# 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?

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It depends. How many of those teams are the Toronto Maple Leafs? – Emily Nov 26 '13 at 5:15

We count the ways in which Team A can win the series, and double the result. To count the ways A can win the series, we make a list like yours.

A wins in $4$: There is $1$ way this can happen.

A wins in $5$: A has to win $3$ of the first $4$, and then win. There are $\binom{4}{3}$ ways this can happen.

A wins in $6$: A has to win $3$ of the first $5$, then win. There are $\binom{5}{3}$ ways this can happen.

A wins in $7$: A has to win $3$ of the first $6$, then win. There are $\binom{6}{3}$ ways this can happen.

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I understand this approach, but am having trouble understanding why my approach is incorrect. I know its incorrect but can't understand why since it seems logical to me. Any help? – Al Jebr Nov 26 '13 at 4:57
Let's look at your calculation for $5$. Your chose the $4$ games A wins. Once you have done that, the game that B wins is determined. So there are $\binom{5}{4}$ ways only (and then we subtract $1$ as you did. If you wish, we could write $\binom{5}{4}\binom{1}{1}-1$. – André Nicolas Nov 26 '13 at 5:01
For $6$, there are two mistakes. First it should be $\binom{6}{4}$. Second, you need to subtract a lot more than $2$, for there are more than $2$ ways for A to have won in $4$ or $5$. You would want $\binom{6}{4}-1-4$. A similar subtracting process would work for the $7$ game case. – André Nicolas Nov 26 '13 at 5:05
For 6, what other cases would we subtract? Team A cannot win the first 4 in a row, nor games 2,3,4,5. What other combinations can it not win? I only see those two. – Al Jebr Nov 26 '13 at 5:09
Actually I see where I went wrong. Thank you. Another possibility that cannot happen is AABAAB. – Al Jebr Nov 26 '13 at 5:10

For another approach, if the series ends before the seventh game, extend it to seven games by having the losing team win the rest. The series will now be four games to three. There are $2$ ways to choose the losing team, and ${7 \choose 3}$ ways to choose which game the losing team wins. The total is then $2{7 \choose 3}=70$

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There are $\begin{pmatrix} 7 \\ 4\end{pmatrix}$ ways of arranging $4$ W's and $3$ L's. This is the answer.

Note that the following sequences are essentially identical:

$$WWLWWLL$$ $$WWLWW$$

Why?

Because once the winning team has amassed four wins, it doesn't matter if we count the remaining un-played games as losses or not. Simply assume that they are. It doesn't change the equation.

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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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