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Probability of Repeating

The Seahawks won the Superbowl! What's Next?

"It is very hard to repeat." I hear this all the time and have always accepted it to be true without much thought. After all, it's only been done once since 2000 (2005 Patriots). Wikipedia even has a blurb about it.

But the more I think about it, the more I question it. This implies that it is easier to win the Superbowl if a team didn't win the previous year. More precisely, it suggests that there is a place of finish outside of first place that gives a team a greater chance of winning it all the next year. If that's the case, I want to find this magical place of finish.

From various internet sources, I compiled results from all Superbowls, conference championships from those years, and divisional playoff games (since 1970). This gives me the top 8 teams from each of the last 44 years, and the top 4 teams from all 48 Superbowl years. By my count (which admittedly may be error prone, I just did this out of curiosity one day while procrastinating at work), out of the 47 tries (no chance of repeat in year 1), there have been 8 successful Superbowl repeats, which puts the success rate at 17%:

  • 1968 Green Bay, 1974 Miami, 1976 Pittsburgh, 1980 Pittsburgh, 1990 SF, 1994 Dallas, 1999 Denver, 2005 New England

Next, I look at Superbowl losers. How often do they go on the win the next year? This happened just twice, or at a rate of 4%. Man, Superbowl losers suck!

  • 1972 Dallas, 1973 Miami

Up next are the conference championship losers (#3 and #4 finishers). To have data for all Superbowl years, I'm including AFL/NFL championship results from the pre-merger years. In total, #3, #4 teams have gone on to win the Superbowl the following year 9 times. Most recently, this happened to the 2011-2012 Ravens, who won it all in 2013. Out of 47 tries, that is a rate of 19%. To simplify things, I assign each team half of the chance of success, or 9.5%, rather than figuring out any tie breakers.

Finally, by my count, since 1970 merger, the divisional playoff games losers (#5, #6, #7, #8 teams) have won the following year a total of 12 times. Out of the 43 tries (post-merger years only), that's a combined rate of 28%. Keeping things simple, divided evenly, each team has a 7% chance.

Tallying it up (Table 1), during the Superbowl era, the top 8 teams have won the following year's Superbowl a whopping 68% of the time, leaving just 32% for the rest of the league combined. Best part is (for Seahawks fans), the Superbowl winner has by far the best chance of winning the next year at 17%.

Table 1

Current Year Place of Finish

Chance of Winning Following Year’s Superbowl

#1 (Winner!)

17%

#2 (The Biggest Loser)

4%

#3

9.5%

#4

9.5%

#5

7%

#6

7%

#7

7%

#8

7%

So while it is not easy to repeat, it is actually a far easier task than what’s faced by any of the other teams. Historically speaking, the Superbowl winner has almost twice as good of a chance of winning next year as any other team! Now, I don't know what's going to happen next year, but you can't tell me that the Seahawks don't have a great shot because they won this year.

Probability of Winning the Superbowl for all 32 teams

So far, I'm just counting past data. The rest of this article is a little more speculative. Based on the past data, I’m putting forth a model for the probability of winning the Superbowl based on each team’s place of finish in the previous year. I would love to test the model against the complete data of each Superbowl winner’s previous year’s place of finish (maybe look at original draft position, before any trades), but I don’t have the time to put that together or know where to find that data easily. Suggestions?

I’m not great at statistics. I have a handle on the basics but mostly I just look up stuff on Wikipedia, so apologies if I'm butchering terminologies. Looking at the data in Table 1, aside from the 4% chance for the Superbowl losers, which is rather inexplicable, the rest of the data look like they would fit some kind of probability distribution. When plotted against a geometric distribution with p=17%, it comes out surprisingly nice (the equation for the curve is f(N)=0.17*(1-0.17)^(N-1)).

Sb_medium

via lh6.googleusercontent.com

A geometric distribution is typically used to describe the chance of one success after N independent trials, where the chance of success in each trial is p. Does it make sense here? Is it just a coincidence that the curve appears to fit a few data points? Maybe, but let me at least attempt to rationalize it. I interpret it to mean that for each team, the probability of playing to their previous year’s ranking is 17%. This isn’t surprising because there are many factors that can derail a team: injuries, age, loss of key contributors, injuries. So the 1st place team from the previous year has a 17% chance of winning the Superbowl again. For the 2nd place team to win, they must play to their ranking AND have the 1st place team fail (17%*83%=14%). Each subsequent team has a 0.83X smaller chance of winning it all. For the Nth place team, the chance of winning is 0.17*0.83^(N-1). That is, they must play to their ranking and have all N-1 teams ahead of them fail.

I don’t know if this makes all that much sense. The rationale isn’t exactly foolproof. However, this is a simple model that fits the data rather well (1).

1. Except for the Superbowl losers, who seem to just fall on their faces into the depth of suck. It's actually very puzzling. I really can’t think of any logical reason why it’s so tough for Superbowl losers to win the next year.

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