Pete Sampras vs Sebastien Lareau — Wimbledon Round of 64
Wimbledon · ATP · Round of 64
3 — 0
Sets
Monday, 21 June 1999 ·
grass
Sampras P.6-4 6-2 6-3
Lareau S.
Sampras P.
6663
Lareau S.
4230
Pete Sampras defeated Sebastien Lareau 6-4 6-2 6-3 in the Wimbledon round of 64 on grass. The result followed the form book — Pete Sampras came in leading the head-to-head 2–0, defending the title. Pete Sampras extended the head-to-head to 3–0.
Key context
- Match story: Pete Sampras closed it out in three sets, 6–4 6–2 6–3.
Head to head
2–0| Date | Tournament | Round | Surface | Winner | Score |
|---|---|---|---|---|---|
| 1999-07-26 | Los Angeles | R16 | hard | Pete Sampras | 7-6 6-3 |
| 1996-04-08 | Hong Kong | R32 | hard | Pete Sampras | 6-1 3-6 6-1 |
Statistics
This season1st serve in·Even
60%
62%
1st serve points won·Slight edge
80%
69%
2nd serve points won·Even
53%
49%
Service points won·Slight edge
69%
62%
Return points won·Even
36%
36%
Break points saved·Slight edge
69%
62%
Aces / match·Dominant
12.3
5.9
Double faults / match·Dominant
5.3
3.4
Averages over each player's completed matches in the selected window (190 vs 55). ATP, WTA & Challenger.
Conditions
Court speed
Medium-fastrewards aggression
SlowFast