Ask The Wizard #209
NFL 13.31 (based on 2000 to 2007 seasons)
College Football 15.72 (based on 1993 to 2007 seasons)
NBA 11.39 (based on 1987 to 2003 seasons)
So, the 2008 Super Bowl finished 15/13.31 = 1.13 standard deviations away from expectations. I’m ignoring the adjustment factor for a discrete distribution, to keep things as simple as possible. The probability of being 1.13 standard deviations or more from expectations, in either direction is 25.85%. This can be found in Excel, using the formula 2 × normsdist(-1.13).
Banker: 45.86%
Player: 44.62%
Tie: 9.52%
The probability of a banker win, given that the bet is resolved is 45.86%/(45.86%+44.62%) = 50.68%. The probability of losing both steps of the progression is (1-0.5068)2 = 24.32%. The banker bet pays 19 to 20, so you will have a 75.68% chance of winning $95 or $90 (depending on whether you win on the first or second bet), and a 24.32% chance of losing $300.