What is the probability that you win the second game given that you won the first if the outcomes of the two games are independent? I’m working on a probability problem, but I’m having a hard time understanding the question exactly.
The problem is:

You are going to play two games of chess with an opponent you have never played against before. Your opponent is equally likely to be a beginner, intermediate, or master. Depending on which, your chances of winning an individual game are 90%, 50%, or 30%, respectively.
a) What is your probability of winning the first game?
b) Given the information that you won the first game, what is the probability that you will also win the second game ( assume that, given the skill level of your opponent, the outcomes of the two games of independent each other)?

I’ve done problem a. $1/3\cdot 0.9+1/3\cdot 0.5+1/3\cdot 0.3=0.567$
For problem b, I am confused how independence works here.
I know that the two games are independent, but if I am playing with the same opponent as the first game for my second game, wouldn’t I have a higher probability for problem b compared to problem a?
Or does the question imply that I’ll get different opponents each time and the answer to problem b would be 0.567?
I’ve searched, and the answer seems to be divided.
Probability questions seem too much like English problems to me:( Could you please help me understand?
Thank you in advance!
 A: 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.
A: b-)probability of winning second given that you won first $=$?
denote probability of winning $1$st game by $A$
denote probability of winning the $2$nd game by $B$
$$p(\text{winning $2$nd, given won $1$st}) = p(B \mid A)= \frac{p(A \cap B)}{p(A)}$$
since both games are independent: $p(A \cap B)= p(A) \cdot p(B)$
thus $p(B \mid A) = p(B) =p(\text{winning second time})$
