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?
It might be better to say:
The probability of getting $a$ runs of one particular side of the coin before getting $b$ runs of the other side of the coin is $$P(E) = \frac{2^b-1}{2^a + 2^b - 2}$$
The situation is symmetric in the sense that it doesn't matter what sides of the coin you're talking about --- heads or tails. But it does matter that the runs of $a$ come before the runs of $b$.
Intuitively, $a$ and $b$ are not symmetric because if you increase $a$ (the number of runs you have to get first), you expect the likelihood of success to go down. If you increase $b$ (the number of runs you have to avoid getting before succeeding at $a$), you expect the likelihood of success to go up:
It is rather likely to get a run of 2 before your first run of 1000. It is rather unlikely to get a run of 1000 before your first run of 2. This is true regardless of whether you're talking about heads/tails or tails/heads.
Let's generalize to weighted coins. Suppose you want to know the probability of getting $a$ runs of one side of the coin before getting $b$ runs of the other. Suppose the one side of the coin comes up with probability $\alpha$, and the other side of the coin comes up with probability $\beta = 1-\alpha$.
Then, following that post:
$$\begin{align*}
P(E) &= P(E|T)P(T) + P(E|T^C)P(T^C)\\
& = \frac{p_1}{1-q_1q_2}\alpha + \frac{p_1q_2}{1-q_1q_2}(1-\alpha)\\
& = [q_2 + (1-q_2)\alpha] \frac{p_1}{1-q_1q_2}
\end{align*}$$
and here $p_1$ and $q_1$ are the probability of succesfully/unsuccessfully getting a run of $a$, and similarly for $p_2, q_2$.
We have $p_1 = \alpha^{a-1}$, $p_2 = (1-\alpha)^{b-1}$, $q_1 = 1-p_1$, $q_2 = 1-p_2$.
$$\begin{align*}
P(E) & = [q_2 + (1-q_2)\alpha] \frac{p_1}{1-q_1q_2} \\
&= [1-(1-\alpha)^{b-1} + (1-\alpha)^{b-1}\alpha] \frac{p_1}{1-q_1q_2}\\
&= [1 + (\alpha-1)(1-\alpha)^{b-1}] \frac{p_1}{1-q_1q_2}\\
&= [1 - (1-\alpha)^{b}]\frac{\alpha^{a-1}}{1-(1-\alpha^{a-1})(1-(1-\alpha)^{b-1})}
\end{align*}$$
Best Answer
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 $0$.)