Game probabilities for the wildcard round are available at the New York Times. This week I take a look at the matchup between KC and IND.
When the Colts host the Chiefs on Saturday in the first of this weekend’s four wild-card games, each team will feel as if it is looking into the mirror. The two teams are nearly statistically identical and tend to play disciplined, low-risk, ball-control football while letting their opponents make mistakes...
The two opponents are even more similar on defense. Kansas City’s and Indianapolis’s net yards per pass attempt allowed are 6.4 and 6.2, both close to the average of 6.2. They both allow 4.5 yards per carry, a bit more than the league average of 4.1. Both teams have grabbed more than their share of interceptions, as the Chiefs intercept the ball on 3.3 percent of all pass plays and the Colts on 2.8 percent...
How do the odds for the packers game change if you consider only packers passing with Aaron Rodgers?
I would like to know what type of advanced math subjects you used to create your models. I have a model, but I think I could make it better if I improved my math skills. Thank you.
you have the colts being more than 56% more likely to win that game than the chiefs (61-39)/39 and yet you still conclude that it is basically a tie between evenly matched teams?
Anon #3, if you read the article, it points out that the expectation is almost entirely down to the fact that the Colts are playing at home. The way to get a coin-flip game is not to match two identical teams (unless it's on a neutral field) but to have a better team playing away at a worse one, as is the case in both NFC games.
paul,
please do not do this "if you read the article" stuff, when the comment was based on what the article says.
the point i am making is that 60/40 is still a coin flip. i.e. that 60/40 is indistinguishable from 50/50. at least according to the author of the article. (based on reading the article)