Alexander Zverev vs Daniil Medvedev: 2023 Cincinnati Masters - Preview & Prediction

| by Evita Mueller

Alexander Zverev is unbeaten in his last seven matches at the Cincinnati Masters, but Daniil Medvedev will try to change that.

This is one fabulous matchup between two really good players who know each other well. We've seen this matchup plenty of times, as Medvedev generally held the upper hand. The Russian is looking very motivated in Cincinnati, which is expected.

He won this event in the past, and he'd like to do so again, especially after the somewhat lacklustre showing from Toronto. He's admitted to feeling more pressure lately, and we've also seen it play out on the courts. Zverev is no different, as he would also like to forget the lousy outing from Toronto and is also a former champion here.

The German won it back in 2021, as he missed last year's edition due to his ankle injury; that's how his winning streak is still intact.

Head-to-head: 

As we mentioned above, they played each other many times. Medvedev leads the matchup 9-6 so far, as he won the previous three matches. All three happened earlier this year, with the one in Monte-Carlos being particularly heated. It's rather known that they aren't the fondest of one another as they share some history.

Prediction:

Predicting what might happen here is hard. Mainly because the win condition for both players will be the serve. If Zverev serves well, he can keep the match as close as possible, and while Medvedev will get the ball back many times, it should give the German enough initiative to win the rallies easier.

It's a similar thing for Medvedev because he should really avoid playing too many long rallies against Zverev, as the German can certainly stall it out if he wants to. Generally, Medvedev has been pretty good and finding solutions against Zverev, but he's not playing that much better right now.

All three of his wins earlier this year came when he played far better tennis. I'll stick with him, though.

Prediction: Daniil Medvedev to win in three sets.

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