Rangers and tiebreakers: Breaking down various baseball playoff scenarios

Posted Saturday, Sep. 21, 2013  comments  Print Reprints

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The final week of the baseball season is setting up to have some drama that should make TNT envious.

The National League playoff teams are essentially set, although the NL Central race could come down to the final days for St. Louis, Pittsburgh and Cincinnati.

The real drama lies in the American League wild-card race. Boston has clinched the AL East, and Detroit and Oakland are close to securing division titles, but the one-game wild card is anybody’s guess.

The Texas Rangers are in the thick of it, along with Tampa Bay and Cleveland. Kansas City, New York and Baltimore also are realistically still in it.

So what happens if there’s a three-way tie for the two wild-card spots? How about a four-way tie for the second wild card?

Following is a look at how various playoff tiebreakers would be settled by MLB. These aren’t all the possible scenarios, but the ones that seem likely to happen (and some that are unlikely).

• How is home-field advantage determined in a two-team tiebreaker?

1. Head-to-head winning percentage

2. Higher winning percentage in intradivision games

3. Higher winning percentage in intraleague games

4. Higher winning percentage in the last half of intraleague games

5. 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

• Two teams tie for second wild-card spot: One tiebreaker game would be played Monday, Sept. 30 (tentatively) to determine the second wild card. Home-field advantage is determined by the rules above (head-to-head winning percentage).

The Rangers would host tiebreaker games in a two-way tie scenario against Tampa Bay, New York and Kansas City. They’d be on the road against Cleveland or Baltimore.

• Three teams tie for two wild cards: The teams are seeded based on how they fared against each other. For instance, going into Saturday, let’s say the Rangers, Rays and Indians all finished with the same record.

They would be seeded as follows: 1. Cleveland (7-5) 2. Tampa Bay (7-6) 3. Texas (5-8).

The Indians would choose their path — A, B or C — followed by the Rays then the Rangers. Path A is the best route, followed by B and then C.

Club A would host Club B, and the winner would become the first wild card. Club C would host the loser of the first game, with the winner becoming the second wild card.

• Three teams tie for second wild card: This is determined similarly to how the three-way tie for both wild cards is. The three teams are seeded, and the top-seeded team chooses its path, A, B or C.

Club A hosts Club B, and the loser is eliminated. The winner then hosts Club C to determine the wild-card team. Club B is the least desired, although the A and C paths are intriguing. Is it easier to win two home games, or sit out the first game and have to win one on the road?

• Four teams tie for two wild-card spots: The teams would be seeded, once again, based on cumulative records among themselves. Club A hosts Club B, and Club C hosts Club D. The winners of each of those games would be declared the wild card clubs.

• Four teams tie for second wild card: Similar to the previous scenario, the teams would be seeded. Club A hosts Club B, and Club C hosts Club D.

The losers are out, and the winner of the A-B game hosts the winner of the C-D game to determine the second wild card.

Another interesting discussion among the teams to determine best route: Club A is the most desirable, but what would the second-seeded team choose? Is it better to be Club C and host the first game, and possibly be on the road for the second game? Or would you rather risk it by being Club B on the road first, and then hosting the second game?

Drew Davison, 817-390-7760 Twitter: @drewdavison

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