HAVANA, Cuba, Feb 2 (ACN) Cuba will seek today its third win in the Chess Olympiad for people with disabilities held in Belgrade, Serbia.
The Cuban players, currently ninth in the standings with four points, will be opposite Bangladesh in the fourth round after beating Romania 3-1 on Wednesday with individual victories by Pedro Morales and Carlos Larduet.
Bangladesh, also with four points, notched up a 3.5-0.5 win against the Czech Republic.
In other results, India, the Philippines and Poland won their third consecutive matches against Israel (3-1), Serbia 2 (3.5-0.5) and a team playing under the FIDE flag (3.5-0.5), respectively.
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