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Original question: In a soccer game, two teams are playing against each other in a best of seven series. The game ends when one team wins four games and each game corresponds to a win for the team. What is the sample space of the best of seven series?

Answer: $2(1+4+10+20)=70$

My question is whether my answer or the answer above is correct. Even though I think my answer makes more sense I also think it is wrong because my answer is very different from the given answer.

My method is to use the following formula and add the total number of ways to win then multiply the sum by $2$ to include the total number of ways to lose:

  • If the number of games $= 4$: $$\frac{n!}{n_1!n_2!...n_k!}$$

  • If the number of games $> 4$: $$\frac{n!}{n_1!n_2!...n_k!}-1$$

where $n$ is the total number of letters and $n_1,n_2,...,n_k$ are possible duplicate letters.

$W$ = {win}

$L$ = {lost}

  • Team wins $4$ games out of $7$: (1 way)

$$WWWW$$

  • Team wins $5$ games out of $7$: (4 ways)

$$LWWWW,WLWWW,WWLWW,WWWLW$$

$WWWWL$ (removed since this is the same as $WWWW$)

$$\frac{5!}{4!}-1=4$$

  • Team wins $6$ games out of $7$: (14 ways)

$$\frac{6!}{4!2!}-1=14$$

  • Team wins $7$ games out of $7$: (34 ways)

$$\frac{7!}{4!3!}-1=34$$

So the sample space (total number of ways of winning/losing) is

$$S=2(1+4+14+34)=106$$

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3 Answers 3

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In a best of seven series, the winning team must win the last game.

Say team A wins. We can double our result afterwards to account for the possibility that team B wins.

Team A wins in four games: This can occur in one way: $W_AW_AW_AW_A$, where $W_A$ means that team A wins the game.

Team A wins in five games: For this to occur, team A must win exactly three of the first four games and the fifth game, which can occur in $$\binom{4}{3}\binom{1}{1} = 4$$ ways. The elements in the sample space are $$W_AW_AW_AW_BW_A,\qquad W_AW_AW_BW_AW_A,\qquad W_AW_BW_AW_AW_A,\qquad W_BW_AW_AW_AW_A$$

Team A wins in six games: For this to occur, team A must win exactly three of the first five games and the sixth game, which can occur in $$\binom{5}{3}\binom{1}{1} = 10$$ ways. I will leave listing the corresponding elements in the sample space to you.

Team A wins in seven games: For this to occur, team A must win exactly three of the first six games and the seventh game, which can occur in $$\binom{6}{3}\binom{1}{1} = 20$$ ways. Again, I will leave listing the corresponding elements in the sample space to you.

Hence, there are $$1 + 5 + 10 + 20 = 35$$ ways for team A to win the series and $$2(1 + 5 + 10 + 20) = 70$$ ways for one of the teams to win the best of seven series.

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  • $\begingroup$ My answer is correct for the first two cases of winning in four and five games but why is the total number of ways when Team A wins in six games $10$ and in seven games $20$ significantly different from my answers of $15$ and $34$? I don't think I overcounted since I have divided by the possible duplicate letters and subtracted $1$ from the result unless the formula I used was wrong $\endgroup$
    – user314
    Feb 6, 2021 at 13:04
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    $\begingroup$ That is because you are choosing any $4$ games to win from $6$ which is not correct. Subtracting $1$ does not take out all cases where the person could have won in first $4$ or first $5$. $\endgroup$
    – Math Lover
    Feb 6, 2021 at 13:08
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    $\begingroup$ In case of $5$ games, it worked out because there is only one way to win in $4$ but in $6$ games, you should do $\frac{6!}{4!2!} - 5 = 10$. By the way $\frac{6!}{4!2!} - 1 = 14$ and not $15$. $\endgroup$
    – Math Lover
    Feb 6, 2021 at 13:11
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    $\begingroup$ You already found that ways to win in $4$ games is $1$ and ways to win in $5$ games is $4$. So if we have to win in $6$, we need to take out cases where we are winning in $4$ or $5$ games that is total of $5$. $\endgroup$
    – Math Lover
    Feb 6, 2021 at 13:22
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    $\begingroup$ No it is not valid. Instead it should be $S(X=n) = \frac{n!}{n_1!n_2!...n_k!}-S(X \lt n)$. $S(X=n)$ is number of ways to win in $n$ games and $SX \lt n)$ is number of ways to win in less than $n$ games. $\endgroup$
    – Math Lover
    Feb 6, 2021 at 13:32
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I completely agree with the other responses, but would like to point out a shortcut. Assume that Team-1 wins, perform the enumeration, and then multiply by 2, re Team-2 could also win.

There is a bijection between each sequence of games that end immediately after Team-1 achieves its 4th win and each 7 game sequence where Team-1 wins exactly 4 games. Simply take each sequence where Team-1 has its 4th win prior to the 7th game and then extend the sequence with Team-1 losing all of the remaining games (if any).

Note that $2 \times \binom{7}{4} = 2 \times 35 = 70.$

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Let us fix a team and try to find the number ways for this team to win.

The answer is:$$\sum_{k=3}^6\binom{k}3=1+4+10+20=35$$where $\binom{k}3$ corresponds with the number of games that end after $k+1$ meetings.

We denote the results as e.g. $WLWWLW$ where the team won at meetings 1,3,4,6.

Note that in that situation where the game ends after $k+1$ meetings a $W$ must stand on the $k+1$-spot and exactly 3 letters $W$ must be among the $k$ former spots. So this results in $\binom{k}3$ possibilities.

More directly again looking at a fixed team we can think of 7 open spots of which 4 must be filled up with a letter $W$. The other spots stay empty if there is no letter $W$ on RHS and are filled with a letter $L$ otherwise. This gives (again):$$\binom74=35$$possibilities.

(actually a proof of the hockey-stick identity).

There are two teams that can win so the final answer is:$$2\times35=70$$

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