(6) Angelique Kerber vs (12) Caroline Wozniacki
Thursdays final match on Centre Court promises to be one of the matches of the day when a rejuvenated Caroline Wozniacki and Angelique Kerber do battle. Both ladies turned their seasons around in Eastbourne and haven’t looked back since. Both will be dark horses at the U.S Open in just over a week’s time.
After receiving a bye into the second round, Angelique Kerber stormed through her second round match in Cincinnati on Wednesday. The German comprehensively defeated the always dangerous Ekaterina Makarova 6-4 6-1. Kerber’s results during the U.S Open series have been impressive. The German reached the final in Stanford before losing to Venus Williams in an epic third round battle in Montreal last week.
Caroline Wozniacki is also in impressive form. The former world number 1 pushed Serena Williams all the way in their quarter-final match in Montreal last week. Wozniacki looked to be in control of the match during the second set but Serena was too good in the critical moments. Wozniacki was hardly tested in her opening two matches in Cincinnati this week. The Dane breezed past Magdalena Rybarikova and Shuai Zhang 6-2 6-3 on both occasions.
Kerber has won 5 of the last 6 meetings between the two and leads the overall head-to-head series 5-3. Their most recent encounter took place in Eastbourne; Kerber came from a set down to win 6-3 in the third. Eastbourne was an important tournament for both ladies, both came in struggling for form and both left with a new sense of confidence after solid runs. It’s always a close match when these two come up against each other; it’s one of the closest match-ups on tour. Their recent head-to-head record doesn't lie; I believe Kerber has a slight edge at the moment. Although Wozniacki has been in good form of late, she hasn't had many notable victories. The same can’t be said about Angie, who has had some impressive victories over tough opponents in the past few months.
Prediction: (6) Angelique Kerber to defeat (12) Caroline Wozniacki in three sets.