By the standard techniques explained here in a similar context, one can show that the first time $T_A$ when the word A=HTHT is produced has generating function
$$
E(s^{T_A})=\frac{s^4}{a(s)},\qquad a(s)=(2-s)(8-4s-s^3)-2s^3,
$$
and that the first time $T_B$ when the word B=TTH is produced has generating function
$$
E(s^{T_B})=\frac{s^3}{b(s)},\qquad b(s)=(2-s)(4-2s-s^2),
$$
The next step would be to develop these as series of powers of $s$, getting
$$
E(s^{T_A})=\sum_{n\ge4}p_A(n)s^n,\quad E(s^{T_B})=\sum_{k\ge3}p_B(k)s^k,
$$
and finally, to compute the sum
$$
P(T_B<T_A)=\sum_{k\ge3}\sum_{n\ge k+1}p_B(k)p_A(n).
$$
An alternative solution is to consider the Markov chain on the space of the couples of prefixes of the words A and B and to solve directly the associated first hitting time problem for this Markov chain, as explained, once again, here.
(+1 to Arturo's and Mike's comments.)
Added later Introduce the decomposition into simple elements
$$
\frac1{(1-s)b(s)}=\sum_{i=1}^4\frac{c_i}{1-s\gamma_i}.
$$
Then, one can show that
$$
P(T_B<T_A)=\sum_{i=1}^4\frac{c_i}{a(\gamma_i)},
$$
by decomposing the rational fractions $1/b(s)$ and $1/a(s)$ into simple elements, by relying on the fact that all their poles are simple, and by using some elementary algebraic manipulations.
At this point, I feel it is my duty to warn the OP that to ask question after question on math.SE or elsewhere along the lines of this post is somewhat pointless. Much more useful would be to learn once and for all the basics of the theory of finite Markov chains, for instance by delving into the quite congenial Markov chains by James Norris, and, to learn to manipulate power series, into the magnificent generatingfunctionology by Herbert Wilf.
If the first toss is a tail, you’re certain to get TH before you get HH. If the first toss is a head, however, you could still get TH before getting HH. Thus, the probability of getting TH first is greater than $\frac12$. (I’m assuming a fair coin.)
Added: This answers the original question, but the reasoning can be extended to get a numerical result. If you get a T before getting HH, you are certain to get TH before HH. Thus, the only way to get HH first is to get it before tossing even one T, which means getting it in the first two tosses. That occurs with probability $\frac14$, so the probability of getting TH first must be $\frac34$. (The only outcome that results in never getting HH or TH is an infinite string of tails, which has probability $0$.)
Best Answer
Since the rvs are discrete and independent, the way I see it you need to find an expression for $P(X-Y>0)$:
$$ \sum_{k=4}^{\infty}P(X=k \cap Y \leq k-1)= \sum_{k=4}^{\infty}P(X=k)P(Y \leq k-1) $$ and since events 'toss $k$ times to get 3 H in a row' are disjoint you immediately have $P(Y \leq k-1) = \sum_{j=3}^{k-1}P(Y=j)$. Can you handle from here?