Every NFL fan knows the 1972 Miami Dolphins are the only team to finish a season perfect. How hard is that, really? A little probability gives a surprisingly clear answer.
The basic math
If a team wins each game with the same probability p, and games are treated as independent, the chance of winning all 17 is simply:
P(17-0) = p^17
Here is what that looks like for different per-game win probabilities:
| Win probability per game (p) | P(17-0) = p^17 | Roughly 1 in... |
|---|---|---|
| 0.60 | 0.017% | 5,900 |
| 0.70 | 0.23% | 430 |
| 0.80 | 2.25% | 44 |
| 0.85 | 6.31% | 16 |
| 0.90 | 16.68% | 6 |
Even a team that is an 85% favorite in every single game (which no real team is) goes 17-0 only about once in 16 tries. Small drops in per-game probability compound fast: moving from 0.90 to 0.80 cuts the perfect-season chance by about 7x.
What history says
- 1972 Dolphins: 14-0 in the regular season, then three playoff wins including the Super Bowl, for a 17-0 record overall. That is the only perfect season, playoffs included, in the Super Bowl era.
- 2007 Patriots: 16-0 in the regular season, but they lost Super Bowl XLII, so they did not finish 18-0.
Today's regular season is 17 games, so a "17-0" regular season is a different (and in some ways easier) target than the Dolphins' 17-0, which included the playoffs.
Caveats
The independence assumption is a simplification. Real games are correlated (injuries, schedule strength, a quarterback who is available or not), and p is different each week. But the p^17 estimate is a good back-of-the-envelope tool for seeing why perfect seasons are so rare.
Try it yourself
If you want to play with the idea, you can build a roster and see how close you can get to 17-0 in 17-0 Game, a free browser NFL roster and season challenge.
What per-game win probability would you assign to the best team you have ever seen?
Top comments (0)