Early on, Oklahoma point guard Trae Young was supposed to be the front-runner. Lately, there has been plenty of chatter about big men Deandre Ayton and Marvin Bagley III from Arizona and Duke, respectively. Some will argue Kansas senior Devonte’ Graham. Young was sensational until teams adjusted to his style, and he got worn down by the bear that is the Big 12. Ayton and Bagley have been incredibly consistent. All three freshmen will likely be very good NBA players, if not great ones. Graham has had a very good year for Kansas, and has finished extremely well, leading the Jayhawks to an unprecedented 14th straight Big 12 crown. But the player of the year in college basketball should be a simple choice. That player has little flash in his game. He may not even be a first-round pick. But he is the best player on the best team, I believe. He is Jalen Brunson, who, barring a shocking vote by the Big East coaches, will be announced as the league’s player of the year on Wednesday night. The son of Rick Brunson, an eight-year NBA veteran, is averaging 19 points per game, 4.8 assists and 3.0 rebounds, all impressive numbers in the rugged Big East. But here’s what’s really impressive: The 6-foot-3 lead guard is shooting 52 percent from the field. Guards don’t shoot that kind of percentage. Graham, having a terrific year, is shooting 40 percent from the field. Brunson, a smart and savvy southpaw who has the ball in his hands almost all of the time, is committing just 1.7 turnovers per game. He has a 2.8 assist-to-turnover ratio, which is absurdly good. In the two Villanova wins over Xavier, Brunson had 28 points, 13 assists and one turnover. He dominated those games without scoring a ton. But the real reason why I would vote for Brunson can’t be quantified by any numbers. It’s his clutch gene, how he performs when the game is on the line. There isn’t another player in the country I would want with the ball in his hands in the final seconds other than him.