Janet Lee vs Samantha Reeves — Australian Open Round of 64
Australian Open · WTA · Round of 64
2 — 0
Sets
Monday, 18 January 1999 ·
hard
Lee J.7-5 6-1
Reeves S.
Lee J.
762
Reeves S.
510
Janet Lee defeated Samantha Reeves 7-5 6-1 in the Australian Open round of 64 on hard. The result followed the form book — Janet Lee came in leading the head-to-head 2–1, defending last year's round of 128. Janet Lee extended the head-to-head to 3–1.
Key context
- Match story: Janet Lee came through a tight passage before pulling clear — 7–5 6–1.
Head to head
2–1| Date | Tournament | Round | Surface | Winner | Score |
|---|---|---|---|---|---|
| 2001-07-13 | Mahwah NJ 50K | R32 | hard | Janet Lee | 3-6 6-4 6-4 |
| 2000-06-26 | Wimbledon | Q64 | grass | Samantha Reeves | 6-2 6-4 |
| 1999-04-12 | Tokyo | R32 | hard | Janet Lee | 6-2 6-2 |
Statistics
This season1st serve in·Even
62%
62%
1st serve points won·Even
59%
56%
2nd serve points won·Even
33%
37%
Service points won·Even
49%
48%
Return points won·Even
42%
42%
Break points saved·Slight edge
43%
49%
Aces / match·Dominant
5.0
1.3
Double faults / match·Dominant
1.0
7.0
Averages over each player's completed matches in the selected window (1 vs 8). ATP, WTA & Challenger.
Conditions
Court speed
Medium-fastrewards aggression
SlowFast