How do you reckon that the state after the waiting period ends is fair, if, as the article states, any point in time is more likely to have the initial side up? It would follow that at the end of the waiting period, the initial side is more likely to be up. And going forward from there, that new initial state (measured from the end of the waiting period) will continue affecting the rest of the flip. This bias will diminish over time, which backs up Premise 7 that more revolutions reduces the bias.

On a tangential note, this behavior reminds me of Benford's law, which is probably based on other statistical reasoning.

https://en.m.wikipedia.org/wiki/Benford%27s_law

I don't really follow the logic presented in the article.

For convenience, let's assume that the coin spins at 2 seconds per revolution. The coin begins in a Heads up position. After 0.5 a second, the coin is on its edge. The coin then takes another second to pass through Tails and again be on its edge. The coin then takes another 0.5 a second to reach its initial state of Heads up.

So it sounds to me like the coin spent 0.5 a second in "Heads", then 1 second in "Tails", then 0.5 a second in "Heads" before starting all over again.

If I borrow the numbering analogy from the article, it sounds like if we examine the coin in 0.5 second increments we get H T T H H T T H H T T H H, etc. which invalidates the article's claim.

Furthermore, the article claims a 51-49 split between sides. Even if we accept its reasoning (i.e. H T H T instead of H T T H H T T H) To get that specific result, (if my math is right) the coin would have to flip around 25 times on average (i.e H T H T H T H T H T H T H T H T H T H T H T H T 0.5*H), which is a specific number not mentioned in the linked article.

All of that to say that I have no intention of refuting the paper behind the article, nor do I think they are wrong, but just that the logic presented by the article (and not the paper which I didn't read!) sounds too simplistic to be correct to me.

More simply:

Imagine a counter ticking 1,2,3,4...

If you call stop at a random time, uniform over the whole tick time, there's a bias toward odds.

But if you choose to call stop at moment 20.5 (and stopping at a fractional time forces the system to round to the nearest whole number), plus random variation due to real-world physics, evens and odds are equally likely.

I think what the parent post meant is that this reasoning is flawed by assuming the coin spends the same time face-up in the very first iteration as in every other one, and that's a fallacy. To put it simply, assuming a 4 second rotation, the first second will be spent in state A, the next second will be spent equalising the time in both states, the next second will actually put state B second ahead in face-up totals, then reequalizing, ad vitam eternum - which invalidates the article's point about always having one with time at most equal, and one with time at least equal, since they alternate in total face-up time.

Edit : Basically, unless you start the coin in the vertical state, the odds-and-even comparison doesn't apply.

Ah, you're right. This is the case if the coin is flipped with no precession. The paper linked takes precession into account based on the angle you flip the coin at, and this significantly changes the time H or T is up. The paper has two excellent graphs depicting this on Page 4.

The paper linked was also based on 27 observable coin flips, all made with the same coin-flipping machine. Out of those 27, 18 were between .500 and .504 in probabilities - and the paper concludes by saying that the "the classical assumptions of independence with probability 1/2 are pretty solid".

Edit: Actually, 21 out of the 27 flips were in the 0.5 - .505 range.

I disagree. "Any point in time is more likely to have the initial side up" if you count the initial state, which for the purpose of a coin toss you wouldn't. After a coin is flipped it's going to spend 50% of its time heads up and 50% of its time tails up.

I'm pretty sure the article is wrong. Hold on while I write a little script to test this out.

Edit: Script below; feel free to critique or make modifications where you think they are more realistic.

// Let 1 represent heads and 0 represent tails. coin_state = 1;

number_of_heads = 0;

number_of_tails = 0;

for (i = 0; i < 1000000; i++)

{

// How fast the coin flips (flips per second)

flip_speed = (Math.random() * 10) + 20;

// How long the coin is in the air (seconds)

catch_time = Math.random() + 2.5;

number_of_flips = Math.floor(flip_speed * catch_time);

/* Odd number of flips means coin ends up in opposite state from start (tails in this case). Even number of flips means coin ends up in same state as start (heads in this case) */

if (number_of_flips % 2) { coin_state = 0; }

if (number_of_flips % 2 == 0) { number_of_heads++; }

else { number_of_tails++; }

}

console.log("Heads:", number_of_heads);

console.log("Tails:", number_of_tails);

The article is wrong in that it assumes the initial state of the coin has as much duration as the subsequent states. If that was the case, the difference in duration would be the same as in the odd-even sequence :

  number       1 2 3 4 5 6 7 8 9
  odd          1 1 2 2 3 3 4 4 5 
  even         0 1 1 2 2 3 3 4 4
  difference   1 0 1 0 1 0 1 0 1

However, what happens here is different - you start at "half a state", meaning that the graph is a see-saw, as such:

  rotations    0.0  0.25 0.50  0.75 1.00 1.25 1.50
  head         0.0  0.25 0.25  0.25  0.5 0.75 0.75
  tail         0.0  0.0  0.25   0.5  0.5  0.5 0.75
  difference     0  .25     0  -.25    0  .25    0

You're right based on the article. It turns out the paper takes precession into account, which, depending on the angle you flip it at, significantly changes the time that the initial face is up.

You're right. My comment doesn't make sense.

Agreed, I think this is an overly-simplified explanation. Skimming the paper it looks like the crucial thing which introduces bias is precession of the coin.