First day pairings have been announced for the 23rd annual Big 12 Men’s Golf Championship April 26-28 at The Greenbrier in White Sulphur Springs, W.Va.
Action kicks off on Friday, April 26 with No. 1 seed Oklahoma State and No. 2 Texas teeing off at 7 a.m. CT / 8 a.m. ET. At 7:45 a.m. CT / 8:45 a.m. ET, No. 3 Oklahoma and No. 4 Texas Tech will begin followed by No. 9 West Virginia and No. 10 Kansas State at 8:30 a.m. CT / 9:30 a.m. ET. Seeds one through four along with nine and 10 will tee off on the first hole.
On the 10th tee, No. 5 Baylor and No. 6 TCU will tee off at 7:22 a.m. CT / 8:22 a.m. ET while No. 7 Kansas will play alongside No. 8 Iowa State beginning at 8:07 a.m. CT / 9:07 a.m. ET.
Five Big 12 squads enter the Championship ranked in the top 22 of Golfweek/Sagarin rankings, led by No. 1-ranked Oklahoma State. No. 8 Texas is followed by No. 9 Oklahoma, No. 19 Texas Tech and No. 22 Baylor.
Teams will consist of five golfers and the lowest four scores per round from the designated five-player team will be used. Scoring shall be stroke play.
Pairings for Day 2 will be based on the 36-hole team totals from Day 1, beginning with the No. 1 and No. 2 ranked teams. Day 3 will be based on the 54-hole team totals with the leaders teeing off last.
The awards ceremony will follow Sunday’s final round. The event is free and open to the public. Live scoring will be available on Big12Sports.com.