[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.