For the benefit of other readers, let’s look at an example from the last Super Bowl, to illustrate what you are asking. The Patriots were a 12-point favorite, but lost by 3. So the game finished 15 points away from the point spread. If I understand the question correctly, you are asking about the standard deviation of this difference. The mean difference was close to zero for each league studied. Here is what I get for the standard deviation:
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).