Intuitively, why does it make sense to go second in dice game to maximize chance of winning? Here's a question from a probability book I am working through:

Let's add more fun to the triplet game. Instead of fixed triplets for the two players, the new game allows both to choose their own triplets. Player $1$ chooses a triplet first and announces it; then player $2$ chooses a different triplet. The players toss the coins until one of the two triplet sequences appears. The player whose chosen triplet appears first wins the game.
If both player $1$ and player $2$ are perfectly rational and both want to maximize their probability of winning, would you go first (as player $1$)? If you go second, what is your probability of winning?

There's $8$ possible triplets sequences for each player:

HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

The players can't have the same triplet, hence there being $64 - 8 = 56$ probability outcomes to calculate for player $2$ winning. After spending half an hour tediously calculating all $56$, it turns out that player $2$ can always choose a triplet, dependent on what player $1$ picked, as to win with probability at least ${2\over3}$. However, I am wondering if there is an intuitive way to see that without tediously doing all $56$ computations.
Or if seeing that player $2$ can always win with probability at least ${2\over3}$ is too much to ask for of an intuitive heuristic, how can we see that player $2$ can always win with probability at least ${1\over2}$?
Edit: Since the problem statement is referring to earlier parts of the problem, I am reproducing those problem statements here as well:

Part A. If you keep on tossing a fair coin, what is the expected number of tosses such hat you have $HHH$ (heads heads heads) in a row? What is the expected number of tosses to have $THH$ (tails heads heads) in a row?
Part B. Keep flipping a fair coin until either $HHH$ or $THH$ occurs in the sequence. What is the probability that you get an $HHH$ subsequence before $THH$?

 A: Edit:*
Yes, the probability can be computed for a general pattern. See below the figure.
Intuitive explanation of winning edge:
Given the other player picks any pattern, say HHH, the player who goes second can pick a pattern which has the beginning of the opponents' pattern as a suffix.
So the second player would fix THH in this case. This means that except for the case that HHH occurs as the first 3 tosses, he can win.
More generally, given an arbitrary pattern, say $X_1,\ldots,X_n$ pick a pattern of the form $A, X_1,\ldots,X_{n-1}.$
By doing this, the first player is forcing the other player to typically win at a depth one more than herself, except if their pattern occurs in the beginning.
Here is a pictorial illustration of another example showing the depth of winning states. The first player (opponent) picks HTH and the second player (you) pick HHT.

There is a neat way of computing the winning probability in this Penney-Ante game which was first mentioned by John H. Conway.
Given two $q-$ary words (from a finite alphabet of $q$ letters) $X=(x_1,\ldots,x_n)$ and $Y=(y_1,\ldots,y_m),$ define the correlation  of $X$ and $Y$ as
$$
C[X,Y;z]=\sum_{i=0}^{n-1} f(n-i) z^i,
$$
where $f(i)=1$ if and only if the partial word $(x_i,\ldots,x_n)$ is a prefix of the word $(y_1,\ldots,y_m),$ otherwise $f(i)=0.$ In general, $[X,Y]\neq [Y,X].$ Then the odds that word $Y$ beats word $X$ in this game is given by
the expression
$$
\frac{C[X,X;q]-C[X,Y;q]}{C[Y,Y;q]-C[Y,X;q]},
$$
though the proof was not given by Conway and was supplied much later by Guibas and Odlyzko.
Note that if the odds are $o$ the probability of winning is $$o/(1+o).$$
A: If I understand the game correctly, here is my intuition for why player 2 has a higher chance of success
Suppose the first two elements of player 1's triplet showed up at some point in the game which is not the second flip (i.e. the game didn't start with the first two elements of 1's  tripple)
Then at any such instances, there must have been a flip before the double, if player 2 chose their triple to end in the two elements with which player 1's triple starts, every time the first two elements of 1's sequence shows up, there is a $50\%$ chance that player 2's sequence just ended, awarding them the victory
However if player 2 chooses their first element so that their first two elements don't match up with the last two elements of 1's sequence, then the reverse is not true
So every time 1 "gets a chance to win", 2 "got a chance to win" but every time 2 gets a chance to win, 1 did not get a chance to win, this is not rigorous, but you are asking for the intuition and I believe this is it
