Is it possible to 'split' coin flipping 3 ways?
If you throw your coin $n$ times you have $2^n$ outcomes, the probability of each of which is $\frac{1}{2^n}$. The larger $n$ is, the better you can divide $2^n$ into three approximately equal parts:
Just define $a_n=[2^n/3]$ and $b_n=[2\cdot 2^n/3]$, where $[\cdot]$ denotes rounding off (or on). Since $\frac{a_n}{2^n}\to\frac{1}{3}$ and $\frac{b_n}{2^n}\to\frac{2}{3}$ as $n\to\infty$, each of the three outcomes
"the number of Heads is between $0$ and $a_n$",
"the number of Heads is between $a_n$ and $b_n$", and
"the number of Heads is between $b_n$ and $2^n$"
has approximately the probability $\frac{1}{3}$.
Alternatively, you could apply your procedure to get four outcomes with the same probability (Heads-Heads, Tails-Tails, Heads-Tails, Tails-Heads) to your problem in the following way:
Associate the three outcomes Heads-Heads, Tails-Tails, Heads-Tails with your three possible choices. In the case that Tails-Heads occurs, just repeat the experiment.
Sooner or later you will find an outcome different from Tails-Heads.
Indeed, by symmetry, the probability for first Heads-Heads, first Tails-Tails, or first Heads-Tails is $\frac{1}{3}$, respectively.
(Alternatively, you could of course throw a die and select your first choice if the outcome is 1 or 2, select your second choice if the outcome is 3 or 4, and select your third choice if the outcome is 5 or 6.)
A simple (practical, low-computation) approach to choosing among three options with equal probability exploits the fact that in a run of independent flips of an unbiased coin, the chance of encountering THT before TTH occurs is 1/3. So:
Flip a coin repeatedly, keeping track of the last three outcomes. (Save time, if you like, by assuming the first flip was T and proceeding from there.)
Stop whenever the last three are THT or TTH.
If the last three were THT, select option 1. Otherwise flip the coin one more time, choosing option 2 upon seeing T and option 3 otherwise.
EDIT Oh dear, Rasmus has extended his answer and rendered this one obsolete! In fact, stop reading this right now and go see what he has done.
I interpret what you are asking as trying to find a way to decide without introducing bias (as would be introduced by counting tails-heads as the same as heads-tails). Rasmus's suggestion of repeating the experiment for a certain configuration seems the best choice.
I drew a little tree and tried to group outcomes into three "equally-likely" sets and then realized this is impossible because $3$ does not divide $2^n$ for any $n$. (The quantity $2^n$ being the number of unique outcomes after $n$ "flips".)