Schaul v Jankovic
WTA Tour 2003
Lineball analyst6–0
Claudine Schaul takes it, 6–0 on categories
Schaul wins on sets lost. WTA Tour 2003: Schaul 16 matches, Jankovic 16 matches.
- Schaul averages 1.2 sets lost a match to Jankovic’s 1.5.
- Schaul has the better sets won %: 47.2 to Jankovic’s 38.5.
- Schaul averages 1.1 sets won a match to Jankovic’s 0.9 (13% more).
- Winners’ edge: Schaul’s team won 50%, Jankovic’s team won 25% of the matches they played.
Built only from games in our stats database. A read of the numbers, not a tip.
Tale of the tape
Per game, with league rank among regular players.
- 61.3First serve in %–
- 61.9First-serve points won %–
- 61.3Break points saved %–
- 48.3Return points won %–
- 0.46 total · best 3Acesper game–
- 11.4183 total · best 18Games wonper game11.1178 total · best 17
- 44.1Break points converted %–
Head to head
13 matches
since 20042
since 20042
- Hard · 3 matchesSchaul 1Jankovic 2
Last meetings
- 3 Jan 2005 · Auckland · Round of 32 · HardJankovic won 7-5 6-2
- 24 Mar 2004 · Miami · Round of 128 · HardJankovic won 6-1 5-7 6-3
- 10 Mar 2004 · Indian Wells · Round of 128 · HardSchaul won 7-6(9) 6-4
Last 10 games
First serve in %, oldest to newest in 2003.
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