Serbian tennis player, Laslo Dere has made it into the third round of the 2019 ATP Indian Wells Open. At this stage of the tournament, Laslo Dere will play against Miomir Kecmanović who is also from Serbia.
Miomir Kecmanović won his previous match in straight sets as he beat Maximilian Marterer from Germany. It was a dominant display by Miomir Kecmanović as he impressed with his serving. During the win, Kecmanović won 80% of his service points which allowed him to win all of his service game. In addition to his serving, Miomir Kecmanović also returned serve well as he broke his opponent’s serve on 3 occasions. This will be a tougher match for Miomir Kecmanović and it may be difficult for the Serb to adjust to this situation as he is more commonly seen in Challenger tournaments rather than ATP events.
On the other side of the coin, Laslo Dere also won his previous match in straight sets. It was a simple victory for Dere as he won 80% of his service games while breaking his opponent’s serve on 4 occasions. Dere has been in very good form over the past few weeks as he made the semifinal of the ATP Brasil Open and won the ATP Rio Open. With such a strong run of form heading into this match, Laslo Dere will be very tough to beat.
This will be the first time that Laslo Dere and Miomir Kecmanović will be playing each other at an ATP event.
Taking everything into consideration, I am going to pick Laslo Dere to win. Laslo Dere has been in exceptional form heading into this match as he recently won an ATP title. Laslo Dere is a much stronger player than Miomir Kecmanović so I am picking him to get the victory.