Let $e$ be the expected number of tosses. It is clear that $e$ is finite.
Start tossing. If we get a tail immediately (probability $\frac{1}{2}$) then the expected number is $e+1$. If we get a head then a tail (probability $\frac{1}{4}$), then the expected number is $e+2$. Continue $\dots$. If we get $4$ heads then a tail, the expected number is $e+5$. Finally, if our first $5$ tosses are heads, then the expected number is $5$. Thus
$$e=\frac{1}{2}(e+1)+\frac{1}{4}(e+2)+\frac{1}{8}(e+3)+\frac{1}{16}(e+4)+\frac{1}{32}(e+5)+\frac{1}{32}(5).$$
Solve this linear equation for $e$. We get $e=62$.
Let $A_{m,n}$ be the probability that player $A$ flips two consecutive heads before player $B$ flips three consecutive heads, given that $A$'s (resp. $B$'s) current run of heads has length $m$ (resp. $n$). Then
$$
A_{m,n}=\frac{1}{4}\left(A_{m+1,n+1} + A_{m+1,0} + A_{0,n+1}+A_{0,0}\right);
$$
the boundary conditions are that $A_{m,n}=1$ for $m\ge 2$ and $n < 3$, and $A_{m,n}=0$ for $m \le 2$ and $n\ge 3$. You want to find $A_{0,0}$. The relevant six equations are:
$$
\begin{eqnarray}
A_{0,0} &=& \frac{1}{4}A_{1,1} + \frac{1}{4}A_{1,0} + \frac{1}{4}A_{0,1} + \frac{1}{4}A_{0,0}\\
A_{0,1} &=& \frac{1}{4}A_{1,2} + \frac{1}{4}A_{1,0} + \frac{1}{4}A_{0,2} + \frac{1}{4}A_{0,0} \\
A_{1,0} &=& \frac{1}{2} + \frac{1}{4}A_{0,1} + \frac{1}{4}A_{0,0} \\
A_{1,1} &=& \frac{1}{2} + \frac{1}{4}A_{0,2} + \frac{1}{4}A_{0,0} \\
A_{0,2} &=& \frac{1}{4}A_{1,0} + \frac{1}{4}A_{0,0} \\
A_{1,2} &=& \frac{1}{4} + \frac{1}{4}A_{0,0},
\end{eqnarray}
$$
or
$$
\left(\begin{matrix}3/4 & -1/4 & -1/4 & -1/4 & 0 & 0 \\
-1/4 & 1 & -1/4 & 0 & -1/4 & -1/4 \\
-1/4 & -1/4 & 1 & 0 & 0 & 0 \\
-1/4 & 0 & 0 & 1 & -1/4 & 0 \\
-1/4 & 0 & -1/4 & 0 & 1 & 0 \\
-1/4 & 0 & 0 & 0 & 0 & 1
\end{matrix}\right)\times\left(\begin{matrix} A_{0,0} \\ A_{0,1} \\ A_{1,0} \\ A_{1,1} \\ A_{0,2} \\ A_{1,2}
\end{matrix}\right)
= \left(\begin{matrix}
0 \\
0 \\
1/2 \\
1/2 \\
0 \\
1/4
\end{matrix}\right),
$$
assuming no typos. Further assuming no typos entering this into WolframAlpha, the result is
$$
A_{0,0} = \frac{1257}{1699} \approx 0.7398,
$$
which at least looks reasonable.
Update: As pointed out in a comment, the above calculation finds the probability that $A$ gets two consecutive heads sooner than $B$ gets three consecutive heads; the original problem asks for the opposite. The correct boundary conditions for the original problem are that $A_{m,n}=1$ for $m<2$ and $n\ge 3$ and $A_{m,n}=0$ for $m\ge 2$ and $n\le 3$. The matrix equation becomes
$$
\left(\begin{matrix}3/4 & -1/4 & -1/4 & -1/4 & 0 & 0 \\
-1/4 & 1 & -1/4 & 0 & -1/4 & -1/4 \\
-1/4 & -1/4 & 1 & 0 & 0 & 0 \\
-1/4 & 0 & 0 & 1 & -1/4 & 0 \\
-1/4 & 0 & -1/4 & 0 & 1 & 0 \\
-1/4 & 0 & 0 & 0 & 0 & 1
\end{matrix}\right)\times\left(\begin{matrix} A_{0,0} \\ A_{0,1} \\ A_{1,0} \\ A_{1,1} \\ A_{0,2} \\ A_{1,2}
\end{matrix}\right)
= \left(\begin{matrix}
0 \\
0 \\
0 \\
0 \\
1/2 \\
1/4
\end{matrix}\right),
$$
The result becomes $A_{0,0}=\frac{361}{1699}\approx 0.2125$. The two results add to slightly less than one because there is a nonzero probability that both players hit their goals at the same time... this probability is $81/1699\approx 0.0477$.
Best Answer
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.