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?

When you flip the third coin, you’ve already observed the first two flips.
Mathematically, $p(\text{flip}_3=H \mid \text{flip}_2=H, \text{flip}_1=H) = 0.5$. This comes from the independence among coin flips.
The value 0.125 is the joint probability $p(\text{flip}_3=H, \text{flip}_2=H, \text{flip}_1=H)$. This is the probability of flipping three heads in a row.

By the way, the two are related. The joint probability can be decomposed into each coin flip:
\begin{align}
p(\text{flip}_3=H, \text{flip}_2=H, \text{flip}_1=H) =& p(\text{flip}_3=H \mid \text{flip}_2=H, \text{flip}_1=H) p(\text{flip}_2=H \mid \text{flip}_1=H) p( \text{flip}_1=H) \\
=& p(\text{flip}_3=H) p(\text{flip}_2=H) p( \text{flip}_1=H) \\
=& 0.5 \times 0.5 \times 0.5 \\
=& 0.125
\end{align}
The first step is the chain rule of probability, and the second comes from independence amongst coin flips.
