Dayton will receive either the No. 2 or No. 3 seed in the A-10 tournament, which starts Wednesday with two games involving the bottom four seeds at Capital One Arena in Washington, D.C. Dayton will play in the quarterfinals at 6 p.m. March 11 if it receives the No. 2 seed or at 8:30 p.m. in the final game of the day if it receives the No. 3 seed.
Here are the scenarios involving the top three teams:
• If Dayton beats Davidson and second-place Virginia Commonwealth (21-7, 14-3) loses at Saint Louis on Saturday, Dayton will tie for second place and earn the No. 2 seed.
The tiebreaker will be Dayton and VCU’s results against Davidson. Dayton will have a 1-0 record against Davidson, and VCU is 1-1 against Davidson.
In every other scenario, Dayton would get the No. 3 seed.
• VCU will get the No. 1 seed if it beats Saint Louis — they play at 4 p.m. — and Davidson loses to Dayton. Davidson would get the No. 2 seed in that scenario.
• If Dayton and VCU both lose, Davidson will be the No. 1 seed, VCU will be No. 2 and Dayton will be No. 3.
The top four seeds get double byes and have to win three games in three days instead of four in four days or five in five days.
St. Bonaventure (19-8, 11-5) can clinch the No. 4 seed if it wins at home against Richmond (19-11, 10-7) on Friday. If the Bonnies lose and Saint Louis beats VCU on Saturday, Saint Louis will be the No. 4 seed.
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