Jan Lennard Struff vs Radu Albot — Indian Wells Masters Qualifying Round 2
GER22
MDA11
Struff J.
Jan Lennard Struff defeated Radu Albot 7-6 6-3 in the Indian Wells Masters qualifying round 2 on hard. The result followed the form book — Jan Lennard Struff came in ranked #41, leading the head-to-head 3–0, defending last year's round of 128. Jan Lennard Struff extended the head-to-head to 4–0.
Key context
- Match story: Jan Lennard Struff closed it out in two sets, 7–6 6–3.
Head to head
3–0| Date | Tournament | Round | Surface | Winner | Score |
|---|---|---|---|---|---|
| 2019-07-01 | Wimbledon | R128 | grass | Jan Lennard Struff | 6-4 6-3 6-2 |
| 2019-05-27 | Roland Garros | R64 | clay | Jan Lennard Struff | 7-6(2) 7-6(3) 6-7(4) 6-2 |
| 2016-10-31 | Paris Masters | Q2 | hard | Jan Lennard Struff | 6-4 6-2 |
Serve vs return
This seasonEach player's serve crossed against the other's return, converted to expected hold of serve. Recent completed matches, ATP, WTA & Challenger. About
Serve & return on hard
Per-surface Serve / Return Rating (0–100, percentile within the hard field, opponent-adjusted over 52 weeks). About
Statistics
This seasonAverages over each player's completed matches in the selected window (191 vs 199). ATP, WTA & Challenger.
Umpiry Score
Pre-matchThe Umpiry Score combines match-context factors — matchup, form, fatigue, travel and motivation — into a single 0–10 score, updated daily. Read how it works.