Conditional Probability simplification First of all, apologize for the long winded description.  So we had a lecture on conditional probability and one example was the following:

Assume a best-of-3 hockey series for some local team,
The probability of the local team winning the first game is 1/2
In subsequent games probability of winning is determined by outcome of previous game:
If they won previous game, probability of winning next game is 2/3
Otherwise, the probability of winning the next game is 1/3
What is the probability of local team winning series, given they won first game?

I personally thought this question was pretty simple, so I solved it intuitively without learning the proper formula for conditional probability.  I thought since the first game has already happened, we can basically solve the rest of the probability based on the fact that first game is already determined, as if P(B) = 1 where B is the conditional event.  So the probability would be 2nd game WIN for home team, and 2nd game loss + third game WIN for home team, which was 2/3 + 1/3*1/3 = 7/9 and I was right.  However the professor said I can't do it that way, and that I had to use the formula P(A|B) = P(A ∩ B) / P(B).  And he said that the fact that I got the right answer was merely by luck (i.e. how the question was constructed helped my way of solving and led me to the correct answer by accident).
I was not so convinced however, because even from the step by step (alot of steps by the way) solution he gave, it was obvious that P(B) could be simplified from P(A ∩ B).
In his solution, we had the following: (explanation of notation: WLW means Win Loss Win for home team, in that order)
={,,,,,}
The event that the local team wins the series is:={,,}
The event that the team wins the first game is: ={,,}
[∩]=[,] 
[]=[,,]

So this becomes quite obvious that [] could be factored out from [∩] since [] is basically the probability of home team winning the first game (the first W in any combination) AND all combinations of [∩] contains home team winning first game.  I even tried to create a spreadsheet and varied the values of probability for winning first game, and subsequent games, and proved that my intuition was correct.
So my question is, who is right?  And if I was right, is there a general formula for simplifying the conditional probability in this situation?  At first I thought this is a case of A and B being independent events and that the formula would simplify to P(A|B) = P(A), but the professor shot down that idea also because he said game 2's win probability clearly depends on the result of game 1, therefore they can't be independent events.
I also like to point out, seems there are many similar situations I can think of where conditional prob could be "simplified".  Like, what is the prob of having a 5 card flush if the first card you draw is a Diamond, etc.  I feel like you can always "ignore" the first condition as long as you account for the change from the initial condition, i.e. there are now 51 cards left in deck, and 12 diamonds left.
 A: I would categorize both methods as "correct". This feels like the mindset of the teacher is "this is the answer in the book, therefore it is the ONLY answer". Or "this is in the section with formula so you have to use the formula". This question I think would be geared more towards the multiplicative rule: $P(A \cap B) = P(B|A)P(A) = P(A|B)P(B)$.
You did not get to your answer purely by luck. You created a logical system consistent with the problem and followed through with it. Honestly, I would probably have done it your way even though I have had that formula burned into my brain at some point.
The concept of conditional probability is about rescaling the original sample space to a subset of the sample space. Your methodology did that by a simple reading of the problem. The original sample space was $=\{_w,_w_l,_w_l,_l_w,_l_w,_l\}$ where a sub-script represents the preceding event, since that affects the probability. You intuited what the rescaled sample space was for the first game being won, $S^* = \{W_w, L_wW_l, L_lL_l\}$.  It's a sample space, therefore add up the probability of the elements corresponding to your event, $P(\text{Win}) = P(W_w) + P(L_wW_l)$.
The whole setup is so simple and procedural that I see absolutely no reason to force the use of the conditional probability formula. The conditional probability formula automatically rescales the sample space for you, you just bypassed using your (correct) intuition. In fact, I'd say your method falls more in line with what conditional probability actually is, a the reformulation of the structure of the new sample space once you have placed a condition imposed on the previous one.
Your professor is right in the fact that your solution does not arise from independence since the outcome of a successive game depends on the previous one. However, you basically used your condition to formulate a new sample space that is built precisely for answering the question on hand.
The general format that I teach for solving this kind of problem is a tree diagram where you map out the possible out using a branching diagram and find the probability at the end of each branch. A general form is given here:

Once you have mapped out the probabilities, you just add the probabilities for each branch that corresponds to the event you want the probability for. Your tree diagram would only have three total end branches. Since winning the second game would end the best of 3 match.
If your teacher wanted to force the use of the conditional probability formula, they should probably have found a situation where you cannot recreate a new and complete sample space based off of the information given. Or I guess tell you that you HAVE to use the formula regardless of other solutions.
