The Kilkennyman falters today at the hands of the very strong Gawain Jones, who is now living in Ennis. I give the game below with some light notes, it was an exciting game to watch.
Full results
1 KARL Mc PHILLIPS [4] 0:1 GAWAIN JONES FM [3] 2 JOE RYAN FM [3] 1/2 MARK ORR IM [2.5] 3 JOSEF MUDRAK FM [2.5] 0:1 JOHN A. JOYCE [2.5] 4 STEPHEN JESSEL [2.5] 1:0 ANTHONY FOX [2] 5 KEVIN WHITE [2] 0:1 STEPHEN BRADY FM [2] 6 PAUL WALLACE [2] 0:1 PHILIP SHORT FM [2] 7 EAMON KEOGH [1.5] 1/2 EDMUND SOH [1.5] 8 PAUL WALSH [1.5] 0:1 GORDON FREEMAN [1.5] 9 RAY BYRNE [.5] 0:1 RORY QUINN [1.5] 10 TOM HEALY [1] 1/2 ALAN BEVERIDGE [1]
The eagled eyed among you will have noticed that this was not the draw I posted yesterday, I'm afraid I had a result posted incorrectly to me and that's what caused the confusion.
The standings below are correct.
Place Name Rtg Score M-Buch.
1-2 KARL Mc PHILLIPS 2172 4.0 10.0
GAWAIN JONES FM 2453 4.0 9.5
3-5 JOHN A. JOYCE 2271 3.5 9.0
JOE RYAN FM 2294 3.5 8.0
STEPHEN JESSEL 2238 3.5 7.5
6-8 MARK ORR IM 2338 3.0 9.5
STEPHEN BRADY FM 2365 3.0 7.5
PHILIP SHORT FM 2298 3.0 6.5
9-11 JOSEF MUDRAK FM 2391 2.5 8.5
GORDON FREEMAN 2066 2.5 7.0
RORY QUINN 2047 2.5 5.5
12-16 ANTHONY FOX 2123 2.0 7.5
KEVIN WHITE 2062 2.0 7.5
EDMUND SOH 2049 2.0 7.5
PAUL WALLACE 2171 2.0 7.0
EAMON KEOGH 2122 2.0 5.0
17-19 PAUL WALSH 2012 1.5 8.0
TOM HEALY 1902 1.5 6.0
ALAN BEVERIDGE 2138 1.5 5.5
20 RAY BYRNE 1941 0.5 6.5
The standings charts I give list players on the same score according to a M-Buch score. Maybe I should explain what that is. In a swiss event, many players will end up on the same score, that's inevitable. We do need some sort of tie-breaking system to rank these players. One of the most often used in days gone by was "The sum of opponents score". You simply add up the score achieved by all the opponents of each of the tied players and that's your Buchholz score. However this has some weaknesses, the main one being that if one of the tied players had met the run-away winner of an event while the other had not, it would be almost impossible for the player who did not meet the run-away winner to win the tie-break.
One way to counter-act this effect is to leave out the highest and the lowest scoring opponents when calculating a players Buchholz. The idea is to eliminate distortions in Buchholz values caused by taking into account games against run-away winners and bottom placed players. This is known as the Median Buchholz score and this is the value you see in the tie-breaking column on the standings. It is widely accepted as the fairest tie-breaking system for swiss events.
Games
1.d4 Nf6 2.c4 c5 3.d5 b5 4.cxb5 a6 5.bxa6 g6 6.Nc3 Bxa6 7.e4 Bxf1 8.Kxf1 d6 9.Nf3 Bg7 10.g3 0-0 11.Kg2 Nbd7 12.Re1 Ra6 13.Qc2 Qa8 14.a4 Rb8 Because White is a pawn up, Fritz 8 prefers his position but only by 0.28. As he normally values a pawn at 1.00, we can assume he can evaluate, at least to some extent, the value to Black of the open queenside files. I think most human players would prefer to play Black here, but maybe I'm wrong. 15.b3 e6 16.dxe6 fxe6 17.Rb1 d5 18.exd5 exd5 19.Bf4 Rb4 20.Nb5 Rb6 21.Nc7 Qb7 22.a5
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John Joyce showed his teeth to our Czech friend today with this attractive game, sorry no notes, I didn't have time;
1.d4 f5 2.h3 Nf6 3.g4 d5 4.Qd3 c5 5.dxc5 Nc6 6.gxf5 e5 7.fxe6 Bxc5 8.Nf3 0-0 9.Bg2 Bxe6 10.0-0 Qc8 11.Ng5 Bf5 12.e4 Bg6 13.Be3 Nb4 14.Qd2 dxe4 15.Bxc5 Qxc5 16.Ne6 Qb6 17.Nxf8 Rxf8 18.Na3 Nbd5 19.Nc4 Qc7 20.b3 b5 21.Ne3 Nf4 22.Ng4 Nxg4 23.hxg4 Qc8 24.f3 Qc5+ 25.Kh1 e3 26.Qe1 e2 27.Rg1 Be8 28.Qg3 Bc6 29.g5 Qe3 30.Kh2 Nxg2 31.Rxg2 Rxf3 32.Qg4 Qe5+ 33.Rg3 Qxa1 34.Rxf3 Qe5+ 0-1
Tomorrow's pairings -
1 GAWAIN JONES FM [4] : JOE RYAN FM [3.5] 2 STEPHEN JESSEL [3.5] : KARL Mc PHILLIPS [4] 3 JOHN A. JOYCE [3.5] : STEPHEN BRADY FM [3] 4 PHILIP SHORT FM [3] : MARK ORR IM [3] 5 RORY QUINN [2.5] : JOSEF MUDRAK FM [2.5] 6 GORDON FREEMAN [2.5] : EAMON KEOGH [2] 7 EDMUND SOH [2] : KEVIN WHITE [2] 8 ALAN BEVERIDGE [1.5] : PAUL WALLACE [2] 9 ANTHONY FOX [2] : PAUL WALSH [1.5] 10 TOM HEALY [1.5] : RAY BYRNE [.5]
Gerry Graham
