Assuming that each team has an equal $\frac{1}{2}$ probability of winning each game, and the probability of winning each game is independent.
I solved this using a tree diagram, and got the answer of $\frac{11}{16}$, but I want to figure out a way of solving this without relying on a tree.
I tried diving this up to cases where the series ends in 2 games, 3 games and 4 games, but am having difficulty.
Any help would be greatly appreciated!
On
You can take a Markov chain approach. The possible states that describe the series are of the form $(m,n)$, where $m$ is the number of games won so far by team A and $n$ is the number of games won so far by team B. In your example, the series is in state $(2,1)$.
Let $P(m,n)$ denote the probability that team A will win the series, given it is in state $(m,n)$. Let's consider your specific example of $(2,1)$. Two outcomes are possible starting from this state (each with probability ${1 \over 2}$. Team A can win game 4 and the series could go to state $(3,1)$ or Team B could win game 4 and the series could go the state $(2,2)$. You can express the probability relation as:
$$P(2,1) = {1 \over 2}P(3,1) + {1 \over 2}P(2,2)$$
By the same logic: $$P(3,1) = {1 \over 2}(1) + {1 \over 2}P(3,2)$$ $$P(3,2) = {1 \over 2}(1) + {1 \over 2}P(3,3)$$ $$P(3,3) = {1 \over 2}(1) + {1 \over 2}(0)$$ $$P(2,2) = {1 \over 2}P(3,2) + {1 \over 2}P(2,3)$$ $$P(2,3) = {1 \over 2}P(3,3) + {1 \over 2}(0)$$
You have the necessary equations above to solve for $P(2,1)$ for the probability of winning from any state, for that matter.
Also, you can use matrix notation to express state transitions instead of the tedious equations I wrote above.
Note that the winning team necessarily wins the last game of the series. So there are three cases to consider:
Therefore the answer is $$\frac{1}{4}+\frac{1}{4}+\frac{3}{16}=\frac{11}{16}$$