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texasdave
09-21-2010, 10:30 AM
Okay, I understand this has nothing to do with the Reds but can someone tell me if this makes any sense? Here is a link to the article. http://mlb.mlb.com/news/article.jsp?ymd=20100921&content_id=14908952&vkey=news_mlb&c_id=mlb



If three clubs are tied for first place in the division with an identical winning percentage at the end of the regular season, then the tie is broken by designating the clubs as "A," "B," and "C," with selection of those designations based on the clubs' records against one another. Tiebreaking games would then be played as follows:

• Club A would host club B on Monday, Oct. 4.
• The winner of the A vs. B game would then host club C on Tuesday, Oct. 5.
• The winner of the A/B vs. C game would be declared division champion.

More info. Colorado has the best record among the three so they are Team A. San Diego and San Francisco still can have either the second or third best records so Team B and Team C are still up in the air.

The different scenarios for the teams to advance are as follows:
Team A has to win two home games.
Team B has to win one road and one home game.
Team C has to win one road game.

Does that seem right? The teams with the best and second best records head-to-head have to win two games to advance.
The team with the worst record head-to-head only has to win one.

Teams A and B are going to use their best available pitchers in the first game and then have to match up their second best available pitcher against Team C's best available.

Is this sort of screwy? It seems like the team with the worst head-to-head record gets the advantage.

RedLegsToday
09-21-2010, 10:55 AM
I am not sure, but, I bet that Team C is the team with the best record against the other two, not the worst.

(edit: of course, I just now read the article and it does designate the Rockies as Team A, but, that can't be right).