Match analytics

Philip Henning vs Kris van Wyk · Match odds & ELO prediction

M15 Stellenbosch • Hard • Dec 7, 2024 • 8:00 AM

Hard

Final score

2 - 0

Winner Philip Henning

Key insights

Player performance profile

Philip Henning

HARDSmall sample

100% · 10 on hard

Games won (last 10)

64%

10 matches tracked

Player Skillset

Based on ~0 points across 1 matches

Serve strengthServe strength (Player serve win % - tour average serve win %) scaled by sample size
Shaky
0.037% Pctl
Return strengthReturn strength (Player return win % - tour average return win %) scaled by sample size
Strong
0.081% Pctl
Pressure IndexPressure Index (Break point performance - baseline point performance) with a small adjustment for tiebreak results
Shaky
0.048% Pctl
Tiebreak win %
Insufficient data

Percentiles compare against tour-level players in TennisTrove.

Kris van Wyk

HARD

45% · 2227 on hard

Games won (last 10)

37%

10 matches tracked

Player Skillset

Based on ~0 points across 0 matches

Serve strengthServe strength (Player serve win % - tour average serve win %) scaled by sample size
Shaky
0.037% Pctl
Return strengthReturn strength (Player return win % - tour average return win %) scaled by sample size
Strong
0.081% Pctl
Pressure IndexPressure Index (Break point performance - baseline point performance) with a small adjustment for tiebreak results
Shaky
0.048% Pctl
Tiebreak win %
Insufficient data

Percentiles compare against tour-level players in TennisTrove.

Match Overview

Philip Henning and Kris van Wyk are set to meet at the M15 Stellenbosch on December 7, 2024 in a hard-court singles match. Henning enters with limited recorded results on hard courts in 2024, while Wyk has limited recorded results on hard courts this season. Elo ratings point to a clear statistical advantage for Henning entering this matchup. In their head-to-head history, Henning leads 2–0 over Wyk, including a win in their most recent meeting.

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