Want a laugh? Here's the Bucks generic Win Probability vs. the Lakers
I knew the Bucks were playing tonight but I forgot whom. I looked on NBA.com and saw they faced the Lakers in Los Angeles. I mentally computed a loss, but then I thought... what are the exact odds that a loss will occur?
Using the Bucks road efficiency numbers and the Lakers home efficiency numbers, and then running them through Professor Berri's winning percentage metric, I computed the Bucks road winning percentage being .370 and the Lakers home winning percentage being .920.
Thus, if you multiply the likelihood that the Bucks will win by the likelihood that the Lakers will lose divided by the likelihood that the Bucks will win (the so-called "multiplication rule"), you come up with a generic win probability for your Milwaukee Bucks. That would be 7.9% for those scoring at home.
I call it a "generic" Win Probability because I didn't adjust for strength of schedule (which I normally do using "Point Value over Average", but those who followed my abortive "NBA Advanced Power Ranking" blog will know I have fallen way behind on my numbers... its hard keeping up with all those calculations). BTW, basketball-reference.com sets the Bucks Win Probability at a somewhat brighter 11%.