Coin tossing game - players 1 and 2 each choose a sequence of 3 tosses

Question

We are playing a fair tossing game. Each player chooses a sequence of 3 outcomes. Whoever's sequence is reached first wins the game. Player 1 chooses first and player 2 chooses second. There are 8 possible choices for player 1 and 7 for player 2.

Should you go first or second to have the highest probability of winning?

The brute force way of solving this problem is to enumerate all possible pairs. There are $\binom{8}{2} = 28$ unique pairs. A lot of pairs. Though I think another set of symmetry arguments could further cut this by half down to 14. For example, considering the case HHH is the same as considering the case TTT, or more generally, consider the case XYZ is the same as considering the case $\bar{X}\bar{Y}\bar{Z}$.

A smarter way to approach this problem is based on the reasoning that player 2 should always choose the first 2 outcomes of player 1's sequence as the last 2 outcomes of its sequence. How can I prove this in a relatively easy way?

For example, say, player 1 chooses HHH. Player 2 should choose THH in this case. In this case player 1 only has a $\frac{1}{8}$ of winning. If the first 3 tosses are not all heads, then player 2 is guaranteed to win because the sequence $HH$ in $THH$ will always occur before $HHH$. If a tail occurs in any of the first 3 tosses, then player 1 can never win. This case was simple for me to reason.

Now, say, player 1 chooses HTT. Based on my logic, player 2 should choose THT or HHT, but it's not clear to me which one is a better choice by inspection, unlike the previous case where we only had 1 possibility. So we have 2 choices this time. How can I quickly think about which choice is a better choice? Or in other words, how do I determine whether the first outcome in player 2's sequence should be a tail or head?