Three-way wild card tiebreaker?

Didn't this actually happen a couple of years back? Granted, I don't remember what they did...

 

They draw letters. Two teams play while the third one, I think the one that draws A, gets to sit. The winner of the first game plays the A team at non team A's ballpark.

 

It looks like in a 3 way tie it is up to your head-head records on who gets A, B, and C. I know the Astros lost the season series to both the Cubs and Giants so they would definitely have to win 2 in order to get the wild card in this situation.


..... From http://www.all-baseball.com/ref/tie.html

b. Tie among three teams, in any division. If the tied teams have identical records against each other, the League President shall supervise a draw which results in the tied teams being designated the A, B and C teams. A will play B at the home field of A, the day after the conclusion of the championship season. The following day, C will play the winner of A-B at the winner’s home field. The winner of the game shall be declared the Wild Card team. When, however, the tied teams do not have identical records against each other, their designation as the A, B or C teams shall be determined in accordance with the procedures set forth in subparagraphs 2(b) (1-4) above.

 

Here are the rules for determining who is A, B, and C ... rules 2b (1-4) as a follow on to my previous post

1) Team 1 has better record against both Team 2 and Team 3, and Team 2 has better record against Team 3: Team 1 chooses designation as A, B or C team, and Team 2 chooses from remaining choices;

2) Team 1 has better record against both Team 2 and Team 3, and Team 2 and Team 3 have same record against each other: Team 1 chooses designation as A, B or C team and League President supervises draw between Teams 2 and 3 for remaining choices;

3) Team 1 and Team 2 have same record against each other but better record against Team 3: League President supervises draw between Teams 1 and 2. Winner chooses designation as A, B or C team and loser chooses from remaining choices;

4) Team 1 has better record against Team 2, Team 2 has better record against Team 3, and Team 3 has better record against Team 1: Rank teams on the basis of overall winning percentage within the three-team group; team(s) with higher percentages select designation as A, B or C teams(s); when two or more teams tied in overall winning percentage, League President supervises draw between teams so tied.

 

I would love to be in a tiebreaker, that means we have a chance.

DD

 

Team 1 = SF (8-4)
Team 2 = Cubs (12-13)
Team 3 = Astros (11-14)

 

Does anyone find it odd that for a two team playoff, the team with the better head-head record does not get the homefield advantage?

Not that I am complaining since the Astros would get a home game against the Cubs, but it just seems a bit strange

 

But they only have to win one game to advance, they need to only burn one top of the line pitcher, and they get one day of rest while the other two teams are playing. Actually, if they wanted, they could pick to play two home games by picking team B and then beating team C in the first game and then they'd host team A in the last game. Either way the decision is easy: two home games (need to win both) or one road game (you need to win it obviously), you take the one game series for all the marbles.

 

No I agree with you .... I am talking about if 2 teams tie

i.e. if we tie the Cubs and Giants lose out, the Cubs would have to play a one game playoff at Houston even though they won the season series 9-8

I like it, but it just seems strange

 

I suspect that scheduling is the driving force behind the rule. For example, assume that the Cubs and Astros still had 3 games left with each other. It would be unknown as to who has the better head to head until after possibly the last game of the season. The coin flip gives teams, fans and networks ample opportunity to prepare for the 1 game playoff.

 

I think we had to play the rockies if i remember correctly

 

No I think we finished a game out once again ... That was the season the crazy Cubs fan ran out of the stands and attacked the Cubs pitcher that last series of the year

 

[MLB.com] - Wild Card Tie Break Information

Okay got this from Astros.com newsletter that came out about tiebreaker tickets etc. here is a link with the different scenarios drawn out from MLB.com

http://mlb.mlb.com/NASApp/mlb/mlb/n...915&content_id=859158&vkey=news_mlb&fext=.jsp

The Dodgers, Giants and Cubs finish with the same winning percentage, leaving the NL West and the NL Wild Card winners undecided. What happens?

The Dodgers and Giants would have a one-game playoff to determine the West champion on Monday. The loser would play the Cubs (or Astros or Padres) in a one-game playoff to determine the Wild Card winner on Tuesday. If that scenario plays out, then both National League Division Series would begin on Wednesday. Under normal circumstances (no Tuesday tiebreaker), three Division Series will begin on Tuesday and the only exception is the one not involving the NL Wild Card.

If three clubs are tied for first place in a division (or Wild Card) with an identical winning percentage at the conclusion of the regular season and the tied clubs do not have identical records against one another in the regular season, how is the tie for the division championship (or Wild Card) broken?

If the three tied clubs have identical records against one another in the regular season, the Office of the Commissioner shall supervise a draw that results in the clubs being designated Club A, Club B and Club C.

If the tied clubs do not have identical records against one another in the champoinship season, they will be designated Club A, Club B and Club C based on their records in head-to-head competition during the championship season as follows:

If Club 1 has a better record against each of Clubs 2 and 3, and Club 2 has a better record against Club 3, then Club 1 shall choose a designation as Club A, Club B or Club C. Club 2 shall choose a designation from the remaining two designations, and Club 3 shall be assigned the remaining designation.

If Club 1 has a better record against each of Clubs 2 and 3, and Club 2 and Club 3 have the same record against each other, then Club 1 shall choose a designation as Club A, Club B or Club C. The Office of the Commissioner shall supervise a draw between Clubs 2 and 3, the winner of which shall choose one of the remaining two designations. The remaining club shall be assigned the remaining designation.

If Club 1 and Club 2 have the same record against each other but each has a better record against Club 3, then the Office of the Commissioner shall supervise a draw between Clubs 1 and 2, the winner of which shall choose a designation as Club A, Club B or Club C. The club losing the draw shall choose a designation from the remaining two designations. Club 3 shall be assigned the remaining designation.

If Club 1 has a better record against Club 2, Club 2 has a better record against Club 3, and Club 3 has a better record against Club 1, then the three clubs shall be ranked on the basis of overall winning percentage within that three-club group, and the club with the highest winning percentage from among that three-club group shall have first choice among designations as Club A, Club B or Club C. The club with the next-highest winning percentage among the group shall have the next choice between the two remaining designations, and the club with the lowest winning percentage shall be assigned the remaining designation. If two or more of the clubs within this three-club group have the same winning percentage among the group, the Office of the Commissioner shall supervise a draw between the clubs so tied to determine priority of selection among the designations.

Club A shall play Club B at the ballpark of Club A on Monday. The following day, the winner of the first game shall be the home club in a second game against Club C. The winner of the game between Club C and the club that won the Club A-Club B game shall be declared the division champion.

What if four contenders in a league finish with the same winning percentage? Let's say the Dodgers, Giants, Cubs and Astros all finish with the same record.

In that scenario, Los Angeles and San Francisco would first play a one-game playoff to determine the NL West champion. Then the losing team would join Chicago and Houston in the three-team playoff tiebreaker. This would all take place between Monday and Wednesday, pushing back the start of the NLDS.

 

• TO WIN WILD CARD
If the Astros go:
3-1: Cubs must go 2-3. Giants must go 3-3
2-2: Cubs must go 1-4. Giants must go 2-4
1-3: Cubs must go 0-5. Giants must go 1-5

• TO TIE FOR WILD CARD
If the Astros go:
3-1: Cubs must go 3-2. Giants must go 4-2
2-2: Cubs must go 2-3. Giants must go 3-3
1-3: Cubs must go 1-4. Giants must go 2-4

http://www.chron.com/content/chronicle/sports/baseball/index.html#scenarios

 

We only have three games left.

 

That's from the Chronicles Web site should have put that on there before i get attacked hehe.

 

Tiebreakers

Following is a listing of tiebreaking scenarios and game locations and times related to potential ties for the National League West
Division Championship and/or National League Wild Card, according to Major League Rules 33 and 34.

Two-team tie for NL West (Los Angeles and San Francisco)
If there is a tie for a Division Championship and the winning percentage of the two tied Clubs is lower than the winning percentage of the second-place Club with the best record among all non-Division winners in the same League, the tie for the Division Championship shall be broken with a one-game playoff. The schedule for this tiebreaker as it relates to the National League West would be as follows:
Monday, October 4 - Los Angeles at San Francisco, 4:05 p.m. (EDT)

Two-team tie for NL Wild Card (Chicago, Houston or San Francisco)
A two-team tie for a Wild Card shall be broken with a one-game playoff. The schedules for the following potential tiebreakers would be as follows:
Monday, October 4 - Houston at San Francisco, 4:05 p.m. (EDT)
Monday, October 4 - Chicago at San Francisco, 4:05 p.m. (EDT)
Monday, October 4 - Chicago at Houston, 8:05 p.m. (EDT)

Three-team tie for NL Wild Card (San Francisco, Houston and Chicago)
If three Clubs are tied for first place in the Wild Card with an identical winning percentage at the conclusion of the championship season and the tied Clubs do not have identical records against one another in the championship season, they will choose a designation of Club "A," "B," or "C" based on their records in head-to-head competition during the championship season as follows:
If Club 1 has a better record against each of Clubs 2 and 3, and Club 2 has a better record against Club 3, then Club 1 shall choose a designation as Club "A," "B," or "C," and Club 2 shall choose a designation from the remaining two designations. Club 3 shall be assigned the remaining designation.

http://chicago.cubs.mlb.com/NASApp/...41001&content_id=877823&vkey=pr_mlb&fext=.jsp

http://chicago.cubs.mlb.com/NASApp/...915&content_id=859158&vkey=news_mlb&fext=.jsp
During a conference call held earlier today, San Francisco (4-2 vs. Chi; 4-2 vs. Hou) chose its designation a Club "C." Chicago (10-9 vs. Hou) chose Club "A" and Houston was assigned Club "B." Therefore, the schedule for this three-team tiebreaker scenario would be as follows:
Monday, Oct. 4 - Houston (B) at Chicago (A), 2:05 p.m. (EDT)
Tuesday, Oct. 5 - San Francisco (C) at Houston (B)/Chicago (A) winner

 

I hope the Stros just win it straight up because SF will be a tough place to win a 1 game playoff.