Game probabilities for week 11 are available at the New York Times. This week I do a back of the envelope analysis of how likely it would be for a team like the Chiefs to finish the regular season undefeated..
...But we know N.F.L. games are not coin flips, so how good would a team have to be to have a 50/50 shot at going undefeated through 16 games? If a team were so good that it had a 90 percent chance of winning any one game, it would only have a 19 percent chance of going 16-0. In fact, a team would need a 96 percent chance of winning any one game before it had better than even odds of going undefeated. So it’s a rare thing for a reason...
I would guess that the Chiefs' chances of going undefeated are higher than the less than 1 percent you listed for the simple fact that each game isn't independent. For example, if the Chiefs win in Denver (and assuming it doesn't happen because of 3 Broncos fumbles returned for touchdowns), their efficiency ratings will likely go up, meaning they would likely be a better team than we estimate now and their chances in their future games likely go up as well. I'm not really nitpicking given that I have no idea how I would incorporate Bayes theory into a model like this, but I thought I would just point this out.
aren't all possible combinations of wins and losses equally rare?
for instance WLWLWLWLWLWLWLWL
one could look at it as there is only one way to go 16-0, but many ways to go 8-8.
and just for fun,
If we simply assumed N.F.L. games were 50/50 independent coin flips, KC chances of of being 9-0 is a somewhat insanely improbable 0.195 percent. But they did it.
Their chances of actually going to 16-0 now would be 0.78%
Dont forget there are 32 teams, and we only get excited about the best ones and worst ones. So yeah, its unlikely that KC should have gone 9-0, but each year we have 32 shots to see how extreme somebody's record could be.