Alycia Parks vs Veronika Erjavec Head to Head Tennis Stats and Match Analysis

Alycia Parks and Veronika Erjavec have not faced each other yet in our recorded match-ups. This head-to-head comparison analyzes their playing styles and recent performance statistics to provide insights into how a potential matchup might unfold.

Alycia Parks: Against top 100 players in the last 52 weeks, Alycia Parks holds a 11-21 record (34.4%). Throughout her career, she has competed against top 100 115 times, winning 43 (37.4%). Alycia Parks has lost her last 4 matches. Over the past 10 matches, she holds a 3-7 record. Over the past 52 weeks, Alycia Parks has posted a 29-36 record (44.6% win rate). Over her career at WTA level, Alycia Parks holds a record of 33-47 (41.2%) on hard, 13-20 (39.4%) on clay, and 2-7 (22.2%) on grass. In the last 52 weeks: 12-17 (41.4%) on hard, 5-6 (45.5%) on clay, and 0-1 on grass at WTA level. Alycia Parks has won one WTA tour-level title on hard. She has also won six titles at Challenger/ITF level (five on hard and one on grass). Alycia Parks serves an average of 6.5 aces per match in the last 52 weeks, consistent with her career average of 6.9.

Veronika Erjavec: Against top 100 players in the last 52 weeks, Veronika Erjavec holds a 5-6 record (45.5%). Throughout her career, she has competed against top 100 18 times, winning 9 (50.0%). Over the past 10 matches, she holds a 5-5 record. Over the past 52 weeks, Veronika Erjavec has posted a 44-28 record (61.1% win rate). Over her career at WTA level, Veronika Erjavec holds a record of 1-5 (16.7%) on hard, 1-1 (50.0%) on grass, and 0-2 on clay. At Challenger/ITF level: 220-117 (65.3%) on clay and 25-28 (47.2%) on hard. In the last 52 weeks: 1-4 (20.0%) on hard, 1-1 (50.0%) on grass, and 0-2 on clay at WTA level. 32-11 (74.4%) on clay and 0-1 on hard at Challenger/ITF level. Veronika Erjavec has won seven titles at Challenger/ITF level (all on clay). Veronika Erjavec serves an average of 1.0 aces per match in the last 52 weeks, consistent with her career average of 1.1.