3 ms·

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 H

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.

gowld 9y ago
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.
- vorathing 9y ago
  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.
wakamoleguy 9y ago
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.
- vorathing 9y ago
  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.

gowld 9y ago

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.

vorathing 9y ago
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.

wakamoleguy 9y ago

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.

vorathing 9y ago
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.