Clockwise: Shane Matthews, Warren Elliott, Russel Porter, Duane Rowe

Going into the final weekend of the Jamaican Championship, there are a lot of scenarios that could play out.  The standings are as follows:

1. Elliott 6½/7; 2. Porter 6½/8; 3. Rowe 5/7; 4. Matthews 5/7; 5. Wheeler 3½/7; 6. Lorne 3½/8; 7. Holness 2½/7; 8. Scott 2/7; 9. Wisdom 1½/8 and 10. Richards 1/8.


Basically one point (in any combination) for Elliott will give him the title. Only Porter can equal the 7½ points that is possible for Elliott, but Porter would lose the tiebreak. Both Rowe and Matthews cannot afford to lose or draw. Porter only has one round left to play. A loss for Elliott (in either round) would make things interesting indeed.

Let's examine a couple of  interesting scenarios for the penultimate round:

• Scenario #1: Porter-bye; Matthews-Wheeler, 1-0; Elliott-Rowe, ½-½. This would mean that Elliott (7) would have to clinch against Wheeler with at least a draw. Porter only has one round left so, he'd have to beat Matthews and hope for a Wheeler upset over Elliott. Rowe and Matthews would both be eliminated.
  
• Scenario #2: Porter-bye; Matthews-Wheeler, 1-0; Elliott-Rowe, 0-1. This would mean that both Matthews and Rowe would be tied with 6-2 and would trail Elliott and Porter by only ½-point. This would make way for quite a tense finish. In this scenario, Porter-Matthews would be the marquee finalé. Matthews can defend his title if he beats Porter, Elliott loses to Wheeler and Holness holds Rowe. Rowe can win a three-way tie on tiebreaks if he beats Holness, Matthews beats Porter and Elliott is held by Wheeler. Porter would win the title  if  he beats Matthews and Elliott is held by Wheeler. If Porter only draws, then it leaves the door open for Rowe (with a win) or Elliott (with at least a draw).
  

Posted by The Chess Drum: 27 March 2004