They said that now, at the midpoint of the NFL season, there are still three undefeated teams. That is the most there have ever been at this point in the season. What is the chance we have at least one of those teams staying undefeated for the whole season?
There are seven games to go for each team. First we would want to know what the chance of winning a single game would be. I'm going to go with an 80% chance. That sounds about right. They're undefeated at this point, so obviously pretty good, yet they wouldn't be invincible.
The chance of any one team of winning their next seven games is (0.8)^7 = 0.21. Now what is the probability of New England, or Cincinnati, or Carolina remaining unbeaten? The easiest way to deal with a three team "or" problem is to look at it negatively. There is a 1-0.21 = 0.790 chance of a team having at least one loss the rest of the way. The chance all three are defeated is 0.493. So, there must be a 1-0.493 = 0.507 of that not being the case. So chances are at least one team will be undefeated - a 50.7% chance.
It could be done the straight forward way, but its a little more cumbersome.
0.21+0.21+0.21-(0.21)(0.21)-(0.21)(0.21)-(0.21)(0.21)(0.21)+(0.21)(0.21)(0.21) = 0.507 = 50.7%
For full disclosure, since I thought about it yesterday, Cincinnati lost on Monday Night Football. I was just giving them the win. I probably shouldn't have done that.
Now we're down to two teams. There is still a (0.8)^7 = 0.21 chance of any one team winning seven in a row. The chance that either Carolina or New England does it is:
(0.21)+(0.21)-(0.21)(0.21) = 0.347 = 34.7%
If you take exception to my 80% single win assumption, it's easy to substitute another value. This makes a bigger difference in the final probabilities than I would have thought. For example, if you figure the chance of a team of this caliber wins a single game is 85%, the chance one of the remains unbeaten is now 53.9%. If 90%, the chance of having an unbeaten is all the way up to 73.8% .
