i will solve a general case , say probability of $A$ winning is $p$ and for a team to win it needs to win $n$ matches. now say $k$ matches have been played , and $A$ won the series , it means $A$ won the last match , thus out of the rest $k-1$ matches $A$ won $n-1$ games. thus for $A$ to win the series in $k$ matches probability is $$\binom{k-1}{n-1}p^{n}(1-p)^{k-n}$$
now $k>=n$ and $k<=2n-1$ ($A$ needs to win before $B$) thus our total probability is $$\sum_{k=n}^{2n-1} \binom{k-1}{n-1}p^{n}(1-p)^{k-n}$$
Two provide a little bit of insight, assume for simplicity that the team who wins two (not three) games wins the series.
Simplification
Since you have three games with two possible outcomes each, it is enough to consider a probability with $2^3=8$ elements. Namely
$$
\Omega = \{\omega_{AAA}, \ \omega_{AAB}, \ \omega_{ABA}, \ \dots, \ \omega_{BBB}\}
$$
For example, $\omega_{ABB}$ denotes the event that $A$ wins the first game and $B$ wins the second and third. Now define a random variable $X$ like this:
$$
X=\begin{cases}1, & \text{A wins the series}, \\0, & \text{B wins the series} \end{cases}
$$
Using the notation from above, we see that
$$
X(\omega) =\begin{cases} 1,& \omega \in \{\omega_{AAB}, \ \omega_{ABA},\ \omega_{BAA}, \ \omega_{AAA} \}=:\Omega_A \\ 0, & \omega \in \{\omega_{BBA},\ \omega_{BAB},\ \omega_{ABB}, \ \omega_{BBB} \}:=\Omega_B\end{cases}
$$
Finally we can calculate the desired probability, therefore we define the set of all the $\omega$, where $A$ wins the first game:
$$
C:=\{\omega_{AAA}, \ \omega_{AAB},\ \omega_{ABA}, \ \omega_{ABB} \}.
$$
And now:
$$
\mathbb{P}(\text{A wins series} \:| \text{A wins first game})= \mathbb{P}(X=1\: |\: C) = \mathbb{P}(\Omega_A\: |\: C) =\frac{\mathbb{P}(C\cap \Omega_A)}{\mathbb{P}(C)} = \\
\frac{\mathbb{P}(\{\omega_{AAB}, \ \omega_{ABA},\ \omega_{AAA}\ \})}{\mathbb{P}(C)},
$$
which can be calculated easily using the additivity of $\mathbb{P}$. For example
$$
\mathbb{P}(C) =\mathbb{P}(\omega_{AAA})+\mathbb{P}(\omega_{AAB})+\mathbb{P}(\omega_{ABA})+\mathbb{P}(\omega_{ABB})=\\
p^3+p^2(1-p)+p^2(1-p)+p(1-p)^2.
$$
General Case
For the general case you need to consider a probability space with $2^5 =32$ elements. Then $\Omega$ looks like this:
$$
\Omega= \{\omega_{AAAAA}, \ \omega_{AAAAB}, \dots, \ \omega_{BBBBB}\}.
$$
The interpretation of the $\omega$s is the same as before: e.g., $\omega_{AABBA}$ describes the event that $A$ wins the first two games and the last one, and $B$ wins games 3 and 4.
The random variable $X$ and the set $C$ can also be defined as before: the definition of $X$ yields a partition of $\Omega$ into two set $\Omega_A$ and $\Omega_B$, where $\Omega_A$ contains all events where $A$ wins more often than $B$ and $\Omega_B$ contains all events where $B$ wins more often than $A$.
$C$ is again the set of all events where $A$ wins the first game.
$$
C=\{\omega_{AAAAA}, \ \omega_{AAAAB}, \dots, \ \omega_{ABBBB}\}
$$
And therefore
$$
\mathbb{P}(\text{A wins series} \:| \text{A wins first game})= \mathbb{P}(X=1\: |\: C) = \mathbb{P}(\Omega_A\: |\: C) =\frac{\mathbb{P}(C\cap \Omega_A)}{\mathbb{P}(C)}
$$
Final Coment
Finally, I want to say that for exercises like this, it is way faster to compute the desired probabilty directly, without going into to much detail about the probability space in the shadows. But it is a nice exercise after all.
Best Answer
Lets think about the full distribution for $X$ and $Y$. What you have done is combined the distributions of $T_1$ and $T_2$. We want the combination of $X$ and $Y$. When $X=6$, $Y=0$. When $X=4$, $Y=1$. When $X=1$, $Y=4$. When $X=3$, $Y=3$. When $X=0$, $Y=6$. If $W$ is win, $D$ is draw and $L$ is loss, we can: $WW, LL, WL, LW, WD, DW, DD$
We want the combinations where $X \gt Y$ strictly. That is $X=6$, $Y=0$ AND $X=4$, $Y=1$. This is represented by $WW$ or $WD$ or $DW$. Sum this:
$2$ Wins: $\frac 12 \times \frac 12$
$1$ Win $1$ Draw: $\frac 12 \times \frac 16$
So $$P(X>Y)=P(X=6,Y=0)+P(X=4,Y=1)$$
$$=P(WW)+2 \times P(WD)$$
$$=\frac 12 \times \frac 12 \space + \space 2 \times\frac 12 \times \frac 16= \frac 14+ \frac 1{6}= \frac 5{12}$$
Edit ### - Looks like my first guess was right.
So 5/12 I hope is your answer.