Wimbledon 2012: Novak Djokovic vs. Florian Mayer Preview

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Wimbledon 2012: Novak Djokovic vs. Florian Mayer Preview
Clive Brunskill/Getty Images
Novak Djokovic

Novak Djokovic is into the Wimbledon quarterfinals once again, and this year he will face Germany's Florian Mayer.

Mayer is the 31st seed at this year's All England Club, and he is a somewhat unlikely quarterfinalist. However, he is in a section of the draw that saw sixth-seeded Tomas Berdych fall in the first round to Ernests Gulbis. Mayer himself also pulled off an upset in his section when the German beat Richard Gasquet in the fourth round.

Mayer, despite his underdog status, has been this far at the All England Club before. Back in 2004 when Mayer was just 20 years old, he survived to the final eight at Wimbledon and actually looked like he was going to have a stronger career than the one he's enjoyed. Since that quarterfinal appearance, Mayer has been a tremendous underachiever in majors, having never made it past the third round in any Grand Slam except for the one currently running.

Djokovic's credentials require little review, as he is the defending champion at Wimbledon this season. He has only faced Mayer one time in what was a straight-sets victory at ATP Dubai in 2011. It would be shocking if Mayer were able to defeat Djokovic given that the Serb has not lost in a Grand Slam to anyone ranked outside of the top 20 since the 2012 French Open.

Additionally, Mayer, now playing in his 27th Grand Slam, has only once beaten a top-five player in a Grand Slam match; that being Guilermo Coria at Wimbledon in 2004.

The odds makers love Djokovic in this match. The longest odds on the Serb available appear to be with 888sport.com, who have him priced at -2500 (1/25) to win straight up.

Mayer's victory over Gasquet is a little bit head-turning, but the German is more of a flash than a consistent player. In my opinion, Djokovic stands a much better chance of winning than what the odds above imply. I think his true odds are probably closer to -5000 (1/50).

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