Probability question - Bernoulli trials vs Binomial distribution I am handed a fair coin. I understand that the probability of H/T is 0.5.
Trial 1.
I flip and observe H. I expect that on my 2nd toss, the probability of H is still 0.5
Trial 2.
I flip and observe H. Again I expect that upon my 3rd toss, p(H) = 0.5.
Why is it that I should expect p(H) = 0.5 on my 3rd trial?  If I use the binomial distribution, I know that observing 3 successes in 3 trials is actually p=0.125.  I did not set out to flip the coin 3 times as a set, yet I've done so.
What am I missing or forgetting?

This is a classic gambler's fallacy. There is no law of averages for i.i.d. processes. See Regression to the mean vs gambler's fallacy
