted area is 10%. It doesn’t matter if the painted area is in one piece or has a bizarre shape. The probability is simply the ratio of the painted area to the total area of the sphere.

Night all.

July 9th, 2011 at 11:37 pm

well, let’s just say we were using quarters…

I would say if you took about 5 quarters and taped them together to were the little creases were cut off,

( No cheating )

then you would have pretty much the same chance of getting any of the sides…

Heads,Tales, or the edge.

Although….

the weight of the bundled quarters would probably make it to were it would mostly just fall to the floor….

but at least you would have 1/3 of every side….

( And the edge, too o////o )

July 10th, 2011 at 4:44 am

Hi Mr. Moo Cow. But isn’t 4 quarters better, or 6, 5.317 quarters better? We want to know exactly how many quarters are needed.

“the weight of the bundled quarters would probably make it to were it would mostly just fall to the floor….” Perhaps you haven’t heard of the experiment allegedly performed by Galileo on the leaning tower of Pisa, or seen the demo done on the moon: http://video.google.com/videoplay?docid=6926891572259784994#

July 10th, 2011 at 5:05 am

Hi again cazayoux. I’m sorry, I didn’t really read your post properly last night, so I jumped to a conclusion.

∫(√(1-x²) – h)² dx = ∫((1-x²) – 2h√(1-x²) +h²) dx

The “hard” bit of that is: ∫√(1-x²) dx = (x √(1-x²) + ArcSin(x))/2

I got that via the substitution x = sin(θ)

Then the integral becomes ∫cos²θ dθ = θ/2 + Sin(2θ)/4

I also Googled on standard integrals to remove all pain.

July 10th, 2011 at 6:25 am

OK – after using some cheap, government funded labour, I have the following results for 2p coin flips.

2.1 represents 1 x 2p coin of 1 coin thickness, 2.2, 2 coin thickness, 2.3, 3 coin thickness etc up to 2.15 – 15 coin thickness. It is 2.13 where the height = diameter.

100 coin tosses per “size” (using a keen intellect and excel, with wolfram alpha to verify the excel calculations I hope to show the results as a percentage at some later stage)

This was not a scientific study – the coins were allowed to land on a level(ish) surface. Different people tossed their own stack, with different tossing styles etc.

It could be argued that the results would be indicative of real world conditions!

Heads Tails Edge

2.1 58 42 0

2.2 51 49 0

2.3 52 48 0

2.4 41 58 1

2.5 50 50 0

2.6 48 51 1

2.7 46 52 2

2.8 49 44 7

2.9 40 38 22

2.10 37 34 29

2.11 29 34 27

2.12 32 36 32

2.13 24 18 58

2.14 12 14 74

2.15 4 2 94

The interesting point is at 9 coin thickness – about 3/4 thickness to diameter ratio.

Observations seemed to indicate that kinetic energy / momentum played a part in either toppling the coins off the edge or not enough to show heads/tails, if you catch my drift.

Anyone fancy analyzing these figures?

And would anyone like to buy some 2 pence coins from me that have been stuck together with araldite? Superglue didn’t work so well.

Sundays on duty can tend towards boredom sometimes…!

July 10th, 2011 at 7:06 am

actually there is still an even smaller probability that the coin falls and stays on an edge…

July 10th, 2011 at 8:27 am

Hi Karl. 11 or 12 coins looks better. Thickness = diameter does seem to be in the right neighbourhood.

As a rule of thumb, after N trials expect deviations of the order of √N. So to get 10% accuracy, you’d need to do 100 trials; for 1% you’d need about 10,000 trials.

July 10th, 2011 at 4:14 pm

Hi cazayoux. Even easier way to do the integral:

http://www.wolframalpha.com/input/?i=integrate%28sqrt%281-x%5E2%29%2Cx%29 or better still:

http://www.wolframalpha.com/input/?i=integrate%28%28sqrt%281-x%5E2%29-h%29%5E2%2Cx%29 or even better: http://www.wolframalpha.com/input/?i=integrate%28%28sqrt%281-x%5E2%29-h%29%5E2%2Cx%2C0%2C1%29

July 10th, 2011 at 4:25 pm

Hi Chris, I was very pleasantly surprised to read your post 49 giving me undeserved credit for arriving at the answer in post 2. Undeserved because I’d already withdrawn this in favour of post 4 (see post 27) and also because I’d made a mistake, as you showed.

However, I think you made a mistake in correcting mine, then I made another mistake on top of that. So I’ll need to think a bit more about all this. (I still favour my d = 2t answer in post 4 but I’ll need more justification)

But I must pay tribute to Karl’s heroic effort to seek experimental verification. More food for thought in his results. It cost him a pretty penny too (sorry).

This is a good puzzle. We’re heading for 60 posts and no end in sight.

July 10th, 2011 at 4:27 pm

One suggestion for making the theoretical answer actually be right, is to coat the table with a very sticky and viscous layer. The coin will stop spinning on contact, and then fall over slowly.

July 10th, 2011 at 4:38 pm

Or something soft and yielding, maybe a bed?

July 10th, 2011 at 4:55 pm

Hang onto those coins, Karl. Why don’t you adopt Chris’s suggestion: spread treacle over your dining table and repeat your earlier experiment.

July 10th, 2011 at 6:14 pm

The diagram is supposed to be representing a coin (edge down) at the moment of contact with the table. The /s represent the faces of the coin. The edge isn’t shown as I can’t find a suitable character to do it (them) with. The coin is shown at an unstable equilibrium state. The smallest disturbance will make it fall one way or another. {Aside: I read a nice argument that a pencil standing on its tip will fall over after something like 20 seconds – entirely due to Heisenberg’s uncertainty principle}.

/¦ t

/ ¦ /

/ ¦ /

/ ¦ /

/ ¦ / d

/ ¦ θ /

/ ¦ /

¦/ φ

¯¯¯¯¯¯¯¯¯¯¯¯¯¯

At this unstable equilibrium point, one edge of the coin is directly above the opposite edge. Both of them and the centre of mass (not shown) are on the indicated vertical line. The angles θ and φ add to 90°. t and d are the thickness and diameter of the coin. So tanθ = t/d. If the coin has angle φ between that shown and φ – 180°, i.e. slanted the other way, then it is deemed to have landed on its edge, otherwise it is deemed to have landed on the appropriate face. Assuming φ is equally likely to be anywhere between 0 and 180°, we need φ between 60° and 120° if it is to land on its edge one third of the time. φ = 60° corresponds to θ = 30° and then we have t/d = tan(30°) = 1/√3 ≈ 0.577, or t ≈ 1.155 r, where r is the radius.

But this isn’t the “famous” answer that is usually given. It is also complete twaddle because it completely ignores dynamics.

July 10th, 2011 at 7:15 pm

The famous answer that I’ve been hinting was given by both John von Neumann (a polymath) and John Leech (a mathematician). I’ve read that it took them about 20 seconds to solve.

The solution assumes that the coin can be spinning around any axis and has no preferred axis. In this case it is easiest to imagine the coin being enclosed by a sphere. The circular edges of the coin touch the sphere above and below the equator and they define the tropics (analogously to the tropics of Capricorn and Cancer on the Earth).

When the sphere “lands”, if it is the region between the tropics that touches the table, the the coin is deemed to have landed on its edge. NB if it landed on a tropic, the coin would be in an unstable equilibrium state. The probability, P, of it landing between the tropics is equal to the area of the tropical belt divided by the area of the entire sphere. Thanks to Archimedes, those areas are a doddle to calculate.

If the diameter of the sphere is D, the diameter of the coin is d and the thickness of the coin is t, then the area of the tropical belt is πDt (ta Archimedes). The surface area of the sphere is πD² (ta Archimedes). Using the diagram in my last post, shoud help you to see that the vertical line has length D and that D² = t² + d² (ta Pythagoras). So P = πDt / (πD²) = t/D => P = t/√(t² + d²). We want P = 1/3 => d = 2t√2 ≈ 2.83 t. If you prefer, t ≈ 0.71 r

You can use the same idea with the head over tail only spin, but it somehow doesn’t seem as natural to me – very strange that.

July 10th, 2011 at 7:47 pm

I’ve started to think about the dynamics a bit. I can’t help but think that the potential energy in the final rest state is more important than the landing angle. The coin is probably going to jump into the air again, and there is a greater range of height that will corrrespond to a final head/tail landing than for an edge landing. For that reason, my guess is that the centre of mass should end up at the same height, regardless of which side the coin comes to rest in. That leads to t = d => cazayoux can have his 45°. That also seems to be consistent with Karl’s experiments and the ridiculously high probabilities associated with normal coins i.e. theory => about 5%, practice => about 0.02%).

On the other hand, if the coin starts rolling, I expect it to stay on its edge, and that seems to be advantageous to that state. I’ve decide that I’d rather use slavy’s die after checking it very carefully. See this Wiki: http://en.wikipedia.org/wiki/Dice There’s some quite amazing ways of loading dice.

July 11th, 2011 at 3:44 am

If D is the height of the vertical line (see post 63), then it is also the diameter of the circumscribing circle. Then D² = t² + d². For θ = 30°, d = t √3, so D² = t² + 3t² => D = 2t. I suspect that Wiz made a mental slip way back in post 2, and so said “2″ rather than “√3″, i.e. I suspect that he got his sines and tangents mixed up.

Note that, by analogy with post 63 and the areas associated with the circumscribing sphere, that the arc length of the edge(s) is 1/3 of the total circumference when θ = 30°.

July 11th, 2011 at 3:58 am

These geniuses (genii?) like von Neumann and Leech can see more in 20 seconds than we mere mortals can see in several hours. They’ve gone straight to thinking of this problem in three dimensions whereas I never got beyond two.

I’ve been thinking of the spinning coin as equivalent to a rectangle so that the problem was merely one of determining its height and width. This would be true if the coin was spinning about a horizontal axis. However, the way I see it now is that if the axis was not horizontal then there would be a spin component at the moment of contact with the landing surface. In other words the landing edge of the coin, because it’s moving sideways, changes the final angle of the coin and hence the probability of ending up on its side or rim.

I may have misunderstood what the genii were getting at but it does seem to make sense of their answer. Having said that, I’ve no idea if their answer is right or wrong, but I’m prepared to go along with it.

I think Karl should test this with a million tosses of each of his 2p stacks and let us know the outcome.

July 11th, 2011 at 4:29 am

Hi Wiz. I have no doubt that for the fully randomly tossed coin, that Neumann and Leech have the correct answer for the initial landing. I have no doubt that for the head over tails flip, that you (despite a minor calculational error) have that correct answer for the initial landing.

But, both are wrong when predicting the final resting state of the coin. I’m reasonably sure that d = t is closer to the truth. I’m also sure that such a coin would readily facilitate cheating.

July 11th, 2011 at 1:37 pm

Basically make every surface the same size including the surface.

July 11th, 2011 at 6:36 pm

How many dimensions, n, would space have to have so that you could fit a thin 5 metre long rod into a n-dimensional box of sides 1 metre?

July 13th, 2011 at 11:23 am

Drop the coin into a sand box

July 15th, 2011 at 3:32 pm

Answer: 2R/root(3)

Solution: think in 2d

The center of gravity line of coin must be with an angle 60 degree to both the thickness side, so that total 120 is equal to 120 degree for bothe faces to make it 1/3 probability. Rest is simple trigonometry for a right angle triangle.

July 15th, 2011 at 3:52 pm

Hi Rajesh. You surprised me there. I thought you were going to go for the famous 3D solution.

July 15th, 2011 at 6:23 pm

Chris, it is 3d , but if you think in 2d it’s easier to comprehend and the solution.

July 15th, 2011 at 6:42 pm

Hi Rajesh. The 3D and 2D solutions are different. They are both very easy to understand and calculate. I say your solution is 2D in the sense that you only have the coin spinning about a diameter (just like an ordinary coin flip) – now I think about it, perhaps that’s only 1D. The 3D solution allows for fully arbitrary spins. See my post 64 for the 3D solution; it gives t = r / √2 ≈ 0.71r rather than t = 2r / √3 ≈ 1.155r

By the way, thank you for posting it. It’s a good problem. I wonder what the real answer is though.

July 15th, 2011 at 6:55 pm

Mini problem, post 70: How many dimensions, n, would space have to have so that you could fit a thin 5 metre long rod into a n-dimensional box of sides 1 metre?

Answer: 25. Using Pythagoras in n dimensions = 1² + 1² + … + 1² = n*1² = 5² => n = 25.

July 16th, 2011 at 4:26 am

If we can change the shape of what we generally consider to be a coin, why don’t we just get a cube and label 2 sides “Heads”, 2 sides “Tails” and the remaining 2 as “Side”

2/6 = 1/3

Done.

July 16th, 2011 at 7:00 am

make the thickness of the coin the same as diameter of the coin’s flt side

July 23rd, 2011 at 9:40 pm

that’s easy! just make a coin that is as thick (measuring between each side) as each side is wide (in diameter)

July 24th, 2011 at 4:47 am

Hi rhiannon. It’s not [that] easy at all. Prove that what you say is right i.e. prove that that thickness does give 1/3 precisely. It happens that I think your answer is close to the right one – except that there cannot be a [single] right answer.

I doubt that it is possible to answer this question without at least carefully defining the nature of the surface that the coin lands on, and without a very good grasp of mechanics and dynamics.

Despite that, the two different, reasoned, solutions given are pretty sweet.

February 22nd, 2014 at 7:50 am

I’ve gone off the 3D solution because it considers the coin to have landed before it touches the surface. The 2D one doesn’t have that issue.

To see my reason, actually hold a coin at an angle on a surface – just by moving your point of view so that you see the coin edge-on should help you see that only the 2D solution works.