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Stacy Kaneshiro | ScoringLiveOctober 6, 2011, 8:21am
One weekend, so many scenarios.The Oahu Interscholastic Association Red West will figure out its final four seedings for the 12-team OIA tournament this weekend. The fun begins Friday when No. 3 Leilehua visits Mililani and when Aiea travels to No. 5 Campbell. It might end after Saturday's Waianae at Radford game.When the dust settles, coin tosses might follow.Only West top seed Leilehua and second-seeded Campbell have secured berths to the OIA tournament. The remaining five West teams are still in the chase for the four remaining tournament berths.In the East, in seeding order, Farrington (5-0), Kahuku (4-2), Kailua (4-2), Moanalua (3-2) and Castle (2-3), have secured berths. The winner of Friday's Roosevelt (0-5) at Kaiser (0-5) game will get the East's sixth seed.No. 3 Leilehua (5-0) at Mililani (2-3), 7 p.m. Friday (OC 16)The Mules have already clinched the division's top seed, but are on a roll and don't want to break momentum into the playoffs.Aiea (2-3) at No. 5 Campbell (4-1), 7 p.m. FridayThe Sabers clinched the second seed, but like Leilehua, would like to keep the momentum going. But Na Alii, winners of two in a row, are hot and their defense has played well all season.Waianae (2-3) at Radford (1-4), 6:30 p.m. SaturdayA must-win for the Rams or their season is over. Radford beat Mililani, so it's not unrealistic for an upset here.Possible scenarios:1. Waianae, Aiea and Mililani all lose. Those three, plus Kapolei and Radford, will be tied at 2-4, setting up an historic five-way coin toss would have to eliminate one of the tied teams. Head-to-head, each team is 2-2, beating and losing to the same teams with same records. That drops the quality of record tie-breaker since none had beaten Leilehua or Campbell.2. Waianae, Aiea and Mililani all win and end with 3-3 records. Coin toss eliminates one, making it fifth seed. Remaining two use head-to-head to determine third and fourth. Kapolei (2-4) becomes sixth seed. Radford eliminated.3. Waianae wins, but Aiea and Mililani lose. Waianae is third; Aiea, Mililani and Kapolei are 2-4. Each beat one, but lost to the other. Coin toss determines fourth through sixth. Radford eliminated4. Waianae and Aiea win, Mililani loses. Waianae (3-3) is third, Aiea (3-3) is fourth, Mililani (2-4) is fifth, Kapolei (2-4) is sixth. Radford eliminated.5. Waianae and Mililani win, Aiea loses. Mililani (3-3) is third, Waianae (3-3) is fourth, Kapolei (2-4) is fifth, Aiea (2-4) is sixth. Radford eliminated.6. Aiea and Mililani win, Waianae loses. Aiea (3-3) is third, Mililani (3-3) is fourth. Waianae (2-4), Kapolei (2-4) and Radford (2-4) each beat one, but not the other, so a coin toss eliminates one team. Head-to-head determines order after one team eliminated.7. Aiea wins, Mililani and Waianae lose. Aiea (3-3) is third. Mililani (2-4), Waianae (2-4), Kapolei (2-4) and Radford (2-4). Among the four tied teams, Radford and Mililani are 2-1, so they are fourth and fifth, respectively. Waianae and Kapolei are 1-2. Waianae is sixth. Kapolei is eliminated.8. Mililani wins, Aiea and Waianae lose. Mililani (3-3) is third. Aiea (2-4), Waianae (2-4), Kapolei (2-4) and Radford (2-4) are tied. Head-to-head among the four, Waianae and Kapolei are 2-1; Waianae is fourth and Kapolei is fifth. Aiea and Radford are 1-2; Na Alii are sixth. Radford is eliminated.Notes: Tie-breaking procedure order: Head-to-head; quality of record (wins against teams with better records; in above case, if someone had beaten Leilehua or Campbell); coin toss. After each tie-breaker differentiates teams, tie-breaking process repeats with remaining teams (i.e. In scenario No. 8, Waianae and Kapolei separate themselves from Aiea and Radford. Tie-breaker goes back to head-to-head.For simplification purposes, the coin-toss scenario involving three teams, the odd team is the lowest seed. (It could be the odd team is highest with remaining two going head-to-head for the last two seeds.)
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