# Math Help - World Series!

1. ## World Series!

Please explain step by step to complete this. Thanks a million!

A play-off championship includes up to five games. Whichever
team wins three games first is the winner of the series. In how
many different ways can a team win the playoff series?

For the baseball World Series, a team must win four out of a
possible seven games. In how many different ways can a team
win the World Series?

2. ## Re: World Series!

Hey SuperMaruKid.

Can you show us some working out? If you have any ideas (even incomplete) can you please demonstrate them for the readers here?

3. ## Re: World Series!

5-3 equals 2×5 = 10 space different ways to win for the first question
5: Games
3: Win
2: Losses

7-4 equals 3×7 = 21 different ways to win for the second question
7: Games
4:Win
3: Losses

4. ## Re: World Series!

Not quite right. What you've calculated is the number of ways to combine 3 wins and 2 losses, which is C(5,2) = 10. But the series does not necessarily last 5 games -for example the team could sweep the first three games. Your method also doesn't take into account the fact that the last game of the series must be a win.

If you write out all possible winning combinations you have:

3 game series (no losses): WWW
4 game series (one loss): LWWW, WLWW, WWLW
5 game series (2 losses): LLWWW, LWLWW, LWWLW, WLLWW, WLWLW, WWLLW

So the total number if ways to win is 10. This happens to be the same answer you got for the case of a best-3-of-5 series, but gives a different answer than what you got for the best-4-of-7 series.

The way to approach this is to consider the number of losses that the team takes combined with the fact that the last game must be a win. For example in a 3-game series (a sweep) there are no losses and so there is one way that can occur - WWW. In a 4-game series there is one loss, and that loss must take place in the first three games, so the number of ways to win a 4-game series is C(3,1)=3. For a 5-game series the team takes two losses in the first four games, and there are C(4,2)=6 ways to do that. Add all these together and you get 10.

Now apply this technique to the question about best 4-of-7 games and see what you get, considering that the series may last 4, 5, 6, or 7 games. Your answer should be a bigger number than what you had previously.

5. ## Re: World Series!

Originally Posted by ebaines
Not quite right. What you've calculated is the number of ways to combine 3 wins and 2 losses, which is C(5,2) = 10. But the series does not necessarily last 5 games -for example the team could sweep the first three games. Your method also doesn't take into account the fact that the last game of the series must be a win.
If you write out all possible winning combinations you have:
3 game series (no losses): WWW
4 game series (one loss): LWWW, WLWW, WWLW
5 game series (2 losses): LLWWW, LWLWW, LWWLW, WLLWW, WLWLW, WWLLW

So the total number if ways to win is 10. This happens to be the same answer you got for the case of a best-3-of-5 series, but gives a different answer than what you got for the best-4-of-7 series.
Actually it does give the same answer in a best of seven series.
Consider this revised list:
3 game series : WWWLL
4 game series: LWWWL, WLWWL, WWLWL
5 game series: LLWWW, LWLWW, LWWLW, WLLWW, WLWLW, WWLLW
That is a complete listing of three W's and two L's.
Here is the calculation.
The winner is determined in the third, fourth, or fifth game.

In the best of seven series, the winner is determined in the fourth, fifth, sixth, or seventh game.
That is four W's and three L's. Here again is the calculation.

Note that is $\binom{7}{3}$.

Here is a bit of related history.

6. ## Re: World Series!

Good point - interesting how that works out. Looks like the OP calculated C(7,2) to get an answer of 21, rather than C(7,3).

7. ## Re: World Series!

Hello, SuperMaruKid!

For the baseball World Series, a team must win four out of a possible seven games.
In how many different ways can a team win the World Series?

The team could win the first four games: $WWWW$
There is ${\color{red}1}$ way.

Five-game series: . $\_\,\_\,\_\,\_\,W$
The first four games must have three W's and one L in some order.
There are: . ${4\choose3,1} \:=\:{\color{red}4}$ ways.

Six-game series: . $\_\:\_\:\_\:\_\:\_\:W$
The first five games must have three W's and two L's in some order.
There are: . ${5\choose3,2} = {\color{red}10}$ ways.

Seven-game series: . $\_\:\_\:\_\:\_\:\_\:\_\:W$
The first six games must have three W's and three L's in some order.
There are: . ${6\choose3,3} = {\color{red}20}$ ways.

Therefore, there are: . $1 + 4 + 10 + 20 \:=\:{\color{blue}35}$ ways.