<s>[INST] Probability that A wins a best of 7 (4 matches) $A$ and $B$ play a series of best of $4$. The probability that $A$ wins a given game is $p$. Assume that the games are independent of each other. What is the probability that $A$ wins?
Let $A_{i}$ be the event that $A$ wins the series in $i$ matches.
Then, the book states, $$P(A) = P(A_{4}) + P(A_{5}) + P(A_{6}) + P(A_{7}) = p^{4} + \binom{4}{3}p^{4}q + \binom{5}{3}p^{4}q^{2} + \binom{6}{3}p^{4}q^{3}.$$
I am having a bit of trouble understanding the calculation. I am trying to be a bit more formal here. As an example, lets concentrate on $A_{5}$. It really is $A \cap G_{5}$ where $G_{i}$ is the event that $i$ matches were played in the series. Then, $$P(A_{5}) = P(A \cap G_{5}) = P(A|G_{5})P(G_{5}) = \binom{5}{4}p^{4}qP(G_{5}).$$ Now, we equate this term to $P(A_{5})$ to get $\binom{5}{4}p^{4}qP(G_{5}) = \binom{4}{3}p^{4}q$, from which we get, $$P(G_{5}) = \frac{\binom{4}{3}}{\binom{5}{4}}.$$
Is this complete nonsense? If not, can someone give me an intuition for $P(G_{5})$? It just doesn't click in my head right now. To rephrase, $P(A_{5}) = \binom{4}{3}p^{4}q$ was given in the answer, but in my attempt, I have this $P(G_{5})$ term which I am not sure how to find without the equation $\binom{5}{4}p^{4}qP(G_{5}) = \binom{4}{3}p^{4}q$.
 [/INST] I don't see any way to go with your approach without utilizing the book's approach but I will do so anyways.
For $G_5$ to occur, either team $A$ needs to win $3$ or the first $4$ games and then win game $5$ with probability $$\binom{4}{3}p^{4}q$$ or team $B$ needs to win $3$ or the first $4$ games and then win game $5$ with probability $$\binom{4}{3}q^{4}p$$ 
so $$P(G_5)=\binom{4}{3}p^{4}q+\binom{4}{3}q^{4}p$$
and $$P(A\mid G_5)=\frac{\binom{4}{3}p^{4}q}{\binom{4}{3}p^{4}q+\binom{4}{3}q^{4}p}$$
Then
$$P(A|G_{5})P(G_{5})=\frac{\binom{4}{3}p^{4}q}{\binom{4}{3}p^{4}q+\binom{4}{3}q^{4}p}\cdot\left(\binom{4}{3}p^{4}q+\binom{4}{3}q^{4}p\right)=\binom{4}{3}p^{4}q$$
Of course it's easier to just note that team $A$ must win $3$ of the first $4$ games and then win the $5^{th}$ game. This is a negative binomial with $n$ trials given $k$ successes where $n=5$ and $k=4$.
We have
$$P(X=n)={n-1 \choose k-1}p^kq^{n-k}={4 \choose 3}p^4q$$
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