[a_c]

I always tell people that result of coin flip is highly start state dependent. Imagine a sequence of H(ead), T(ail), H, T, H, T, ... if the sequence starts with H first, in no way can the number of T exceed that of H, but the number of H might be 1 greater that that of T. I never tested my self, but I hypothesize that the propability will be more skewed if the number of revolutions is less, i.e. having a shorter Head-Tail sequence.

Edit: The sequence was meant to represent the sequence of head and tail facing up during the rotation. A sequence of [H, T] denotes one full rotation, starting with head. [T, H, T, H] denotes two full rotation, starting with a tail, ends with a head. I didn't mean the result of a flip. So the result of a flip is the final element of the sequence.

[roflmaostc]

Of course that is true, if you wait not significantly long enough.

Imagine the situation that you flip the coin (starting at H) and you grab it in air immediately. Of course, you will get H as result.

But let's say, the time to stop the coin can be a relatively long time T. Then, I think the probability is some kind of sum. Let's choose \Delta T= 10ms as time discretization:

P(H) = 1 / T * (10ms-0ms) + (30ms-20ms) + (50ms-40ms) + ... = 1/T \sum_{i=0}{floor(T / (2 * \Delta T))} \Delta T

P(T) = 1 / T ((20ms-10ms) + (40ms-30ms) + (60ms-50ms) + ... = 1/T \sum_{i=0}{floor(T / (2 * \Delta T)) - 1} \Delta T

For T -> \infty P(H) and P(T) getting more similar.

But, in practice you wouldn't wait equally distributed in time but more like a Gaussian distributed time period. Hence, each term of the sum would get weighted differently. And the variance and the offset of the Gaussian distribution can shift the probability in favor of H or T. It's really dependent of the concrete parameters. If you grab always after 35ms, then you'll always get T for example.

[Adverblessly]

If I were to dig through my comment history I would find I have already responded to this exact sort of comment before, so let me regurgitate :)

If the coin is resting on your hand waiting to be flipped, it is currently mid-way through being on side up. This is because the switch between being e.g. heads up to tails up is done when the coin is vertical. If it isn't a clear explanation, try imagining catching the coin and "flattening" it at different angles, while 50% of angles will match either side, at the moment the coin is flipped, it is already half-way through the angles representing the current side.

This means that the correct sequence you describe it not THTHTH but rather THHTTHHTTHH. Taken at even intervals, both sides will appear the same number of times. Taken at odd intervals, at half of the intervals there are more Ts and the other half have more Hs.

[cabirum]

Can you toss a coin without it flipping at least once (changing state)? I find it quite unlikely to happen, so if the starting state was H, your sequence will be THTH...

[a_c]

I can't quite define what counts as start of sequence. Maybe a sequence should always have at least one element, and toss straight up is allowed. But if some flipping is mandatory, then the start of sequence would mean the other side of the coin. All I could deduce before this paper is the probabilty of coin flip is skewed.

[mucle6]

I would have never considered this. Such an interesting way to think about the problem