Bracketology - The myth of the five-twelve upset

Since 1985, the number of wins from the 5^th seed over the 12^th seed in the first round of the ncaa tournament are 65 wins to 96 total games played (that is a 67.77% win prediction).  Considering how “experts” seed teams, one could expect that that the number 5 seed would beat the number 12 seed… but not all of the time.  If we give the seeds weight, then the experts would predict that the the number five seed would lose 5/16 times (this is giving them the benefit of the doubt) and win 11/16 times in their matchups.  That is an expected win percentage of 68.75%.  Is this significant?  I could be.

<WARNING... The rest  of this argument makes use of Statistics>

Calculating a z score we find that z = (observed value- expected value)/(standard deviation of the expected values)

Observed value = .6777

Expected value = .6875

Standard deviation of the expected values can be calculated using the binomial formula (s = sqrt(np(1-p))) since our data only considers wins versus losses and 96 games provides a large enough sample to satisfy the conditions of the equation.

Thus, s  = sqrt(96*.6875*(1-.6875)) = 4.541475

And z = (.6777-.6875)/ 4.541475 = -0.0021578892

Now lets set up a hypothesis test.

The null hypothesis would reflect no difference between the observed and the expected, ie., = .  That is, = .6875.  whereas the alternate hypothesis would indicate that < , or    < .6875. 

To be sure lets set a level our of significance to rather conservatively, say 5%.

 To determine significance, we need only compare the observed z = -0.0021578892 with the 5% critical value z* = -1.645 from a standard z table.  Because -0.0021578892 is not farther than -1.645 from zero (not by a long shot!) we cannot conclude that the alternative hypothesis is correct to a 5% level (or even much less) and must accept the null hypothesis. 

Furthermore, to those who would make the argument, “there is always a 5/12 upset in every tournament”, I respond with, “is that really surprising?”  With an expected win percentage of approximately 69%, and assuming that one 5/12 game does not have an effect (that is… is independent of) any other 5/12 game in the tournament.  Then the probability that all 5/12 splits have 5 as the winner over 12 is (.69)^4 = .2234 or about 22% of the time… restated78% of the time at least one of the will win in the first round. 

Add on the normal fluctuations of a low number of samples (by law of large numbers standards) and a recent “streak” of #12 wins is easily explained.