Comments on: There is only a 10% chance Oregon, Kansas State, and Notre Dame all go undefeated http://www.footballperspective.com/there-is-only-a-10-chance-oregon-kansas-state-and-notre-dame-all-go-undefeated/ NFL History and Stats Sun, 06 Jan 2013 21:28:44 +0000 hourly 1 https://wordpress.org/?v=6.9.4 By: George http://www.footballperspective.com/there-is-only-a-10-chance-oregon-kansas-state-and-notre-dame-all-go-undefeated/#comment-6284 Sun, 06 Jan 2013 21:28:44 +0000 http://www.footballperspective.com/?p=3844#comment-6284 I think both formulas are right (the principle behind them anyway I haven’t copied this into Excel so I wouldn’t know if that works but they look virtually the same as what I have). I do this but do it slightly differently, and change my spread number from positive to negative and vice versa dependent on the %’s (I expect I could work out how to not have to do that but it doesn’t worry me). With a spread of 0, one formula above would essentially give you 100% – 50%, the other would give you 50% (so one is effectively Team A, the other would be Team B it’s just a question of how you would lay it out).

If you want to get into this, I’d recommend the book from Wayne Winston in the link at the top (it makes things a lot clearer – and then you could end up like me posting random queries on Excel forums when you want to take it a bit further).

Good luck.

]]> By: Chris Povirk http://www.footballperspective.com/there-is-only-a-10-chance-oregon-kansas-state-and-notre-dame-all-go-undefeated/#comment-6238 Sat, 05 Jan 2013 17:22:06 +0000 http://www.footballperspective.com/?p=3844#comment-6238 One typo I noticed when lazily copy-pasting the given formula into Excel:

Above you give this formula:
p(W) = 1-(1-NORMDIST(0.5,SPREAD,13.86,TRUE))+0.5*(NORMDIST(0.5,SPREAD,13.86,TRUE)-NORMDIST(-0.5,SPREAD,13.86,TRUE))

This differs from Neil’s formula at the link you gave in that yours starts with an extra “1-“. Here’s Neil’s:
p(W) = (1-NORMDIST(0.5,SPREAD,13.86,TRUE))+0.5*(NORMDIST(0.5,SPREAD,13.86,TRUE)-NORMDIST(-0.5,SPREAD,13.86,TRUE))

It seems likely that the “1-” was intended to invert the probability so that, e.g., a spread of “3” means that a team is +3, i.e., an underdog, the opposite of how Neil’s formula treats it. However, even if that’s the case, it’s not working quite right. It needs an extra set of parentheses:
p(W) = 1-((1-NORMDIST(0.5,SPREAD,13.86,TRUE))+0.5*(NORMDIST(0.5,SPREAD,13.86,TRUE)-NORMDIST(-0.5,SPREAD,13.86,TRUE)))

You can see that the formula given in this post is off by plugging in a SPREAD of 0. Instead of getting a 50.0% chance of victory, you’ll get a 52.9% chance. The 2.9% difference for SPREAD=0 is small, and the error is probably never any larger than that, regardless of SPREAD. Still, it may matter to someone, especially if he uses the formula independently on both teams, producing two win probabilities that sum to over 100%.

Just an FYI for the other lazy copy-pasters out there. (I can’t tell whether the numbers in this post were affected, thanks to rounding and perhaps to mistakes on my part in trying to reproduce them.)

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