Here's another method: consider H to be 0 and T to be 1. Consider the results that you get from your coins as successive bits (after the binary point) in a binary number. Assign numbers in the interval [0, 1/3) to the first choice; [1/3, 2/3) to the second choice; [2/3, 1) to the third choice. Flip coins until you know, no matter what happens on the remaining flips, in which of these intervals the resulting real number in [0, 1) will lie.
More concretely, this means that if you flip two coins and observe HH, make the first choice; no matter what happens the result will be between .0000... = 0 and .001111... = 1/4. Similarly if you observe TT, make the third choice; the result will be between 3/4 and 1. If you observe HT, you can't commit to a choice yet; the result will be between 1/4 and 1/2, which overlaps two of the intervals. Something similar is true for TH.
In either case flip a third time. HTT corresponds to the interval [3/8, 1/2) which lies entirely in [1/3, 2/3), so HTT corresponds to the second choice; similarly for THH. If you see HTH you're still undecided between the first and second choices; THT is still undecided between the second and third.
Continuing in this way, HTHH corresponds to the first choice, HTHT and THTH are still undecided, THTT corresponds to the third choice.
In general, for the cut-points 1/3 and 2/3, you flip coins until you get either two heads or two tails in a row. If you first get two heads in a row, this corresponds to the first choice if the initial flip was a head, the second if the initial flip was a tail. If you first get two tails in a row, this corresponds to the second choice if the initial flip was a head, the third choice if the initial flip was a tail.
You might wonder how many flips this takes, on average, to give a result. With probability 1/2 you get a result on the second flip; with probability 1/4, on the third flip; with probability 1/8, on the fourth flip, and in general with probability $1/2^{n-1}$ for $n \ge 2$. The sum $\sum_{n \ge 2} n/2^{n-1}$ has value 3, so on average this method takes three flips.
This is a bit slower on average than the method Rasmus suggested, which takes on average 8/3 flips. However, the same method works even if the three choices are not equally likely! (It won't have such a clean way of being expressed as "flip until you get two in a row", though.)