Pomeroy Big 12 Tournament Percentages
| 2nd Round | Semis | Finals | Champs | |
| Kansas | 100.0% | 86.1% | 77.8% | 50.5% |
| Texas A&M | 100.0% | 89.6% | 68.1% | 35.3% |
| Texas | 100.0% | 75.7% | 24.8% | 8.0% |
| Oklahoma | 87.4% | 13.5% | 8.6% | 2.4% |
| Kansas St. | 100.0% | 58.8% | 8.5% | 1.7% |
| Texas Tech | 89.7% | 40.3% | 5.0% | 0.9% |
| Missouri | 74.6% | 21.3% | 3.8% | 0.7% |
| Oklahoma St. | 69.5% | 8.6% | 2.8% | 0.5% |
| Iowa St | 12.6% | 0.3% | 0.1% | 0.0% |
| Colorado | 10.3% | 0.9% | 0.0% | 0.0% |
| Nebraska | 30.5% | 1.8% | 0.3% | 0.0% |
| Baylor | 25.4% | 3.0% | 0.2% | 0.0% |
These were done, as always, with data from www.kenpom.com. Each column is the team’s chance of advancing to that round. They take into account the fact that these games are being played at the Ford Center in Oklahoma City- all Ok. St.’s games are considered “semi-home”.
A couple specifics of interest- Kansas beats Texas A&M 57.5% of the time, and beats Texas 77.9% of the time. They beat Colorado 98.9% of the time by an average of 29.2 points (although that may or may not be of interest).
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5 Responses to “Pomeroy Big 12 Tournament Percentages”
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March 4th, 2007 at 9:31 pm
My totally-pulled-out-of-my-[CENSORED] numbers are:
Kansas 40%
Texas A&M 37%
Texas 16%
Texas Tech 5%
Kansas St 1%
all others combined 1%
March 4th, 2007 at 9:40 pm
I would agree with those (roughly). I think Pomeroy gives KU too much credit for some of their blowout wins, and is down on UT a little- back predicting the last 7 games, KP has UT winning by an average of 6.6 pts, when in reality they have won by an avg of 10.9.
March 5th, 2007 at 2:21 am
Also, there is just no way to statistically measure the insane impact that Acie Law IV can have in clutch situations, as he has proven time and again. Pomeroy’s “Luck Factor” may atone for this discrepancy somewhat, but I don’t really know how he calculates it. Either way, looks we’ll get a rematch against the Aggies either way, because I don’t see Texas beating them twice. There is a good chance we will play K-State for the third time this year, which is tough, it is hard to beat anyone three times in one season, but sure would be fun if we did. I’d prefer a rematch with Tech.
March 5th, 2007 at 8:14 am
KU will have a depth advantage over either A&M or UT should they advance to the finals against either team.
One of the less significant implications of Saturday’s game was that as the one seed, KU is playing the winner of the projected KSU / TT matchup instead of A&M or UT.
That game will most likely be a struggle for either team.
In my ideal scenario, we play ISU in our second round game, TT in the third round to exorcise the proverbial demons, and A&M in the championship game for a little revenge.
By doing that, KU would also have beaten every team in the B12.
March 5th, 2007 at 9:52 am
I think luck factor works like this, essentially.
1) Take all of a team’s individual game “points scored” values and put them in a hat.
2) Put the “points allowed” in another hat.
3) Pick one number from each hat.
4) If the “points scored” that you picked is greater than the “points allowed” then the team gets a win. Vice versa, they get a loss.
5) Set those two value aside and draw again.
6) Repeat 4 and 5 until you run out of numbers.
7) Fill the hat back up and do it all again, millions of times.
8) The winning percentage you end up with over time is the expected winning percentage.
9) Find the difference between the actual winning percentage and the expected winning percentage - that is the luck factor.
Of course, I think this is actually calculated using statistical functions, instead of doing the random sampling over and over again, and there may be a couple extra calculations thrown in to account for the different pace of games, but theoretically the result should be similar.
That said, I can’t figure out what a real ability to win close games would do to the luck factor.