Who has the best chance to be 7-0?

September, 23, 2013
Sep 23
3:15
PM ET
Four weeks into the season, only Baylor, Oklahoma, Oklahoma State and Texas Tech remain undefeated from the Big 12. Looking at their schedules, however, all four teams have a reasonable chance to remain unbeaten before they begin to play one another.

Who has the best chance to reach 7-0? I've tried to answer that question using conjecture and a little bit of math.

First, I assigned a chance of victory for the next four games for each of the remaining undefeated teams. This was pure conjecture. After that, I inputted the mathematical formula for multiplying percentages (convert to decimals, multiply and then multiply the total by 100) to arrive at the overall chance of each team getting to 7-0. Here are the results:

Baylor

Oct. 5, West Virginia: West Virginia won last year 70-63. Only Baylor looks capable of scoring that much again this year. Chance of victory: 95 percent

Oct. 12, at Kansas State: K-State looks to be limited offensively, and it’s looking like the only way to beat the Bears will be to outscore them. Chance of victory: 65 percent

Oct. 19, Iowa State: Iowa State can’t move the ball, either. At least not well enough to give Baylor much problem in Waco. Chance of victory: 97 percent

Oct. 26, at Kansas: The Jayhawks haven’t won a conference game in three years. Chance of victory: 92 percent

Overall chance of getting to 7-0: 55 percent. Of the four unbeatens, Baylor has the easiest path to 7-0, with its most challenging game being a road trip to Kansas State.

Oklahoma State

Sept. 28, at West Virginia: The Cowboys are a three-touchdown favorite in Morgantown for a reason. Chance of victory: 85 percent

Oct. 5, Kansas State: This year, OSU regains the advantage at QB with J.W. Walsh. Chance of victory: 80 percent

Oct. 19, TCU: The Cowboys get TCU in Stillwater for the second year in a row. OSU coasted past the Frogs last season. Chance of victory: 75 percent

Oct. 26, at Iowa State: Could the 2011 upset be in OSU’s head at all? Chance of victory: 80 percent

Overall chance of getting to 7-0: 41 percent. The Cowboys will be decent-to-heavy favorites in all four games, but they have to avoid the kind of slip-up they suffered in Ames two years ago.

Texas Tech

Oct. 5, at Kansas: The Red Raiders have been a little unsteady at quarterback the last two weeks. Could Michael Brewer return from his back injury in time for this game? Chance of victory: 75 percent

Oct. 12, Iowa State: The Cyclones could still be winless by the time they visit Lubbock. Chance of victory: 85 percent

Oct. 19, at West Virginia: Texas Tech crushed the Mountaineers in Lubbock last year. Chance of victory: 60 percent

Overall chance of getting to 7-0: 38 percent. The Red Raiders haven’t been quite as impressive as the top three teams, but they already own four wins and have three winnable games coming.

Oklahoma

Sept. 28, at Notre Dame: The Sooners are slight favorites according to Vegas, but this is basically a coin-flip matchup. Chance of victory: 50 percent

Oct. 5, TCU: The three-game gauntlet of Notre Dame-TCU-Texas doesn’t look nearly as daunting as it did a month ago. Chance of victory: 75 percent

Oct. 12, Texas: You can throw out the records in the Red River Rivalry. Chance of victory: 65 percent

Oct. 19, at Kansas: The Sooners have won eight games in a row by double digits over Kansas. Chance of victory: 92 percent

Overall chance of getting to 7-0: 22 percent. The Sooners have the toughest road to 7-0, due to the road tilt at Notre Dame and the neutral-site rivalry game with Texas.

Final thought
Baylor, OSU, Oklahoma and Tech all have better than a 20 percent chance of getting to 7-0. But the chances all four teams get there are not good. Using the percentages above, there is only a 1.9 percent chance that all four reach 7-0.