Over on Ken Pomeroy's website kenpom.com, Ken generates a set of pythagarean projected winning percentages for every Division I Men's Basketball team which then can be used to predict the outcome of games.
This week, he's solicited readers to submit calculations for the Conference Tournaments, and I thought it would be fun to do those calculations for the Pac-10 Tourney. Yes, that's very geeky, I know.
Having been to the Pac-10 Tourney at Staples Center before and observed and participated in the rooting, it seems fair to treat both UCLA and USC as home teams in the calculations. So I adjusted their pythagarean winning percentage and used that number in my calculations.
So here's what happens to UCLA in the Pac-10 Tournament....
In the quarterfinals, if UCLA is facing Cal, they have a 93.1% chance of winning, against Oregon State, the chance is 98.1%.
In the semifinals, UCLA has an 83.1% chance of beating Oregon, or a 76.4% chance of beating Arizona.
In the finals, UCLA could face any of these five teams -
Southern Cal 77.9%
Washington State 80.2%
Arizona State 96.2%
Overall, UCLA is a heavy favorite to win the whole thing. The odds of that happening are 60.4%
I really think we have a much better than 78% chance of beating justSC, but you know, they do have that Pac-10 third place banner to put up in the Galen Center.
Chances of being
Team Quarters Semis Finals Champion
1. UCLA 100% 94.1% 74.4% 60.4%
2. WSU 100% 71.2% 36.9% 10.1%
3. Southern Cal 100% 65.2% 38.3% 11.5%
4. Oregon 100% 39.7% 8.1% 3.8%
5. Arizona 100% 60.3% 16.3% 9.4%
6. Stanford 100% 34.8% 15.3% 3.0%
7. Washington 76.1% 25.5% 9.0% 1.5%
8. Cal 79.5% 5.5% 1.2% 0.3%
9. Oregon State 20.5% 0.4% 0.06% 0.005%
10. Arizona State 23.9% 3.3% 0.5% 0.03%
I'm looking forward to the games.