A selection of AL wild card tiebreaker scenarios (September 13th)

USA TODAY Sports

There's a very real possibility that we could be seeing a tiebreaker game (or games) before we even get to a wild card game.

So what if the Indians don't earn a wild card spot outright? What if they tie one of their AL competitors for the the second spot? Or what if an even more crazy possibility happens?

First, here's the standings as of this morning:


American League Wild Card Standings

W L PCT GB STRK
Texas 81 64 .558 0 Lost 3
Tampa Bay 79 66 .544 2 Won 1
New York 79 68 .537 3 Won 3
Cleveland 78 68 .534 3.5 Won 1
Kansas City 77 69 .527 4.5 Won 2
Baltimore 77 69 .527 4.5 Lost 3

(updated 9.13.2013 at 6:34 AM EDT)



The three division races are opening up now, with the closest race being the AL West (Oakland 3.5 games ahead of Texas). So these scenarios will not assume the Indians would tie with a team who also was involved in a tie for a division title.

Scenario 1: Indians tie one other team for the second wild card

Wild Card tiebreaker to be played on Monday, September 30

So who would get home field advantage for this game? Here's the MLB tiebreaker rules:

Determining Home-Field Advantage in Two-Team Tiebreakers

1. Head-to-head winning percentage during the 2013 regular season.
2. Higher winning percentage in intradivision games.
3. Higher winning percentage in the last half of intraleague games.
4. Higher winning percentage in the last half plus one intraleague game, provided that such additional game was not between the two tied clubs. Continue to go back one intraleague game at a time until the tie has been broken.

HFA vs other AL teams:

  • Texas: To be played in Cleveland (5-1)
  • Tampa Bay: To be played in Tampa Bay (2-4)
  • New York: To be played in New York (1-6)
  • Baltimore: To be played in Cleveland (4-3)
  • Kansas City: To be played in Cleveland or Kansas City (9-7)

If the Indians get swept in Kansas next week and end up tying the Royals for the second wild, the game would be held in Kansas City. Otherwise, the game would be held in Cleveland.

So what would need to happen for each of those scenarios to take place?

Scenario 1a: Tie with Texas - Game at Cleveland, September 30

(keep in mind that this is just one possibility)

  • Cleveland (13-4) (90-72)*
  • Texas (9-8) (90-72)
  • Tampa Bay - 12-5 or better
  • New York - 10-5 or worse
  • Baltimore - 12-4 or worse
  • Kansas City - 12-4 or worse

This is probably the most unlikely scenario, as it would involve a team leapfrogging Texas (most likely Tampa Bay), and the Rangers tying the Indians for the second spot.

Scenario 1b: Tie with Tampa Bay - Game at Tampa Bay, September 30

  • Cleveland (12-4) (89-73)
  • Tampa Bay (10-7) (89-73)
  • New York: 9-6 or worse
  • Baltimore: 11-5 or worse
  • Kansas City 11-5 or worse

Scenario 1c: Tie with New York - Game at New York, September 30

  • Cleveland (12-4) (89-73)
  • New York (10-5) (89-73)
  • Tampa Bay 9-8 or worse
  • Baltimore: 11-5 or worse
  • Kansas City 11-5 or worse

Scenario 1d: Tie with Baltimore - Game at Cleveland, September 30

  • Cleveland (11-5) (88-74)
  • Baltimore (12-4) (88-74)
  • Tampa Bay 8-9 or worse
  • New York 8-7 or worse
  • Kansas City 11-5 or worse

Scenario 1e: Tie with Kansas City - Game at Cleveland or KC, September 30

  • Cleveland (11-5) (88-74)
  • Kansas City (12-4) (88-74)
  • Tampa Bay 8-9 or worse
  • New York 8-7 or worse
  • Baltimore 10-6 or worse

Scenario 2: Indians tie two other teams for second wild card

Game 1 (Team 2 at Team 3) to be played Monday, September 30

Game 2 (Team 2/Team 3 winner at Team 1) to be played Tuesday, October 1

The first order of business is to look at the head-to-head records. MLB has a scenario in which all each team has an identical record against the other two, but as you can see with the Indians' head-to-head record, that isn't possible.

Here's the HFA determination if the three teams do not have identical records against each other:

• If Club 1 has a better record against Clubs 2 and 3, and Club 2 has a better record against Club 3, then Club 1 chooses its designation, followed by Club 2.

• If Club 1 has a better record against Clubs 2 and 3, and Club 2 and 3 have identical records against one another, then Club 1 chooses its designation. Clubs 2 and 3 would follow the two-Club tiebreak rules to break their tie to pick the next designation.

• If Club 1 and 2 have identical records against one another, but each has a better record against Club 3, then Clubs 1 and 2 would follow the two-Club tiebreak rules to break their tie to pick the first 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; OR Club 1 has a better record against Club 2, Club 2 and 3 have identical records against one another and Club 3 has a better record against Club 1; OR Club 1 and 2 have identical records against one another, Club 1 has a better record against Club 3 and Club 2 and 3 have identical records against one another, then:

a. The Clubs will be ranked by their overall winning percentage amongst the other Clubs combined. The Club with the highest overall winning percentage in that group chooses its designation, followed by the team with the next highest overall winning percentage.

b. If two of the Clubs have identical winning percentages, then they would follow the two-Club tiebreak rules to break their tie to pick their designation.

c. If all three teams have identical winning percentages, then the tiebreak rules above (No. 1) for three clubs having identical records against one another should be followed.

Whew.

Let's for the moment assume that Texas will not be part of a three-team tie for the second wild card. It's possible that happens, but not likely.

All these scenarios assume that the team with best head-to-head record chooses to be Team C, as per this schedule:

Three-Club Tie for One Wild Card Spot:
After Clubs have been assigned their A, B and C designations, Club A would host Club B on Monday, Sept. 30 (tentatively). The winner of the game would then host Club C on Tuesday, Oct. 1 (tentatively) to determine the Wild Card Club.

Scenario 2a: Tie with Tampa Bay and New York

Team 1: Tampa Bay (9-7 vs NY*, 4-2 vs CLE)

Team 2: New York (6-1 vs CLE, 7-9 vs TB*)

Team 3: Cleveland (1-6 vs NY, 2-4 vs CLE)

*Tampa Bay and New York could swap places if the Yankees sweep Tampa Bay Sep 24-26

Game 1: Cleveland at New York, September 30

Game 2: Cleveland/New York winner at Tampa Bay, October 1

  • Cleveland (11-5) (89-73)
  • Tampa Bay (10-7) (89-73)
  • New York (10-5) (89-73)
  • Baltimore 11-5 or worse
  • Kansas City 11-5 or worse

Scenario 2b: Tie with Tampa Bay and Baltimore

Team 1: Tampa Bay (9-6 vs BAL*, 4-2 vs CLE)

Team 2: Cleveland (2-4 vs TB, 4-3 vs BAL)

Team 3: Baltimore (6-9 vs TB*, 3-4 vs CLE)

*If Baltimore sweeps Tampa Bay Sep 20-23,  then the order would go: Tampa Bay, Baltimore, Cleveland

Game 1: Baltimore at Cleveland, September 30

Game 2: Baltimore/Cleveland winner at Tampa Bay , October 1

  • Cleveland (11-5) (89-73)
  • Tampa Bay (10-7) (89-73)
  • Baltimore (12-4) (89-73)
  • New York 9-6 or worse
  • Kansas City 11-5 or worse

Scenario 2c: Tied with Tampa Bay and Kansas City

Team 1: Kansas City (13-10 vs CLE*/TB)

Team 2: Cleveland (11-11 vs KC*/TB)

Team 3:Tampa Bay (5-8 vs CLE/KC)

*If Indians sweep Kansas City next week, Cleveland is Team A, Kansas City is Team B. If the Royals sweep the Indians next week, Kansas City is Team A, Tampa Bay is Team B, Cleveland is Team C.

Game 1: Tampa Bay at Cleveland, September 30

Game 2: Tampa Bay/Cleveland winner at Kansas City, October 1

  • Cleveland (10-6) (88-74)
  • Tampa Bay (9-8) (88-74)
  • Kansas City (11-5) (88-74)
  • New York 8-7 or worse
  • Baltimore 10-6 or worse

Scenario 2d: Tied with New York and Baltimore

Team 1: New York (10-9 vs BAL, 1-6 vs CLE)

Team 2: Cleveland (1-6 vs NY, 4-3 vs BAL)

Team 3: Baltimore (9-10 vs NY, 3-4 vs CLE)

Game 1: Baltimore at Cleveland, September 30

Game 2: Baltimore/Cleveland winner at New York

  • Cleveland (11-5) (89-73)
  • New York (10-5) (89-73)
  • Baltimore (12-4) (89-73)
  • Tampa Bay 9-8 or worse
  • Kansas City 11-5 or worse

Scenario 2e: Tied with New York and Kansas City

Team 1: New York (6-1 vs CLE, 5-2 vs KC)

Team 2: Cleveland (1-6 vs NY, 9-7 vs KC*)

Team 3: Kansas City (2-5 vs NY, 7-9 vs CLE*)

*If Kansas City sweeps Cleveland next week, they would be Team B and Cleveland would be Team C

Game 1: Kansas City at Cleveland, September 30

Game 2: Kansas City/Cleveland winner at New York, October 1

  • Cleveland (10-6) (88-74)
  • New York (9-8) (88-74)
  • Kansas City (11-4) (88-74)
  • Tampa Bay 8-9 or worse
  • Baltimore 10-6 or worse

Scenario 2f: Tied with Kansas City and Baltimore

Team A: Cleveland (9-7 vs KC*, 4-3 vs BAL)

Team B: Kansas City (7-9 vs CLE*, 4-3 vs BAL)

Team C: Baltimore (3-4 vs CLE, 3-4 vs KC)

*If Kansas City sweeps Cleveland next week, then Kansas City would be Team A, Cleveland Team B

Game 1: Baltimore at Kansas City, September 30

Game 2: Baltimore/Kansas City at Cleveland, October 1

  • Cleveland (10-6) (88-74)
  • Kansas City (11-5) (88-74)
  • Tampa Bay 8-9 or worse
  • New York 8-7 or worse
  • Baltimore 10-6 or worse

Scenario 3: Indians tie three other teams for second wild card

Game 1 (Team B at Team A) to be played Monday, September 30

Game 2 (Team D at Team C) to be played Monday, September 30

Game 3 (Team D/Team C winner at Team A/Team B winner) to be played Tuesday, September 31

HFA determination in this crazy scenario:

Determining A, B, C, D Designations in Four-Team Tiebreakers

1. The Club with the highest winning percentage in games among the tied Clubs chooses its designation first, followed by the Club with the second-highest winning percentage and the Club with the third-highest winning percentage. If two Clubs have identical winning percentages, then the two-Club tiebreak rules shall apply to determine which team selects its designation first. If three Clubs have identical winning percentages, then the three-Club tiebreak rules shall apply to determine which teams select their designation first. If all four Clubs have identical winning percentages, then;

2. The Club with the highest winning percentage in intradivision games chooses its designation first, followed by the Club with the second-highest winning percentage and the Club with the third-highest winning percentage. If two Clubs have identical winning percentages, then the two-Club tiebreak rules shall apply to determine which team selects its designation first. If three Clubs have identical winning percentages, then the three-Club tiebreak rules shall apply to determine which teams select their designation first. If all four Clubs have identical winning percentages, then;

3. The Club with the highest winning percentage in the last half of intraleague games chooses its designation first, followed by the Club with the second-highest winning percentage and the Club with the third-highest winning percentage. If two Clubs have identical winning percentages, then the two-Club tiebreak rules shall apply to determine which team selects its designation first. If three Clubs have identical winning percentages, then the three-Club tiebreak rules shall apply to determine which teams select their designation first. If all four Clubs have identical winning percentages, then;

4. The Club with the highest winning percentage in the last half plus one intraleague game, provided that such additional game was not between any of the tied Clubs, chooses its designation followed by the Club with the second-highest winning percentage and the Club with the third-highest winning percentage. Continue to go back one intraleague game at a time until any ties have been broken.

If I am reading this correct, then it appears that the third team can choose its opponent, which seems strange to me. Assuming of course that the second team doesn't choose to play the first team on the road. For this exercise I'll assume Team 3 choose to play Team 2.

Here's a quick chart for the HFA for the Games 3s:

Tampa Bay: New York*, Baltimore*, Cleveland

New York: Cleveland, Baltimore, Kansas City

Cleveland: Baltimore, Kansas City*

Baltimore: None

Kansas City: Baltimore

*Could change because there still games between the two teams

Assuming that Texas would clinch the first wild card, there would be four different possible combinations:

Scenario 3a: Cleveland ties with Tampa Bay, New York, and Baltimore

Team 1: Tampa Bay (22-15)*

Team 2: New York (23-19)*

Team 3: Baltimore (18-23)*

Team 4: Cleveland (6-14)

*Could change based on remaining head-to-head games

Game 1: Cleveland at Tampa Bay, September 30

Game 2: Baltimore at New York, September 30

Game 3: Game 1 winner vs Game 2 winner, October 1

  • Cleveland (10-6) (88-74)
  • Tampa Bay (9-8) (88-74)
  • New York (9-6) (88-74)
  • Baltimore (11-5) (88-74)
  • Kansas City 10-6 or worse

Scenario 3b: Cleveland ties with Tampa Bay, New York, and Kansas City

Team 1: New York (18-12)*

Team 2: Kansas City (15-15)*

Team 3: Tampa Bay (14-15)*

Team 4: Cleveland (12-17)*

*Could change based on remaining head-to-head games

Game 1: Cleveland at New York, September 30

Game 2: Tampa Bay at Kansas City, September 30

Game 3: Game 1 winner vs Game 2 winner, October 1

  • Cleveland (10-6) (88-74)
  • Tampa Bay (9-8) (88-74)
  • New York (9-6) (88-74)
  • Kansas City (11-5) (88-74)
  • Baltimore 10-6 or worse

Scenario 3c: Cleveland ties with Tampa Bay, Baltimore, and Kansas City

Team 1: Kansas City (17-13)*

Team 2: Cleveland (15-14)*

Team 3: Tampa Bay (14-14)*

Team 4: Baltimore (12-17)*

*Could change based on remaining head-to-head games

Game 1: Baltimore at Kansas City, September 30

Game 2: Tampa Bay at Cleveland, September 30

Game 3: Game 1 winner vs Game 2 winner, October 1

  • Cleveland (10-6) (88-74)
  • Tampa Bay (9-8) (88-74)
  • Baltimore (11-5) (88-74)
  • Kansas City (11-5) (88-74)
  • New York 8-7 or worse

Scenario 3d: Cleveland ties New York, Baltimore, and Kansas City

Team 1: New York (21-12)*

Team 2: Cleveland (14-16)*

Team 3: Baltimore (15-18)*

Team 4: Kansas City (13-17)*

*Could change based on remaining head-to-head games

Game 1: Kansas City at New York, September 30

Game 2: Baltimore at Cleveland, September 30

Game 3: Game 1 winner vs Game 2 winner , October 1

  • Cleveland (10-6) (88-74)
  • New York (9-6) (88-74)
  • Baltimore (11-5) (88-74)
  • Kansas City (11-5) (88-74)
  • Tampa Bay 8-9 or worse

Now this is only part of the possibilities, and as you can see, they aren't crazy possibilities. You could have three teams tied for both the wild card spots, which could create a scenario in which two teams play each other in two different elimination games in a three-day period. You could have four teams tied for the both the wild card spots, which would actually be simpler than Scenario 3 (the two winners get the wild card spots, and play each other in the wild card game). And if things break right, you could have this scenario:

  • Tampa Bay (9-8) (88-74)
  • New York (9-6) (88-74)
  • Cleveland (10-6) (88-74)
  • Baltimore (11-5) (88-74)
  • Kansas City (11-5) (88-74)

That would be a 5-way tie for the second wild card spot. MLB hasn't released a tiebreaker for this.

Or, God forbid, this scenario:

  • Texas (7-10) (88-74)
  • Tampa Bay (9-8) (88-74)
  • New York (9-6) (88-74)
  • Cleveland (10-6) (88-74)
  • Baltimore (11-5) (88-74)
  • Kansas City (11-5) (88-74)

Yes, that's a 6-way tie for both wild card spots. I don't want to think about that happening, but it's possible.

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