<s>[INST] How many ways are there for $2$ teams to win a best of $7$ series? Case $1$: $4$ games: Team A wins first $4$ games, team B wins none = $\binom{4}{4}\binom{4}{0}$
Case $2$: $5$ games: Team A wins $4$ games, team B wins one = $\binom{5}{4}\binom{5}{1}-1$...minus $1$ for the possibility of team A winning the first four.
Case $3$: $6$ games: Team A wins $4$ games, team B wins $2$ = $\binom{6}{4}\binom{6}{2}-2$...minus $2$ for the possibility of team A winning the first four games; and the middle four (games $2,3,4,5$), in which case there would be no game $6$.
Case $4$: $7$ games: Team A wins $4$ games, team B wins $3$ = $\binom{7}{4}\binom{7}{3}-3$...minus $3$ for the possibility of team A winning the first four games; games $2,3,4,5$; and games $3,4,5,6$.
Total = sum of the $4$ cases multiplied by $2$ since the question is asking for $2$ teams.
Is this correct?
 [/INST] Let the best of $n$ series be decided after $k$ games. This will happen if in the preceding $k-1$ games $A$ also wins $\lfloor n/2 \rfloor$ and wins the $k^{th}$ game. Hence,
$$C(k) = \dbinom{k-1}{\lfloor n/2 \rfloor}$$
Hence, the total number of ways is
$$\sum_{k=\lfloor n/2 \rfloor+1}^n C(k)$$
In your case, set $n=7$ to get the answer.
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