Match analytics

Ivan Gakhov vs Hans Rehberg Max · Match odds & ELO prediction

Oeiras • Clay • Jan 8, 2025 • 1:15 PM

Clay

Final score

0 - 2

Winner Hans Rehberg Max

Key insights

Player performance profile

Ivan Gakhov

HARD

53% · 87 on hard

CLAY

61% · 4630 on clay

Games won (last 10)

49%

10 matches tracked

Player Skillset

Based on ~389 points across 2 matches

Serve strengthServe strength (Player serve win % - tour average serve win %) scaled by sample size
Shaky
-0.333% Pctl
Return strengthReturn strength (Player return win % - tour average return win %) scaled by sample size
Solid
-0.871% Pctl
Pressure IndexPressure Index (Break point performance - baseline point performance) with a small adjustment for tiebreak results
Strong
+4.283% Pctl
Tiebreak win %
Insufficient data

Percentiles compare against tour-level players in TennisTrove.

Hans Rehberg Max

HARD

100% · 50 on hard

Games won (last 10)

62%

10 matches tracked

Player Skillset

Based on ~91 points across 1 matches

Serve strengthServe strength (Player serve win % - tour average serve win %) scaled by sample size
Vulnerable
-1.715% Pctl
Return strengthReturn strength (Player return win % - tour average return win %) scaled by sample size
Shaky
-2.142% Pctl
Pressure IndexPressure Index (Break point performance - baseline point performance) with a small adjustment for tiebreak results
Shaky
-1.237% Pctl
Tiebreak win %
Insufficient data

Percentiles compare against tour-level players in TennisTrove.

Match Overview

Ivan Gakhov and Hans Rehberg Max are set to meet at the Oeiras on January 8, 2025 in a clay-court singles match. Gakhov enters with a 46–30 record on clay courts in 2025, while Max has posted a 1–3 mark on clay courts this season. Elo ratings point to a clear statistical advantage for Gakhov entering this matchup. In their head-to-head history, Max leads 2–0 over Gakhov, including a win in their most recent meeting.

Recent singles form slightly favors Max, who has won 5 of his last five matches, while Gakhov has gone 1–4 over the same span.