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Stevenson Kelsey (1789) / Canada Anand Kunal (1612) / India
Kelsey Stevenson
wins: 1
losses: 0
percentage: 100 %
Kunal Anand
wins: 0
losses: 1
percentage: 0 %
Stevenson Kelsey (1789) / Canada vs. Anand Kunal (1612) / India
Nov 9, 2013
00:00		 1. Round Stevenson Kelsey (Kan.) 2 6 7 7 Delhi hard I
Anand Kunal (Ind.) 1 7 (5) 6 (5) 6 (5)
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