Who has the best chance to be 7-0?

September, 23, 2013
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:


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.


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.



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