The sample space of the best of seven series

combinatoricsdiscrete mathematicspermutationsprobability

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

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

Best Answer

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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