Dtyjlui

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!

Note that the winning team necessarily wins the last game of the series. So there are three cases to consider:

• The series ends in $5$ games. This occurs if the current leading team wins games $4$ and $5$, which occurs with probability $frac{1}{4}$
  


• The series ends in $6$ games. Since they must win game $6$, they also must win exactly one of games $4,5$. The outcomes look like WLW, LWW. There are $2$, and each has probability $frac{1}{8}$ so this outcome occurs with probability $frac{1}{4}$
  


• The series ends in $7$ games. Since they must win game $7$, they also must win exactly one of games $4,5,6$. The outcomes look like WLLW, LWLW, LLWW. There are $3$ outcomes and each has probability $frac{1}{16}$ so this situation occurs with probability $frac{3}{16}$
  

Therefore the answer is
$$frac{1}{4}+frac{1}{4}+frac{3}{16}=frac{11}{16}$$

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.