Schaul v Jankovic
WTA Tour 2013
Lineball analyst6–0
Claudine Schaul takes it, 6–0 on categories
Schaul wins on sets lost. WTA Tour 2013: Schaul 2 matches, Jankovic 67 matches.
- Schaul averages a sets lost every 2 matches to Jankovic’s 0.9.
- Schaul averages 2 sets won a match to Jankovic’s 1.5 (32% more).
- Schaul has the better sets won %: 80 to Jankovic’s 63.4.
- Longevity: Jankovic has 67 matches, Schaul 2 — the averages matter more than the totals here.
- Small sample: Schaul (2 matches) — treat the averages with care.
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.
- –Titlesper game0.01 total · best 1#29 of 31
- –First serve in %62.1#41 of 87
- –First-serve points won %63.5#42 of 87
- –Break points saved %53.9#45 of 87
- –Return points won %47.1#14 of 87
- –Acesper game2.4160 total · best 9#31 of 87
- 1428 total · best 16Games wonper game12.4
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
Titles, oldest to newest in 2013.
Schaul––
Jankovic––––––––––