NBA Meets Combinations and Permutations

I've only had NBA on the brain for the last 2 weeks.

How many possible outcomes can there be in a 7-game series?

You might assume this is a straightforward combination/permutation type of question, but it's a bit more complicated than that. 

Usually, to determine the difference between a permutation and a combination, you'd have to ask yourself, "Does order matter?" If yes, you're dealing with a permutation. If no, it's a combination. 

In the case of a Best-of-X series, does order matter? The answer is... sort of. 

In a best-of-7 series, if one team wins 4 games and the other wins 3, you may assume the order in which these wins happened doesn't matter.

But, if one team wins all of the first 4 games, the series ends there, and never makes it to 7 games. In that case, the order of the wins affects the possible outcomes. 

To answer the question of how many possible outcomes exist for a 7-game series, you have to consider that it might not end up being a 7-game series after all. 

A best-of-7 series can end in either 4, 5, 6, or 7 games, considering each of those outcomes individually is necessary for determining the possible outcomes. 

Let's say the series ends in 4 games. There are only two ways that happen, with team A winning all 4 games, or with team B winning all 4. 

4-game series outcomes:
AAAA
BBBB

With 5 games it gets a little more complicated:
AAABA
AABAA
ABAAA
BAAAA
ABBBB
BABBB
BBABB
BBBAB

There are only 4 ways each team can win a 5-game series. Here, you see another reason why using a straight combination doesn't work: the team that wins the series MUST win the final game. 

Using a combination approach would lead you to believe that there are 5 ways each team can win a 5-game series. 

5C4 = 5!/(4!•1!) = 5

But that would give us a result like AAAAB, which can't happen in a 5-game series, as team A would have swept after the 4th game. The way around that hitch is to assume a win in the 5th game for the winning team and make the combination 4C3, which equals 4. That's the number of ways each team can win a 5-game series, equally, 8 total outcomes that end in 5 games (4 with team A winning, and 4 with team B winning). 

Here are the ways a 6-game series can conclude:
AAABBA
AABABA
AABBAA
ABAABA
ABABAA
ABBAAA
BAAABA
BAABAA
BABAAA
BBAAAA
BBBAAB
BBABAB
BBAABB
BABBAB
BABABB
BAABBB
ABBBAB
ABBABB
ABABBB
AABBBB

For each team, there are 10 possible ways of winning in 6 games. Following the pattern from before, we assume that game 6 goes to the winning team, then apply 5C3 to each team, which gives 20 possible outcomes. 

Finally, for a 7-game series, here are the possible outcomes:
AAABBBA
AABABBA
AABBABA
AABBBAA
ABAABBA
ABABABA
ABABBAA
ABBAABA
ABBABAA
ABBBAAA
BAAABBA
BAABABA
BAABBAA
BABAABA
BABABAA
BABBAAA
BBAAABA
BBAABAA
BBABAAA
BBBAAAA

20 possible results just for team A winning, reverse the letters to get the additional 20 results for team B winning. 

To use a combination approach for a 7-game series, assume the winning team wins game 7, and use 6C3 to calculate the outcome of the other 6 games. 6C3 = 20, and then multiply that by two to get results for either team winning. 

So in total, there are 2 possible outcomes if the series ends in 4 games, 8 possible outcomes if the series ends in 5, 20 possible outcomes if the series ends in 6, and 40 possibly outcomes if the series ends in 7.

Add them all up and there are 70 possible outcomes for a best-of-7 series.

 

-John P, Instructor