 Super Bowl game probabilities are available now at the nytimes.com Fifth Down. This week I also review the model's accuracy this season and discuss the how the concept of hindsight bias misleads us about our ability to predict future events.
Super Bowl game probabilities are available now at the nytimes.com Fifth Down. This week I also review the model's accuracy this season and discuss the how the concept of hindsight bias misleads us about our ability to predict future events.
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Super Bowl XLIV Game Probabilities
By
Brian Burke
published on 1/29/2010
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predictions
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Congrats on your great work and season. My question: what's the alternative to working with applied stats? There's not much else to go on.
With the over/under at 56.5, your 52/48 odds say the
Colts should be favored by only 1. I like the call... I think this game could easily go either way.
I'm surprised how the Vegas line has moved, up at least 2 points for the Colts (to 5.5) since the opening line. Wonder why this is? Good comment by Jonathan on the nytimes post, I had no idea teams with recent SB experience won that much more than newcomers. Maybe that's why...
Hey Brian, I see that your model predicted NO to lose by probability = .52. Now have you ever done any stats on breaking down that .52 into lose by <= 3 points, lose by <= 7 points, lose by <= 10 points, lose by <= 14 points, etc? With those probabilities, one could estimate the expected value of taking adjusted NO point spreads at various numbers (e.g., +3, +7, +10, +14, etc). Thanks!
johnny,
not sure if this is what you're asking, but it might help. the way i got the colts -1 number from my post above is using bill james's pythagorean formula for baseball, modified for the nfl. you can find the approximate winning % (or in this case, chance of winning 1 game) from the point spread & over/under:
(PF^x) / ((PF^x) + (PA^x))
in the nfl, x would be around 2.5. so you could determine what the percentages would be for various spreads, and do it that way. the current line, at 5.5, gives the colts about a 62% chance to win.
A shortcut I like is win% = Point differential / 34.5 + .500. At 5.5-pt favorites, the Colts have a 66% chance to win; and at 52% favorites by Brian's system, they would be 0.69-point favorites.
Ryan, where in the equation:
(PF^x) / ((PF^x) + (PA^x))
is the variable for point spread and over/under total? I assume PF = points for and PA = points against and since x = 2.5, x is a constant.
Colts are -5.5 and total is 56.5. So where do these two variables fit in the equation?
Thanks.
Johnny, if the Colts are 5.5 point favorites and the o/u is 56.5, then Vegas predicts a final score of 31.5-26. Plugging those two values in gives the Colts a 62% chance to win.
Brian, i know you have done a lot of work on home field advantage - especially how it relates to familiarity and how it deteriates quickly. Have you ever done any research to see if the super bowl is truly neutral or would a team that plays in a specific stadium more often hold some type of home field advantage?
For example would the Colts have a slight home field advantage from playing in Miami more often than Saints (granted they both played their this year). I imagine it could be even bigger if a division rival made the super bowl (maybe if the Pats would have made it this year). Just something i was thinking about.
Interesting thought, but the sample size of SBs is tiny.
WHO IS GONNA WIN SUPER BOWL 44????
oh never mind