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The question is fairly simple. Consider the following parameters which are known to you:

1.) Mass of the coin

2.) The force applied on the coin

3.) The point where the force is applied on the coin

4.) $g$ of the place where the coin is tossed (acceleration due to gravity)

5.) Radius of the Coin

Based on this data, how would you calculate the result of the coin toss?

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  • $\begingroup$ I guess you would need the initial position of the coin as well... $\endgroup$ Commented Oct 30, 2014 at 9:55
  • $\begingroup$ Your question tacitly assumes that this is even possible to do in principle. A statistical probability of a certain outcome (a la quantum mechanics) may very well be the best we can do. But if it were possible, you would also need to know how long the force was applied to the coin, not just the magnitude of the force. $\endgroup$
    – Bryson S.
    Commented Oct 30, 2014 at 9:57
  • $\begingroup$ Related: Is it really impossible to calculate in advance the result of throwing dice? $\endgroup$ Commented Oct 30, 2014 at 10:25
  • $\begingroup$ You left out air resistance, which I suspect has some impact on the outcome. $\endgroup$ Commented Oct 30, 2014 at 13:33
  • $\begingroup$ The simplest way to predetermine the result of a coin toss is to have a coin with two 'heads' and one with two 'tails'. You then choose the one you wish to use, and the result is predetermined (except if it lands on edge). $\endgroup$
    – Jon Custer
    Commented Oct 30, 2014 at 22:57

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At the very least you would also need the dimensions of the coin (mass alone is not enough) and the elastic properties of both the coin and of the surface on which you land... that has a big impact (since the coin can bounce or "stick", depending on the surface). It may be possible to predict for certain sizes of coin and height of drop (a carefully executed half-flip of a huge coin) - but as the number of revolutions completed increases, and particularly if the calculated position at the time of landing approaches the metastable position (landing on edge in such a way that it is equally likely to fall to either side) the uncertainty in the result also increases, until you end up with a resounding "no".

In general, then, the answer is "no". For specific cases the answer can be "yes". It all depends...

But if I were asked to do the calculation - which was your question - I would just solve the equations of motion. Assuming that you have the following information

m = mass of coin
I = moment of inertia about length axis
g = acceleration of gravity
h = initial height
d = distance from center of mass where impulse is applied
p = impulse applied to coin

you can then compute the initial vertical velocity from

$$v = \frac{p}{m}$$

and thus the time to landing by first computing the maximum height (which will be reached after $t=v/g$ seconds, and be at $h_{max}=h+\frac12 g t^2$), and then the time from this height to the ground, given by $t_2=\sqrt{2 h_{max}/g}$. You would probably want the time until you reach "ground minus radius of coin" so you can compute the time when you actually strike the surface - this depends on the angular position of the coin.

Next you would compute the angular velocity of the coin from

$$\omega = \frac{p d}{I}$$

With these two numbers you can calculate how many revolutions the coin has completed when it gets close to the ground, and then you would have to figure out when the coin first hits the surface - this is not when the center of mass hits h=0, but rather when a rim of the coin does - and that depends on the angular position.

Finally, you would have to take into account the properties of the surface to determine how the coin behaves when it hits the ground. This can be the trickiest part of the calculation - which is why I started out by saying that the surface properties absolutely need to be part of the input.

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  • $\begingroup$ I think coin tosses are not chaotic and not very sensitive to small perturbations in the initial conditions. $\endgroup$ Commented Nov 14, 2023 at 13:56
  • $\begingroup$ @ThomasKahle I think that depends on how they land. If they land (and bounce) on a hard surface, the subsequent motion is likely chaotic. But I agree that a coin "caught on a soft surface" is not-so-chaotic, as that 2023 study you reference in your Nov 2023 answer showed (the bias is not big, but it's statistically significant in the experiment they did). $\endgroup$
    – Floris
    Commented Dec 1, 2023 at 20:17
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    $\begingroup$ The problem has an interesting history. There is a 'standard model of coin tossing' with soft landing, but then also more intricate computation and experiment on bouncing on hard surfaces. This work is also cited in the DHM paper. But yes, bouncing certainly does not decrease randomness. $\endgroup$ Commented Dec 4, 2023 at 9:02
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    $\begingroup$ @ThomasKahle I was reminded of this recently when I read about an infamous "gang" who figured out that the way the ball bounces in Roulette is not completely random ... they were able to improve their odds of getting the number right, and cleaned up before casinos threw them out. Some information in this article: onlineroulettesites.org.uk/players/the-ritz-laser-cheaters (though other stories claim no lasers were used). $\endgroup$
    – Floris
    Commented Dec 27, 2023 at 17:34
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The answer to the question is a simple “yes”. This has been investigated a few times in the 20th century, but in 2007 Diaconis, Holmes and Montgomery even built a machine that executes exactly the same coin flip with exactly the same result every time.

Dynamical bias in the coin toss (DHM)

Curiously they also found that human tosses are more likely to land on the same side they started on due to the axis of rotation not lying inside of the plane of the coin. Their predictions were empirically verified in 2023 in this study:

Fair coins tend to land on the same side they started: Evidence from 350,757 flips

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Nov 14, 2023 at 15:18
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There's different levels on which you can answer this question:

First, because the coin is a large enough macroscopic system that quantum-mechanical effects are negligible (i.e. we don't have to worry about the wave nature of the coin). The variables that will affect the final outcome of the coin toss are all classical in nature, and so can be measured and calculated classically without a problem. Therefore, if you know the properties of the system that Floris mentioned in his answer (moment of inertia $I$ of coin, height $h$ of coin, etc) you can in principle calculate the outcome of the coin toss.

The real problem here, then, is that a coin toss can quickly become chaotic (in the physical/mathematical sense of the word). Very small changes in initial conditions can produce different results. Very small changes in the distance the flick is applied from the axis of rotation can produce a different result. Because coins also do not have uniform mass distribution, the coin will experience precession and wobble as it rotates through the air. And although I'm not certain about this, I would not be surprised if we even need to consider the fluid dynamics of the air around the coin. Due to the sensitivity to initial conditions that exceeds the amount of precision in our measurements, it's easy to see that the task becomes unmanageable in practice.

Please note, however, then when other answers say this task is impossible, I think this gives the wrong impression. Determining when a single atom of a radioactive isotope will decay is impossible. Due to quantum mechanical effects, there is indeterminancy built into the system on the quantum level that absolutely cannot be avoided. This is very different than a coin toss, where we could always say "If only I could measure ($x$, $I$, $p$, $h$ or whatever) more precisely..."

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To do it completely without error would be impossible. There are simply too many parameters to be determined to exact precision in order for there to be no error.

One of the answers in John's comment links to this research that provides further details.

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In principle, yes, with enough knowledge about the coin, the environment, and the force (more than is stipulated in the question) it is of course possible to predict which side a classical coin will land on. This is because classical mechanics is totally deterministic.

In practice it is much more challenging. I don’t know if it possible to gather enough knowledge to predict coin tosses or not. My guess is that is possible. The system is not that complicated. A simple physics engine should be able to handle it.

If you’re asking only about the 5 items listed in the question are sufficient then the answer is no. You need to know the width of the coin in addition to the radius. You need to know how long the force is applied, exactly how the coin is initially positioned and oriented. And you also need to know about air resistance.

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