Kasidokostas or Caudron to semifinals

NICE - The main players of the Crystal Kelly tournament last year, will also play the semi-finals this year. Marco Zanetti, the winner in 2010, has seeded after a hard fought victory against Eddy Merckx for the final round as the winner of group A. In group B Torbjörn Blomdahl (last year losing finalist against Zanetti) has secured second place behind Dick Jaspers and therefore is qualified for the semi-finals.

The Swede started against world champion Daniel Sánchez with a run of fifteen (equaling the highest run of Dick Jaspers so far), but he wasted to many innings in the second part of the match. Blomdahl needed thirty attempts to beat Sánchez 50-36. The Swede showed himself happy, because he left Peter Ceulemans at two matchpoints, but has a mucht better overall average, so he can't be passed in the last session.

Peter Ceulemans had no chance against Semih Sayginer 50-26 (25). Torbjörn Blomdah plays in the final round against the unbeaten Dick Jaspers, who had a day off on the fourth day.

Marco Zanetti's match against Eddy Merckx was full of excitement. Zanetti was about ten points behind during a long part of the match, came along his opponent at 38-38. And when the Belgian champion scored only four points in the final innings, Zanetti ended with 4-2-6 at 50-42 in 24 innings.

The final group matches in this group are exciting, because Filipos Kasidokostas and Frédéric Caudron both has a chance to accompany Zanetti to the semifinals.

Frans van Kuijk has the same matchpoints as Kasidokostas now, but has finished his group matches, while the young Greek meets Eddy Merckx in his last match. Kasidokostas made a very good impression. He played a beautiful match and discouraged Frans van Kuijk, who was missing many balls at a millimeter.

Kasidokostas kept scoring constantly, he missed only in one inning and even had a chance to equalize Dick Jaspers' shortest match in 18. Kasidokostas made one little mistake, when he asked the referee to leave the touching balls in the same position. He missed a round the table shot at a millimeter. Therefore, the Greek now needed 19 innings and left Frans van Kuijk at 50-25. The overall average of Kasidokostas is also impressive: 2.365.