OU is not mathematically eliminated from the Big 12 championship. With so many of the Big 12 front-runners playing each other down the stretch, the losses for leaders still can mount.
But no one – not academic whiz Gabe Ikard, not program defender Bob Stoops – is talking about the possibility.
For good reasons. The odds are astronomical.
The Sooners still can finish in a three-way tie atop the Big 12, but I’ve found no way that OU can win a tiebreaker in that scenario.
So OU’s only hope is a two-way tie atop the Big 12 with OSU. That would require OU to win out (beating Iowa State, Kansas State and OSU), and OSU to otherwise win out (beating Texas and Baylor). And Baylor beating Texas.
And, oh yes, Texas Tech sweeping Baylor and Texas. And TCU beating Baylor.
Good luck with that.
But it’s fun to see just how high are the odds. So I gave each of those games a percentage chance of going OU’s way. You can argue any of these, but I’m in the ballpark:
OU-Iowa State: 95 percent (heck, it might be 99 percent; the Cyclones stink).
OU-Kansas State: 50 percent (actually, KSU might be 55 percent, but I’ll stick 50-50).
OU-OSU: 35 percent (the Cowboys are playing better, and they’re at home).
OSU-Texas: 50 percent (I like the Cowboys better, but the game’s in Austin).
OSU-Baylor: 35 percent (Baylor’s really good).
Baylor-Texas: 75 percent (I might be a little low; the ‘Horns are headed for a rattlesnake den).
Texas Tech-Texas: 20 percent (the Red Raiders are tanking).
Texas Tech-Baylor: 10 percent (should be over by halftime).
TCU-Baylor: 5 percent (it might be 1 percent, .35 but let’s stay on the fives).
OK, given those percentages, let’s do the math. You just multiply the percentages.
You better sit down.
OU has a 0.0000218 percent chance of reaching the Fiesta Bowl.
That’s two in 100,000. One in 50,000.
Stoops and Ikard are right. Better to not even bring it up.