With the Super Bowl coming up, we're excited to launch the first edition of our Sports Math blog series, where we'll take a look at some of the goings-on in the world of sport from a mathematical lens. As anyone who follows sports is well aware, data and statistics are becoming more and more important to the administration, playing and enjoyment of sport, so why not take some time to understand and appreciate the math behind the games. We hope you enjoy!
In this edition, I’ll explore a question that you may have heard asked throughout the NFL playoffs which relates to the mathematical idea of conditional probability.
How hard is it to beat a team three times in the same season?
This question came up during the divisional round, when the New Orleans Saints and the Tampa Bay Buccaneers were getting ready to face off for the third time this season, and a quick internet search shows that this is a question that has been around for years. But before I get into answering this question, let me give a brief introduction to conditional probability.
The Math - Conditional Probability
Imagine that I hold up a jar full of numbered balls and ask you to select one. Before you do, I ask you "What is the probability that the ball you choose has an even number on it?"
For most people, a reasonable assumption here is that there would be about a 50% chance that the selected number is even (yes, yes, I know, I should have specified whether the numbers were real or complex, or that they were integers, but just roll with me on this).
However, just before you select your ball, I give you a bit more information. All of the balls have prime numbers on them! How might this affect your chances of selecting an even number? Well, since 2 is the only even prime number, your chances of choosing an even number have drastically decreased.
The probability of choosing an even number, with no other information, is 50%, but the probability of choosing an even number, given that all the possibilities are prime, is far smaller.
This is conditional probability. Essentially, we add a bit of extra information (a condition) to an initial scenario, and we consider how this extra information may impact our understanding of the scenario, and our calculation of the odds. This is often phrased as “What is the probability of X being true, given that Y is known to be true.”
But now, how on earth does this apply to football?
As mentioned, the Buccaneers and the Saints faced one another for the third time this season during the playoffs. The Saints had won the first two games, and as the old saying goes, “It’s hard to beat a good team three times in the same season.” Did this mean that the Bucs should be favored in the playoff matchup, because the Saints had already won twice?
We’re going to approach this question in two different ways: one that is more theoretical, and one that is experimental, using the evidence from the last 50 years of the NFL.
Approach #1 - Theoretical
We’ll start with an assumption here that for any given game, each team has a 50% chance of winning. Essentially, each game is a coin toss.
There are only a few different ways that the results of three games can occur (we’ll ignore ties for the purposes of this exercise), so I’ll list them below, in terms of wins and losses for one team, the Saints.
Out of the 8 possibilities here, two of them (WWW and LLL) involve one team winning all three games. So mathematically, the conventional wisdom is correct. It is difficult to beat a team three times in a season, with only a 2 in 8 (25%) chance. The Saints have a 1 in 8 chance of winning all three games, as do the Bucs.
However, this absolutely does not mean that the Saints had a 1 in 8 chance of beating the Bucs in the playoffs. That’s because there is more information available to us, namely that the Saints won the first two games! Of the 8 possibilities above, there are only two outcomes where the Saints win the first two games, WWW and WWL, giving a 50% chance that Saints beat the Bucs all three times, given that they won the first two games.
This is the power of conditional probability. It’s the difference between “What are the odds of a team beating another team 3 times in a season?” and “What are the odds of beating a team 3 times in a season, given that you already won the first two games?” If you already beat the other team twice, it stands to reason that you’re more likely to beat them 3 times than if you know nothing about the first two results.
Now, let’s see if our assumptions match up with the reality of NFL history.
Approach # 2 - Experimental
Check out this article from Full Press Coverage for some of the numbers that I use below.
Since the start of the 1970 season, the Saints and the Buccaneers are the 70th pair of teams to have met 3 times in one season.
Of those 70 occurrences, only 14 of them (20%) resulted in one team winning all three games, which is "similar" to our calculation of there being a 25% chance that one team wins all three games. Either way, I’d say the data agrees with the statement that it’s difficult to beat a team three times in one year.
Once again though, there’s a big however. In the case of the Saints and the Bucs, we know that one team won the first two games already! This was only the 22nd time in NFL history that this occurred, meaning that for the other 48, the teams split the first two games, so there was already no chance of one team winning three times before the third game even began.
Of the previous 21 occurrences, only 14 times did the team who had won the first two games win the third (a 2/3 or approximately 67% probability). This is much higher than our theoretical calculation of 50%. This still seems logical, as our theoretical model assumed each team had a 50% chance of winning every game, which is definitely not reality. It stands to reason that if one team has beat another team twice already, there’s a good chance that they are the better team, and they should be more likely to prevail a third time.
Either way though, both our theoretical model and our experimental model dispel the idea that it’s hard to beat the same team three times in a season, given you’ve already beaten them twice. Context is important, and it’s good to take into account all information you have at your disposal.
That being said, unfortunately for Saints fans, Tom Brady and the Bucs 'bucked' the probability and beat the Saints to move onto the conference championships - and then Super Bowl LV - reducing our experimental probability to 14/22 or about 64%. I guess that’s part of the joy of sports. The data and predictions can tell us one thing, but they still have to play the game!