2018 Queensland Championships: Pairing of round 9, at 16:00 of 01/10/2018
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1 = Maguire,Tom ( 4.5) – Liu,Yi ( 7.5) = 5 – 4 = …
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2 = Zhong,Tony (Junhao) ( 4.5) – Stojic,Dusan ( 7.0) = 11 – 2 = …
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3 = Parle,Hughston ( 5.5) – Harris,Timothy ( 4.5) = 6 – 14 = …
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4 = Esmaili,Ali ( 4.5) – Solomon,Stephen J ( 4.5) = 26 – 1 = …
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5 = Nakauchi,Gene ( 4.5) – Lam,Ross ( 4.5) = 3 – 10 = …
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6 = Slater-Jones,Henry ( 4.5) – Ostapenko,Michael ( 4.5) = 7 – 13 = …
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7 = Spuler,David ( 4.5) – Chan,Jacob Joseph ( 4.0) = 12 – 24 = …
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8 = Paevskiy,Igor ( 4.0) – Ghodosi,Hossein ( 3.5) = 9 – 21 = …
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9 = Morris,Byron ( 3.5) – Wang,Jason ( 3.5) = 25 – 22 = …
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10 = Au,Alexander ( 3.0) – Post,Martin ( 2.5) = 19 – 20 = …
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11 = Hamilton,Don ( 2.0) – Holliday,Bruce M ( 3.0) = 23 – 18 = …
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12 = Salahuddin,Mohammad ( 3.0) – ( bye ) ( 0.0) = 17 – 0 = 1
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13 = Cuellar,Gabriel ( 3.0) – ( not paired ) ( 0.0) = 8 – 0 = 0
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14 = Renjith,Sravan ( 1.5) – ( not paired ) ( 0.0) = 15 – 0 = 0
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15 = Holt,Stuart ( 1.0) – ( not paired ) ( 0.0) = 16 – 0 = 0
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2018 Queensland Championships: Pairing of round 9, at 16:00 of 01/10/2018
===================================================================================
1 = Maguire,Tom ( 4.5) – Liu,Yi ( 7.5) = 5 – 4 = 1 – 0
———————————————————————————–
2 = Zhong,Tony (Junhao) ( 4.5) – Stojic,Dusan ( 7.0) = 11 – 2 = 0 – 1
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3 = Parle,Hughston ( 5.5) – Harris,Timothy ( 4.5) = 6 – 14 = ½ – ½
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4 = Esmaili,Ali ( 4.5) – Solomon,Stephen J ( 4.5) = 26 – 1 = 0 – 1
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5 = Nakauchi,Gene ( 4.5) – Lam,Ross ( 4.5) = 3 – 10 = 0 – 1
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6 = Slater-Jones,Henry ( 4.5) – Ostapenko,Michael ( 4.5) = 7 – 13 = ½ – ½
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7 = Spuler,David ( 4.5) – Chan,Jacob Joseph ( 4.0) = 12 – 24 = 0 – 1
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8 = Paevskiy,Igor ( 4.0) – Ghodosi,Hossein ( 3.5) = 9 – 21 = 1 – 0
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9 = Morris,Byron ( 3.5) – Wang,Jason ( 3.5) = 25 – 22 = 1 – 0
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10 = Au,Alexander ( 3.0) – Post,Martin ( 2.5) = 19 – 20 = 0 – 1
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11 = Hamilton,Don ( 2.0) – Holliday,Bruce M ( 3.0) = 23 – 18 = 0 – 1
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12 = Salahuddin,Mohammad ( 3.0) – ( bye ) ( 0.0) = 17 – 0 = 1
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13 = Cuellar,Gabriel ( 3.0) – ( not paired ) ( 0.0) = 8 – 0 = 0
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14 = Renjith,Sravan ( 1.5) – ( not paired ) ( 0.0) = 15 – 0 = 0
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15 = Holt,Stuart ( 1.0) – ( not paired ) ( 0.0) = 16 – 0 = 0
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