Mathematical analysis: Michigan has a 81.45 percent chance of winning the NCAA tournament

Michigan has an 81.45 percent chance of winning the 2013 National Championship.  Yes, you heard right - the maths prove it:

Michigan's record by City's first letter:

A (Ann Arbor, Auburn Hills):  19-1 (95 percent)

B (Brooklyn, Bloomington):  1-1 (50 percent)

C (Columbus, Champaign, Chicago):  2-2 (50 percent)

E (Evanston, East Lansing): 1-1 (50 percent)

M (Minneapolis, Madison):  1-1 (50 percent)

N (New York City): 2-0 (100 percent)

P (Peoria): 1-0 (100 percent)

S (State College):  0-1 (0 percent)

W (West Lafayette): 1-0 (100 Percent)

Now, the south regional will be hosted in Arlington, Texas - which, incidentally, starts with an A.  Therefore, Michigan has a 95 percent chance of winning each game, meaning they have a 90.25 (.95x.95) percent chance of winning the region and heading to the final four....in Atlanta, Georgia.  Atlanta, it also starts with an A. 

The Wolverines will then have a 85.74 (.9025 x.95) percent chance of advancing to the title game and a 81.45 (.8574 x .95) percent chance of winning the National Title! 

Hooray!  And even though it worked in '89, thank God the Final Four isn't in Seattle.  We'd have no shot based on these advanced statistics.   Take that, Mathlete!

Downvoters be damned, I'm not letting your collective lack of humor get me out of the happy zone. 

Edited to add West Lafayette, somehow I forgot to add that to the list the first go around.