# A coin will land on its edge?

1. Mar 11, 2014

### chromomancer

I've have just registered here in order to post this question. I know it's a good idea to read a
forum for weeks and months (years!) before daring to post, in case a question has already been
discussed in detail -- but it's a burning question! (And it's not homework.)

If it is off-topic, or has been discussed elsewhere, I'd appreciate a link, but I am looking for a serious discussion of what factors might be involved and how the problem might be tackled (I've studied physics at degree level, but that was many years ago).

Question follows:

What is the probability that a coin will land on its edge?

No, it's a serious question. I know that, assuming Newton's
laws, the behaviour is predictable and it's supposed to
come up heads more often than tails if flipped by a human
being, for example. I'm not interested (here) in that kind
of smart answer. Take the question as intended: you know what
I mean, but I'll spell out some reasonable assumptions
below.

It is usually assumed that there is a small, but non-zero,
chance that it can end standing up on its edge.

If the coin is a perfectly symmetrical flat cylinder, and
dropped vertically (in a vertical orientation) onto an
ideal smooth, flat, hard surface, then since there is
nothing to break the symmetry it must end up on its edge.

But what about a real coin?

Assume it is ejected into the system by a sufficiently
chaotic "coin tossing" mechanism that it starts out
from effectively a random velocity, angular momentum,
position and orientation, and drops onto a smoothish,
fairly hard (but not perfectly smooth) surface and the
bounces are not perfectly elastic. Moreover, the initial
horizontal component of velocity is small, but not zero.

The exact form of the initial random distributions probably
don't matter -- but I'll leave you to decide if they do.

I can see two possibilities: either the initial sideways
motion can never be cancelled exactly so the chance of the
coin ending up on its edge are exactly zero (landing on
a carpet, or a sticky surface would be diferent). Or there
really is a small chance that it will stop upright.

My hunch is that it is possible, but the chances are much
less than one in a million. I'm not good (no humans are)
at estimating the probability of very low frequency events
by "common-sense" or intuition. But it may be possible to
do a BOTE calculation to get some idea.

I'd guess it's less than the difference in probability between
heads and tails (real coins are not perfectly symmetrical).

So what I want is an order of magnitude estimate: is it
one in a million, one in a trillion, one in 10^30 or what?

Thanks,
Jonathan

2. Mar 11, 2014

### phinds

I can only comment on this part of your post

In the real-world scenario you have described, your first possibility above is invalid. Given a random toss onto a non-ideal surface there is always the possibility the things can work out so that friction will cancel the sideways motion.

The way it is possible for the coin to land on its edge reminded me of a mechanics thingy that I saw when I was a kid. I had to REALLY think about this one for a while to get that it's true. It seems fairly obvious now, but it sure didn't then. It's this:

The setup:
There is a rod mounted on a swivel on a railway flatcar such that the rod can move 180 from forward, through upright, to backward. There is no friction in the swivel joint. The car can be given any non-infinite acceleration at any time and for any duration and changed at will. It is assumed to ride on a rail line that is straight/flat (not curved with the Earth) under a uniformly perpendicular gravity field. [note: turns out this gravity caveat is probably not necessary, just makes the problem sound slightly less complicated than it might be]. The flatcar is initially at rest and the rod is initially at some angle that is not 0 degrees, 90 degrees, or 180 degrees.
The statement:
There is always a way that the car can be moved so that the rod will end up at 90 degrees with the flatcar stationary.

Maybe it will be obvious to you, but I just had a hard time back then getting my head around how that could be true. (It is, but I wouldn't know how to prove it). It may not even be clear to you why I see this as a similar problem to your coin, but I do.

3. Mar 11, 2014

### chromomancer

Before I reply more fully (I probably won't have time today), a quickie question about your flatcar:

My question: is that in finite time?

It's obvious to me that you can get the rod abritrarily close to 90 degrees, moving arbitrarily slowly: it's the same problem us humans -- well, me anyway :) -- have in standing up, except only in one dimension instead of two (I can only fall over in two directions: it's not a three dimensional problem).

Given infinite time, I can see how the flatcar can get closer and closer to 90, moving slower and slower. (Now I'm wondering if it is obvious, after all.)

Hmmm. I think I just convinced myself it's obvious it can be done in finite time, so I withdraw my question. My reasoning is sort of along the lines like this:

It's a bit like stopping my car when the lights turn red. I know I can slam on the brakes, but can I stop not just somewhere before the lights (and without hitting the car in front), but at a particular position? And I know I can.

Flick the rod so it will go up and over, then it becomes the same problem as me stopping at the lights, the only minor complication is that the rod is slowing of its own accord as well, but that's like stopping on the brow of a hill. I just don't brake quite so hard.

To make the problem a little more challenging, we can now see that not only can you stop with the rod at 90 degrees, and the flatcar stationary, but you can also choose where it will stop!

(That's because if you move the traffic lights, I can still stop at them - that's probably not obvious without a bit of thought. For the rod, you let the rod slope... no, either that makes sense already, or it needs more explanation than I have time for at this moment.)

Bottom line: it's not obvious at all, at first glance, but I believe I could prove it.

Jonathan

4. Mar 11, 2014

### jbriggs444

You've already argued yourself into agreeing that this is possible. I just want to suggest what is (to me) a simpler argument.

Start with the rod vertical and motionless on a motionless car. Clearly you can move the car and knock the rod out of this equlibrium position. So play that sequence backward and you have a recipe for bringing the rod to a vertical and motionless position atop a motionless car.

5. Mar 11, 2014

### CWatters

Perhaps of interest...

http://journals.aps.org/pre/abstract/10.1103/PhysRevE.48.2547

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.

6. Mar 11, 2014

### phinds

Nope. Won't work. Do this: move the car fairly slowly in one direction or the other and wait until the rod is at 1 degree. Now move it in the exact reverse motion and the rod will still just fall over. You would have to accelerate like crazy to overcome it, completely different than the original motion.

7. Mar 11, 2014

### jbriggs444

You misunderstand. By "reversing the sequence", I am talking about starting with a rod in motion [in the opposite direction to the way it ended] and a cart in motion [in the opposite direction to the way it ended] and ending with a rod upright.

Classical physics is time-reversible. It is guaranteed to work.

[edit: Clarified that this is not continuing action from the end of the push]

Last edited: Mar 11, 2014
8. Mar 11, 2014

### Khashishi

I think a recent "Physics Today" had a fairly in depth article on this.

9. Mar 11, 2014

### phinds

Gotcha. You are right in that I misunderstood what you meant. I had to think a bit but I believe that you are right. I was sure at first that you weren't because the initial condition of my problem statement was that the cart was not IN motion.

Do you think your method works if I modify my initial condition to "the rod is held at a 1 degree angle for 10 minutes with the cart motionless, then released and as it is released you can begin whatever motion you wish"? I think it MAY but I'm not as convinced as the case where the cart is simply stopped at the instant that the rod is at 1 degree with an unknown history. I may be overthinking this.

10. Mar 11, 2014

### dauto

I don't believe it. That probability seems too high.

11. Mar 11, 2014

### phinds

I agree w/ you. That seems ridiculously high.

12. Mar 11, 2014

### jbriggs444

If I understand what you are getting at, it seems too simple. So I may be misunderstanding.

You are considering a rod that is momentarily at rest at a slight angle and which then proceeds to move in whatever fashion is consistent with the force of gravity and the motion of the cart beneath it. This naturally results in a [fairly vaguely specified] final configuration at a moment of your choosing.

You then question whether this then demonstrates the possibility of having a rod that is initially in that resulting final configuration, but with its state of motion reversed, and playing back a time-reversed sequence of cart motions to get the rod momentarily at rest at that initial small angle.

Yes, of course that works. You're just playing the movie backwards.

13. Mar 13, 2014

### paisiello2

Why do you say that? Seems a reasonable estimate to me assuming a reasonably flat surface. I suppose that a reasonably flat surface is not a common thing where coin tosses are generally made in which case the probability would then be zero.

14. Mar 13, 2014

### paisiello2

duplicate

Last edited: Mar 13, 2014
15. Mar 13, 2014

### bahamagreen

I can't tell if "probability too high" about the 1:6000 means you think the number is too large (too improbable) or that the probability estimate is too likely...?

I can report that I had a nickel land on its edge about a month ago.

- one casual toss (not an experiment or trial)
- small parabolic arc about a half meter high
- nickel was tossed as part of a handful of small junk in a drawer (screws, paper clips, a few more coins)
- landing surface was Formica counter top
- landing orientation was sideways to the direction of motion (heads facing me)

As far I could tell, the nickel pegged the landing - no rolling or quivering - looked like it just stuck there. I don't see how it could do that unless it got bumped by one of the other things to bring it to rest,,,?

16. Mar 13, 2014

### dauto

1:6000 is equivalent to 1/6000. Probability is too high means that that fraction is too large - The denominator is not large enough.

17. Mar 13, 2014

### dauto

Well, good for you. I think it's too high.

18. Mar 13, 2014

### paisiello2

Again, why do you think that? Just your intuition?

19. Mar 13, 2014

### phinds

I guess I am over-thinking it and confusing myself

20. Mar 13, 2014

### Khashishi

I think I've seen it happen before, so the probability can't be super low.