Let $X$ be the random variable that counts the number of tosses until we get two successive tosses of the same type. Although you did not say so explicitly, I assume that you want the probability mass function of $X$.
What is the probability that $X>n-1$? We need to have a sequence of tosses of length $n-1$ and of type HTHTHT and so on (alternating heads and tails, starting with head), or THTHTH and so on. The probability that the first type of sequence is followed up to and including the $(n-1)$-th toss is $\frac{1}{2^{n-1}}$. The second type of sequence has the same probability, for a total of $\frac{2}{2^{n-1}}$.
The number $X$ of tosses is equal to $n$ if $X>n-1$ and the last toss matches the next to last toss. The probability of that is $\frac{1}{2}$, so
$$P(X=n) =\frac{2}{2^{n-1}}\cdot\frac{1}{2}.$$
This simplifies to $\dfrac{1}{2^{n-1}}$. Note that $n$ ranges over the integers that are $\ge 2$.
Another way: The first toss doesn't matter. After that, it is just a matter of matching the previous toss. So it is very much like tossing a fair coin and winning if you get (say) heads. The only difference is the "wasted" toss at the beginning. If we are tossing a fair coin, and $Y$ is the number of tosses until the first head, then $P(Y=n)=\frac{1}{2^n}$ ($n=1, 2, \dots$). But $X$ has the same distribution as $Y+1$. So $P(X=n+1)=P(Y=n)=\frac{1}{2^n}$, or equivalently $$P(X=n)=P(Y=n-1)=\frac{1}{2^{n-1}}\quad (n=2,3,\dots).$$
Generalization: Suppose that the probability of a head is $p$, and the probability of a tail is $q=1-p$. Either analysis of the problem can be extended to the more general situation. We win on the $n$-th toss if we did not win on the first $n-1$, and then the $n$-th toss matches the previous one.
It is convenient to split the analysis into two cases, $n$ odd and $n$ even. We do a full analysis of the odd case. So let $n=2m$. As is the first analysis, there are two ways to win on the $(2m+1)$-th toss. Either (i) we went HTHTHT and so on, with a tail on the $(2m)$-th toss, and then a tail again on the $(2m+1)$-th toss, or (ii) we went THTHTH and so on, with a head on the $(2m)$-th toss, and a head again on the $(2m+1)$-th toss.
The probability of (i) is $p^mq^m q$ and the probability of (ii) is $q^mp^mp$. Add them together. We get $p^mq^m(p+q)$, which is $p^mq^m$ ($m=1,2,\dots)$.
Next we deal with the case $n$ is even, say $n=2m$. The calculation is very similar to the previous one, so it will be omitted. The probability is the slightly more complicated-looking $p^mq^{m-1}p+q^mp^{m-1}q=p^{m-1}q^{m-1}(p^2+q^2)$, where $m$ ranges over the positive integers.
This can be seen as an MC with 4 states. Denote for convenience $S=\{HHH, TTT\}$
One thing to notice first, is that only before you have tossed any coins at all you need 3 matching tosses till $S$. After that, regardless of the outcome, you have no more than two matching tosses from $S$. Hence we have the following MC:
$S_0$: 3 matching tosses till $S$
$S_1$: 2 matching tosses till $S$
$S_2$: 1 matching tosses till $S$
$S_3$: 0 matching tosses till $S$
Denoting $m_{i,j}$ the mean hitting time of state $j$ starting in state $i$, we obtain the following set of recurrent equations:
$$
m_{0,3}=1+m_{1,3}\\
m_{1,3}=1+0.5 m_{1,3} +0.5m_{2,3}\\
m_{2,3}=1+0.5 m_{1,3} +0.5m_{3,3}
$$
And the boundary condition is of course $m_{3,3}=0$. Solving this set of equations we get
$$
m_{0,3}=1+6=7
$$
Best Answer
If the first toss is a T, then the event $Y=1$ happens if the first $4$ tosses are THHH. This has probability $\frac{1}{16}$.
If the first toss is a H, then $Y=1$ can happen in $2$ ways: (i) the second toss is a T and then we get HHH or (ii) the second toss is a H, the next is a T, and then we get HHH. Note that (i) has probability $\frac{1}{32}$ and (ii) has probability $\frac{1}{64}$.
Add up our three probabilities, and simplify.
Remark: Your idea was right, there was just a little error in computing the probabilities.