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
This seems like a straightforward problem in conditional probability. Let $W_1$ be the event of winning the first game, $W_2$ be the event of winning the second game, and $F_b, F_i, F_m$ be the events of facing a beginner, an intermediate player, and a master.
We are given the unconditional probabilities $P(F_b) = P(F_i) = P(F_m) = \frac13$, and the conditional probabilities
$$ P(W_1 \mid F_b) = P(W_2 \mid F_b) = \frac{9}{10} $$ $$ P(W_1 \mid F_i) = P(W_2 \mid F_i) = \frac{5}{10} $$ $$ P(W_1 \mid F_m) = P(W_2 \mid F_m) = \frac{3}{10} $$
We are asked to find $P(W_2 \mid W_1)$. You have already observed that $P(W_1) = \frac{17}{30}$. From Bayes's theorem, we can write
$$ P(F_b \mid W_1) = \frac{P(W_1 \mid F_b)P(F_b)}{P(W_1)} = \frac{9}{17} $$ $$ P(F_i \mid W_1) = \frac{P(W_1 \mid F_i)P(F_i)}{P(W_1)} = \frac{5}{17} $$ $$ P(F_m \mid W_1) = \frac{P(W_1 \mid F_m)P(F_m)}{P(W_1)} = \frac{3}{17} $$
By total probability, we have
\begin{align} P(W_2 \mid W_1) & = P(W_2 \mid W_1, F_b) P(W_1 \mid F_b) \\ & + P(W_2 \mid W_1, F_i) P(W_1 \mid F_i) \\ & + P(W_2 \mid W_1, F_m) P(W_1 \mid F_m) \end{align}
Because $W_1$ and $W_2$ are conditionally independent, given the same opponent, we can rewrite this as
\begin{align} P(W_2 \mid W_1) & = P(W_2 \mid F_b) P(W_1 \mid F_b) \\ & + P(W_2 \mid F_i) P(W_1 \mid F_i) \\ & + P(W_2 \mid F_m) P(W_1 \mid F_m) \\ & = \frac{9}{10} \times \frac{9}{17} + \frac{5}{10} \times \frac{5}{17} + \frac{3}{10} \times \frac{3}{17} \\ & = \frac{115}{170} = \frac{23}{34} \end{align}
P.S. I realize it's just a framing device, but I can't believe a single chess player could win $90$ percent against a beginner, $50$ percent against an intermediate player, and $30$ percent against a master.