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Botic van de Zandschulp will take on Tallon Griekspoor in the second round of the 2023 Australian Open on Wednesday (January 18).
Zandschulp is seeded 32nd and is the Dutch No.1. He has qualified for the second round after beating Ilya Ivahska 6-3, 3-6, 7-5, 6-3 in the first round. The Dutchman hit 12 aces but leaked six double faults.
His winning percentage on first serve was lower to Ivashka's but made it up for it with his second serve winning percentage (56-49). Zandschulp converted six of 14 break chances and won five more total points (149-144) than his opponent.
Meanwhile, Griekspoor is coming off a 6-3, 7-6, 6-3 win against Pavel Kolov. In a sensational and dominating performance, the Dutchman bossed in all departments. Griekspoor led his opponent 103-84 in points and now faces his country's top-ranked player.
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Botic van de Zandschulp vs Tallon Greikspoor: Betting Odds
Zandschulp reached a career-high ranking of 22 in 2022. This is his third appearance at the Australian Open, having reached the Round of 32 last year.
Meanwhile, Griekspoor is ranked 63rd in the ATP rankings but reached a career-high of 44 in 2022. This is also his third appearance at the Australian Open, where he reached the Round of 64 last year.
Botic van de Zandschulp vs Tallon Griekspoor: Match Details
Fixture: Botic van de Zandschulp @ Tallon Griekspoor
Date & Time: Wednesday, January 18; 12:30 am ET
Venue: Court 8, Melbourne Park
Botic van de Zandschulp vs Tallon Griekspoor: Betting Prediction
In three previous meetings, Zandschlup has won twice, while Griekspoor has won once, which was on clay. Expect the encounter between the two Dutchmen to be a competitive and fiery one.
Final Prediction: Total Over 38.5 (-115)
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