The odds of a coin landing vertically + 51/49 theory

[Lord Antares]

Is it possible or easy at all to calculate mathematically?

 

If you considered only the initial landing position and ignored the physics that happened thereafter, would it just be surface area of the sides X and surface area of the width Y where the odds would be 1 in (X:Y)?

Of course, if you included actual physics, that would reduce the odds as there is a considerable chance that the coin would flip over after landing vertically due to momentum, angle, etc.

 

Also, while I'm at it, another question (but pls answer the first):

 

I've read a few times a supposedly true fact that the coin has a 2% higher chance of landing on the upper side due to that side being upwards the same or one more amount of times compared to the lower side.

However, this does not make sense to me as no one would consider 0 flips to be a valid coin toss, obviously. Neither would they accept 1 flip as valid, most likely.

So it depends on the minimum amount of flips one deems to validate the coin toss. I would think that number would be 2, if I had to guess, so that would bring us again to the upper side having the advantage; but wouldn't it have a higher advantage than 2% then?

 

Surely, if you consider the minimum amount of flips that makes a coin toss valid + the realistic amount of possible coin flips in a real toss, wouldn't it be hard to arrive at the value of +2%? If I'm wrong, please explain why that specific value.

 

I have zero education in mathematics, so forgive me if I'm missing some major, well known equations. Odds and probabilities fascinate me and I like to think of them and try to apply them in games/uncertain situations etc. so answers to questions like these are most welcome.

[Prometheus]

Possible, but not easy.

[swansont]

Is it possible or easy at all to calculate mathematically?

 

If you considered only the initial landing position and ignored the physics that happened thereafter, would it just be surface area of the sides X and surface area of the width Y where the odds would be 1 in (X:Y)?

Of course, if you included actual physics, that would reduce the odds as there is a considerable chance that the coin would flip over after landing vertically due to momentum, angle, etc.

 

 

 

In this description, the initial landing position will often not be the final orientation ofd the coin, since the coin will bounce. Landing flat on an edge or surface have a vanishingly small probability — virtually all flips will have the rim hit at some angle. So you would need to define what counts as landing as heads or tails, and what counts as landing on the edge. Maybe within 1º? Then you might assume that the odds are 1 in 180.

 

But that wouldn't necessarily be correct. You would also need to account for the dynamics of the toss. You could engineer it so that the toss time and rotation speed never allow it to hit close to being edge-on, or possibly always have it hit edge-on. It's straightforward kinematics to find the toss time, related to the height, and the pick the rotation speed accordingly.

[Lord Antares]

Yes, that's why I meant. That's why I said it would be different if you accounted for physics. I was pondering the question in a more mathematical manner, i.e. counting the initial hit position, disregarding what happens afterwards. Those odds would be significantly higher than if you included physics, no?

 

And you are correct, I need to define what counts as landing on the edge or the sides. I guess the corner edge of the coin would be the boundary between the head/tails side and the width side. Then 1 in 180 couldn't be correct as you have to account or the width of the coin. If 1 in 180 is correct, then it would have to apply the same for both narrow and wide coins, which cannot be right.

Are the odds then simply the ratio of the surface of the sides of the coin divided by the surface of the width side of the coin?

 

I understand that the odds in a real coin toss are significantly lower to land vertically as it needs to be at the right angle; completely vertical.

 

@Prometheus - I am sorry, I said I was illiterate in mathematics so I don't understand what these symbols and signs mean so I got lost in the equations.

[Prometheus]

@Prometheus - I am sorry, I said I was illiterate in mathematics so I don't understand what these symbols and signs mean so I got lost in the equations.

 

I thought the first link was OK, only gets 'mathsy' in the figures.

 

But if you're interested in these kind of problems why not learn some maths? It's hard work but it will allow to look at these problems in far more detail, and can also help look at other, apparently unrelated, problems.

[Lord Antares]

Yes, I was able to follow the text with no problem but the solutions in the boxes contained unknown symbols to me. Maybe a quick google search would help that, to be honest.

You are right, learning math is the most logical solution to this, but I am in such a position that if I need to learn something extensively, I need to focus on something that will directly help my life (pogramming, re-entering college that I am qualified for etc.)

[swansont]

Yes, that's why I meant. That's why I said it would be different if you accounted for physics. I was pondering the question in a more mathematical manner, i.e. counting the initial hit position, disregarding what happens afterwards. Those odds would be significantly higher than if you included physics, no?

 

And you are correct, I need to define what counts as landing on the edge or the sides. I guess the corner edge of the coin would be the boundary between the head/tails side and the width side. Then 1 in 180 couldn't be correct as you have to account or the width of the coin. If 1 in 180 is correct, then it would have to apply the same for both narrow and wide coins, which cannot be right.

Are the odds then simply the ratio of the surface of the sides of the coin divided by the surface of the width side of the coin?

 

That's why I asked what counted as landing on the edge. It will depends on details you haven't provided.

[DrKrettin]

That's why I asked what counted as landing on the edge. It will depends on details you haven't provided.

 

 

Surely the probability of landing on an edge is in fact 1 because it can't spin and fall and land on a face. When that edge hits the floor, the coin falls over onto a face. Probably.

[swansont]

Surely the probability of landing on an edge is in fact 1 because it can't spin and fall and land on a face. When that edge hits the floor, the coin falls over onto a face. Probably.

I was drawing a distinction between the edge (flat) and the rim where the edge meets the head or tail.

[Lord Antares]

No, I guess I wasn't clear enough in my description. I don't mean it landing on the edge, I mean on the width surface, i.e. the narrow area between two edges of the coin, the one that has horizontal stripes over it (at least in my country).

 

I realized the probable solution after I posted this, the previous one was wrong. It is simply to draw a circumscribed circle around the coin with the said width area directly facing you and measure the degrees out of 360 that the width takes up, divide by 2 (for the 2 width sides) and the answer is - the odds are 1 in (360/result), right?

But that implies perfect 2-dimensional spin for the coin. You would have to include physics for 3d calculation.

 

Also, now that I think of it, the real physical answer (expanding beyond initial impact position to real coin toss rules) is closer to 1 in 180, as swan said, because it would require the coin to land perfectly vertically in order not to tip off.

[Janus]

Yes, that's why I meant. That's why I said it would be different if you accounted for physics. I was pondering the question in a more mathematical manner, i.e. counting the initial hit position, disregarding what happens afterwards. Those odds would be significantly higher than if you included physics, no?

 

And you are correct, I need to define what counts as landing on the edge or the sides. I guess the corner edge of the coin would be the boundary between the head/tails side and the width side. Then 1 in 180 couldn't be correct as you have to account or the width of the coin. If 1 in 180 is correct, then it would have to apply the same for both narrow and wide coins, which cannot be right.

Are the odds then simply the ratio of the surface of the sides of the coin divided by the surface of the width side of the coin?

 

I understand that the odds in a real coin toss are significantly lower to land vertically as it needs to be at the right angle; completely vertical.

 

@Prometheus - I am sorry, I said I was illiterate in mathematics so I don't understand what these symbols and signs mean so I got lost in the equations.

If you want to factor in the thickness of the coin, then you can try the following definition for what "side" you consider the coin as landing on. If you are going to ignore the rotational momentum of the coin on impact, then this is the same as balancing the coin on one of its corners and then letting go. Which way does it tip? To determine this, you draw a line from the corner of the coin to its center of gravity. If the lean of the line away from vertical is towards one of the sides, then the coin when released will tip in that direction, if the lean is towards the edge it will tip in that direction. Or put another way, does a line drawn down vertically from the center of gravity pass through an edge or side?

 

If this is what you use to decide between landing on a edge vs. a side, then the odds could be calculated by atan(h/D)/90. Where h is the thickness of the coin and D is the diameter and atan uses degrees. So for example, a penny, with a diameter of 19.05mm and thickness of 1.55 mm would have a 5.17% chance of "landing" on a edge, and a nickel at a diameter of 21.21 mm and thickness of 1.95mm would have a 5.84% chance of doing so.

[DrKrettin]

I think it very appropriate that somebody with the username Janus takes part in this thread.

[Acme]

Probability of a tossed coin landing on edge by Murray, Daniel B.; Teare, Scott W., Physical Review E (Statistical Physics, Plasmas, Fluids, and Related Interdisciplinary Topics), Volume 48, Issue 4, October 1993, pp.2547-2552

 

Abstract

An experiment is reported in which an object which can rest in multiple stable configurations is dropped with randomized initial conditions from a height onto a flat surface. The effect of varying the object's shape on the probability of landing in the less stable configuration is measured. A dynamical model of the experiment is introduced and solved by numerical simulations. Results of the experiments and simulations are in good agreement, confirming that the model incorporates the essential features of the dynamics of the tossing experiment. Extrapolations based on the model suggest that the probability of an American nickel landing on edge is approximately 1 in 6000 tosses.

[Lord Antares]

Thank you, Janus, for the explanation. I was sure you only need to measure the degree value of the width of the coin and divide 180 by the result but you managed to confuse me with atan

 

@Acme - So this is the actual result when you account for real-life physics, instead of just mathematics? That's lower than I thought compared to the 5+% result Janus gave just for the impact odds.

[Acme]

...

@Acme - So this is the actual result when you account for real-life physics, instead of just mathematics? That's lower than I thought compared to the 5+% result Janus gave just for the impact odds.

What I gather from the abstract is that the researchers did experiments -real-life- and the math. I have had the page bookmarked for a number of years because I once saw a nickel land on edge at a bar and I was curious about the odds. (There was quite a squabble about how to settle the 'bet'.) As far as I was able to determine, the actual paper is behind a pay wall.

[Janus]

Thank you, Janus, for the explanation. I was sure you only need to measure the degree value of the width of the coin and divide 180 by the result but you managed to confuse me with atan

 

@Acme - So this is the actual result when you account for real-life physics, instead of just mathematics? That's lower than I thought compared to the 5+% result Janus gave just for the impact odds.

atan is how you find the degree value of the width of the coin. atan is the reciprocal function of the tangent. Draw a triangle that uses one face of the coin for one side, the edge of the coin for another and the diagonal as the third. Dividing the edge width by the diameter of the coin gives you the tangent of the angle between the face side and the diagonal. The atan of this, is the angle value.

 

Granted, with a thin coin, you can get pretty close to the same answer by taking pi times the radius of the coin, dividing this into the edge width. If you do this for the nickel you get a 5.85% chance vs. the 5.84% chance I gave above. If you can live with this 0.01% difference, fine, But be aware that as the thickness of the coin grow with respect to the diameter of the coin, the difference gets larger.

 

For example, if you had a disk with an edge width equal to its diameter, and balanced it on its edge, what are the chances of it falling in one direction over the other. it is 50% however, if you use the "divide the width by pi times the radius method, you calculate a 63.66% of it falling towards the edge vs. the face. However the atan of 1/1 is 45 degrees and 45/90 = 0.5 = a 50% chance.

[Lord Antares]

Thank you for this. I undestand now.

Out of curiosity, could you as well just measure the degree with a protractor and resume calculations?

 

Also, why divide by 90 instead of 180?

[Janus]

Thank you for this. I undestand now.

Out of curiosity, could you as well just measure the degree with a protractor and resume calculations?

 

Also, why divide by 90 instead of 180?

The triangle created by drawing a diagonal from corner to corner and using one face and one edge is a similar triangle to one drawn with two of the sides starting at the CoG of the coin as shown here. Both have the same angle of alpha at the top corner. But as measured from the Cog, this only represents 1/2 of the angle measurement of the edge. ( the angle between red and blue lines. )

 

 

Now you can take the angle arrived at by taking the atan of the ratio of edge to diameter, double it and then divide by 180, or you can just directly divide it by 90.

[Raider5678]

I have zero education in mathematics, so forgive me if I'm missing some major, well known equations. Odds and probabilities fascinate me and I like to think of them and try to apply them in games/uncertain situations etc. so answers to questions like these are most welcome.

High school students have it down pat.

They calculated the probabilities with some rough estimates and immediately applied it to a real life situation.

 

They flip a coin.

 

Heads, they go to sleep.

Tails, they browse the internet.

Stands on edge, they study.