Bucks road trip not THAT horrible
I heard Dennis Krause describe the Bucks recent 0 for 3 West Coast road trip as a "disaster". You know me, I don't respect statements, I respect proof... so I decided to put DK's assertion to the test.
I'm currently reading a fabulous book called Baseball Between the Numbers written by the authors of the website baseballprospectus.com. Great book. The chapter exposing how the promises made by those who push for publicly funded stadiums are empty shams had me screaming "Amen" at every other sentence (I have to do a post on that subject someday soon).
But for the purposes of the instant post, the book contains the famous "Log 5" formula somehow created by Bill James in his spare time back in the early 80s to determine the likelihood that Team A would defeat Team B given each team's winning percentages. I won't bore you with it here, its long... but its a little slice of genius. And I applied it to the Milwaukee Bucks and their opponents on their recent road trip to determine how likely it was that the Bucks would lose all 3 games on the trip in the manner that they did. It turns out the chances of that happening were better than a toss-up, which makes the trip a little less than a disaster.
To be specific, the odds of the Bucks losing all 3 road games in succession to the Lakers, the Suns, and the Warriors was 52%. Here's how I came up with that figure.
The Lakers "pure" home winning percentage (meaning their winning percentage as derived from running the team's home efficiency differential through this expected winning percentage formula) . Then I did the same for the Suns and Warriors. Then I used the same method to calculate the Bucks "pure" road winning percentage. Then I plugged all those numbers into Bill James formula. Here's what I came up with.
The probability that the Bucks would beat the Lakers in LA was 6.2%, the probability that the Bucks would beat the Suns in Pheonix was 27%, and the odds that the Bucks would beat the Warriors in Oakland was 23.5%. (Note: I thought the Warriors game was the "gettable" game of the bunch... turns out it was the Suns game. The Suns are a mediocre home team.).
So, in order to figure the likelihood that the Bucks would lose all three, I simply multiplied together the seperate likelihoods that the Bucks would lose each. Thus the formula was .935 x .765 x .730... which equals about 0.522. So, the Bucks had a slightly better than even shot at losing all three.
Thus, while the road trip was disappointing, I don't think I would refer to it as a "disaster".