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Big Rapids Chess

Our goal is to promote chess in the Big Rapids area by creating a safe environment where members, regardless of experience or identity, can meet to learn about and play chess on a regular basis.

# Big Rapids Chess Tie Breaks

This is an ordered list of riules that we will use to break all ties in rated events.

### Background

Big Rapids Chess will use this ordered list of tiebreak systems in all rated events. If a player who scored in actual play ties with one whose entire point total is due to unplayed games, tiebreak systems will not be used; the player who scored in actual play wins automatically. In team events, the total game points scored by all team members will be the first tiebreak method used. The following list conforms to all USCF requirements - see the U.S. Chess Federation’s Official Rules of Chess, 7th Edition, pages 207-214. A comparable ordered list is also used by the Michigan Chess Association to break ties in their events.

### SWISS SYSTEM EVENTS

The following tiebreak rules are standard for Swiss events. All Swiss events run by Big Rapids Chess will use these tiebreak methods, in this order, unless otherwise posted in the event’s information pages. Tiebreak rules for Round Robin events, like quads, are at the bottom of this page.

#### SS Tiebreak 1: Modified Median Method

A player’s Modified Median score is found by adding the scores of the player’s opponents, but not including the least important of those opponents. For players tied with more than half the maximum score - 3 points or higher in a 5 round tournament - the lowest-ranked opponent is removed from the Modified Median score. For players tied with less than half the maximum score - 2 or less in a 5 round tournament - the highest-ranked opponent is removed from the Modified Median score. For players tied with exactly half the maximum score- like 2.5 in a 5 round tournament - the highest and lowest-ranked opponents are removed from the Modified Median score. For the purposes of determining the Modified Median, any opponent who had an unplayed game, as a bye, forfeit win, etc., shall have that round counted as 1/2 point towards the tied player’s Modified Median, even if it was a full point bye.

Example 1: Adam and Barbara are tied with 4 points in a 5 round tournament. Adam’s opponents’ scores are 1, 2, 3, 3.5, and 4. Barbara’s opponents’ scores are 2, 2.5, 3, 3, and 4. Adam’s Modified Median score is $2+3+3.5+4=12.5$ - the opponent with a score of 1 is dropped. Barbara’s Modified Median score is $2.5+3+3+4=12.5$ - the opponent with a score of 2 is dropped. In this example, the Modified Median does NOT break the tie, and therefore we must go to the second tiebreak method.

Example 2: Calvin and Debrah are tied with perfect scores: 7 points in a 7 round tournament. Calvin’s opponents have scores of 2, 2, 3.5, 4, 5, 5, 6, but the 6-point opponent had a full-point bye in the first round. Debrah’s opponents have scores of 1, 2.5, 4, 4, 5, 5.5, 6. Calvin’s Modified Median score is 2+3.5+4+5+5+5.5=25 - one of the 2-point opponents is dropped, and the 6-point opponent is counted as 5.5 because of the bye. Debrah’s Modified Median score is 2.5+4+4+5+5.5+6=27 - the 1-point opponent is dropped. Debrah wins on tiebreaks because her Modified Median is higher than Jane’s.

#### SS Tiebreak 2: Solkoff Method

A player’s Solkoff score is calculated the same way as the Modified Median, except no scores are dropped.

Example 1: Adam and Barbara - the same players from the Modified Median example - are tied with 4 points in a 5 round tournament. Adam’s opponents’ scores are 1, 2, 3, 3.5, 4. Barbara’s opponents’ scores are 2, 2.5, 3, 3, 4. Adam’s Solkoff score is $2+2.5+3+3+4=14.5$. Barbara’s Solkoff score is $1+2+3+3.5+4=13.5$. Adam wins the tiebreak based on the Solkoff method.

Example 2: Eric and Francis tie in a three-round tournament with 2.5 points each. They played each other in the final round and got a draw. Eric’s opponents have scores of 1, 2, and 2.5. Fancis’s opponents also have scores of 1, 2, and 2.5, but her first opponent had a full-point bye. Eric’s Solkoff score is $1+2+2.5=5.5$. Frances’s Solkoff score is $0.5+2+2.5=5$ - the opponent with the bye counts as 1/2 point instead of full point. Eric wins the Solkoff tiebreak.

#### SS Tiebreak 3: Cumulative Score Method

The easiest to calculate, a player’s cumulative score is found by adding his/her scores from each round, from the start to the finish of the tournament. After adding the round-scores of a player, one point is subtracted if that player had a full-point bye or unplayed win.

Example 1: In a 3 round tournament, George won his first game, drew his second, and won his third game. George’s score at round one was 1, round two was 1.5, and round three was 2.5. His Cumulative score is $1+1.5+2.5=5$. Helen drew her first game and won her second and third. Helen’s score at round one was 0.5, round two was 1.5, round three was 2.5. Her Cumuluative score is $0.5+1.5+2.5=4.5$. In this example, George wins on tiebreaks.

Example 2: In a 5 round tournament, Jennifer gets a bye, wins, loses, wins, wins. Her round scores are 1 as a bye, 2, 2, 3, amd 4. Kelly plays all five rounds: win, win, loss, win, win. Her round scores are 1, 2, 2, 3, and 4. Although Jennifer and Kelly have the same round scores, Jennifer had the bye in the first round, so their Cumulative scores will be different. Jennifer’s Cumulative score is $1+2+2+3+4-1=11$. Kelly’s Cumulative score is $1+2+2+3+4=12$. Kelly wins on tiebreaks.

#### SS Tiebreak 4: Result Between Tied Players Method

If the tied players played each other, the one who wins the game wins this tiebreak. If more than two people tie, all results among tied players should be considered, with rank according to plus or minus, not percentage; 3-1 is “plus two” and beats 1-0 which is “plus one”.

#### SS Tiebreak 5: Most Blacks Method

The player who plays as Black the most times wins this tiebreak.

#### SS Tiebreak 6: Coin Flip Method

The coin toss is used only for non-divisible prizes; it shall not be used to break a tie for any Big Rapids Chess title.

### ROUND ROBIN EVENTS

The following tiebreak rules are standard for Round Robin events. All Round Robin events run by Big Rapids Chess will use these tiebreak methods, in this order, unless otherwise posted in the event’s information pages.

#### RR TieBreak 1: Sonneborn-Berger Method

A player’s Sonneborn-Berger score is determined by adding the final scores of opponents that the player beat, adding half the final score of opponents that the player drew, and adding nothing for games lost or for games not played - byes, forfeit wins, etc.

Example 1: In a four-player Round Robin event, Alexa lost to Betty, beat Cathy, and beat Diane earning 2 points total. Betty beat Alexa, drew Cathy, and drew Diane for 2 points total. Cathy drew Betty, lost to Alexa, and beat Diane giving 1.5 points total. Diane drew Betty, lost to Alexa, and lost to Cathy with just 0.5 points total. Alexa and Betty are tied with 2 points. Alexa’s Sonneborn-Berger score is $0+1.5+0.5=2$. Betty’s Sonneborn-Berger score is $2+(1.5/2)+(0.5/2)=2+0.75+0.25=3$. Betty wins on tiebreaks.

Example 2: In a four-player Round Robin event, there is a clear winner but a tie for second place. Ian drew Jill and Kelly, and lost to Larry to earn 1 point total. Jill drew Ian, beat Larry, and lost to Kelly go get 1.5 points total. Kelly drew Ian, beat Jill, and lost to Larry also getting 1.5 points total. Larry lost to Jill, beat Kelly, and beat Ian for 2 points total and first place. For the second place tiebreak, Jill’s Sonneborn-Berger score is $(1/2)+2+0=2.5$. Kelly’s Sonneborn-Berger score is $(1/2)+1.5+0=2$. Jill is awarded second place on tiebreaks.

Example 1: In a four-player Round Robin event, Evan lost to Faith and Hillary, but beat Gus for 1 point total. Faith drew against Hillary and won against Gus and Evan earning 2.5 points total. Gus lost all three games and 0 points total. Hillary drew against Faith and won against Gus and Evan to also earn 2.5 points total. Hillary and Faith are tied with 2.5 points. Their Sonneborn-Berger scores are identical at $(2.5/2)+1+0=2.25$. The game that Hillary and Faith played against each other was a draw, therefore their tie remains unbroken.