Dejan Petrovic vs Jordan Kerr — New Zealand F3 Round of 32
New Zealand F3 · ITF · Round of 32
2 — 0
Sets
Wednesday, 21 April 1999 ·
hard
Petrovic D.7-6 7-6
Kerr J.
Petrovic D.
772
Kerr J.
660
Dejan Petrovic defeated Jordan Kerr 7-6 7-6 in the New Zealand F3 round of 32 on hard. The result followed the form book — Dejan Petrovic came in leading the head-to-head 1–0. Dejan Petrovic extended the head-to-head to 2–0.
Key context
- Match story: Dejan Petrovic won all two sets in tiebreaks against Jordan Kerr.
Head to head
1–0| Date | Tournament | Round | Surface | Winner | Score |
|---|---|---|---|---|---|
| 1999-02-28 | Australia 1 Masters 1 | R16 | clay | Dejan Petrovic | 4-6 6-3 6-2 |
Statistics
This season1st serve in·Clear edge
51%
65%
1st serve points won·Slight edge
68%
63%
2nd serve points won·Slight edge
43%
35%
Service points won·Even
56%
53%
Return points won·Slight edge
34%
44%
Break points saved·Dominant
43%
17%
Aces / match·Clear edge
5.6
3.0
Double faults / match·Slight edge
3.4
4.0
Averages over each player's completed matches in the selected window (8 vs 1). ATP, WTA & Challenger.