tag:blogger.com,1999:blog-38600807.post6913879929762526254..comments2018-06-02T14:19:34.554-04:00Comments on Advanced Football Analytics (formerly Advanced NFL Stats): Super Bowl XLIV Game ProbabilitiesUnknownnoreply@blogger.comBlogger11125tag:blogger.com,1999:blog-38600807.post-14620118032983816562010-02-03T20:08:03.106-05:002010-02-03T20:08:03.106-05:00oh never mindoh never mindAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-9186579644520937892010-02-03T20:06:25.184-05:002010-02-03T20:06:25.184-05:00WHO IS GONNA WIN SUPER BOWL 44????WHO IS GONNA WIN SUPER BOWL 44????Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-52133039819109943412010-01-31T21:47:59.713-05:002010-01-31T21:47:59.713-05:00Interesting thought, but the sample size of SBs is...Interesting thought, but the sample size of SBs is tiny.Brian Burkehttps://www.blogger.com/profile/12371470711365236987noreply@blogger.comtag:blogger.com,1999:blog-38600807.post-18561742204257269632010-01-31T21:26:53.878-05:002010-01-31T21:26:53.878-05:00Brian, i know you have done a lot of work on home ...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? <br /><br />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.Buzznoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-25947455846482365402010-01-31T13:42:36.665-05:002010-01-31T13:42:36.665-05:00Johnny, if the Colts are 5.5 point favorites and t...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.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-25723785925736699502010-01-30T20:23:27.242-05:002010-01-30T20:23:27.242-05:00Ryan, where in the equation:
(PF^x) / ((PF^x) + (...Ryan, where in the equation:<br /><br />(PF^x) / ((PF^x) + (PA^x))<br /><br />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. <br /><br />Colts are -5.5 and total is 56.5. So where do these two variables fit in the equation?<br /><br />Thanks.johnnyjohnnywuhttps://www.blogger.com/profile/09579082738626978081noreply@blogger.comtag:blogger.com,1999:blog-38600807.post-10501588906601490812010-01-30T17:15:40.499-05:002010-01-30T17:15:40.499-05:00A shortcut I like is win% = Point differential / 3...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.Zachnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-49686078175210446872010-01-30T16:16:38.355-05:002010-01-30T16:16:38.355-05:00johnny,
not sure if this is what you're asking...johnny,<br />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: <br />(PF^x) / ((PF^x) + (PA^x))<br />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.Ryannoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-30696472945873965442010-01-29T21:42:02.602-05:002010-01-29T21:42:02.602-05:00Hey Brian, I see that your model predicted NO to l...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!johnnyjohnnywuhttps://www.blogger.com/profile/09579082738626978081noreply@blogger.comtag:blogger.com,1999:blog-38600807.post-11659313943924078202010-01-29T11:39:04.608-05:002010-01-29T11:39:04.608-05:00With the over/under at 56.5, your 52/48 odds say t...With the over/under at 56.5, your 52/48 odds say the <br />Colts should be favored by only 1. I like the call... I think this game could easily go either way.<br /><br />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...Ryannoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-62009476543387372472010-01-29T09:47:30.735-05:002010-01-29T09:47:30.735-05:00Congrats on your great work and season. My questi...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.Anonymousnoreply@blogger.com