Brian Scott's answer is good. Here's another point of view. First consider what happens on the first three tosses: \begin{array}{lll} \text{HHH} & & \text{HH first} \\ \text{HHT} & & \text{HH first} \\ \text{HTH} & & \text{TH first} \\ \text{HTT} \\ \text{THH} & & \text{TH first} \\ \text{THT} & & \text{TH first} \\ \text{TTH} & & \text{TH first} \\ \text{TTT} \end{array} In four cases you get $\text{TH}$ first and in only two you get $\text{HH}$ first, and in the other two, you've got a $50\%$ chance of getting $\text{TH}$ on the next trial and no chance of getting $\text{HH}$ on the next trial.
What happens after that? You've got a four-state Markov chain whose states are $\text{HH}$, $\text{HT}$, $\text{TH}$, $\text{TT}$, and if after $\text{TH}$ (for the first and second trials) you get $\text{T}$, then you've moved to state $\text{HT}$, (for the second and third trials) etc. The transition probabilities are not all equal, but if you raise the transition matrix to a higher power than $1$, then they're always equal to $1/4$. So in further steps beyond that, the four states are always equally likely (although of course they're not conditionally equally likely given where you are on the immediately preceding trial).