Super Bowl XLIV Game Probabilities

Super Bowl game probabilities are available now at the 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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11 Responses to “Super Bowl XLIV Game Probabilities”

  1. Anonymous says:

    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.

  2. Ryan says:

    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...

  3. johnnyjohnnywu says:

    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!

  4. Ryan says:

    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.

  5. Zach says:

    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.

  6. johnnyjohnnywu says:

    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?


  7. Anonymous says:

    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.

  8. Buzz says:

    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.

  9. Brian Burke says:

    Interesting thought, but the sample size of SBs is tiny.

  10. Anonymous says:


  11. Anonymous says:

    oh never mind

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