Ken Davidoff

Ken Davidoff

MLB

Wild playoff scenarios lurk in AL

Before we get into the Friday Five, we’ll start, fittingly, with an update on the playoff seeds:

AL: Boston (1) vs. winner of Texas (WC1) and Tampa Bay (WC2), Oakland (2) vs. Detroit (3)

NL: Atlanta (1) vs. winner of Pittsburgh (WC1a) or St. Louis (WC1b) and Cincinnati (WC2), Dodgers (2) vs. Pittsburgh (3a) or St. Louis (3b)

Note: If the season ended last night, the Pirates and Cardinals, both 85-61, would play a 163rd game to determine the winner of the National League Central. The game would be at PNC Park because the Pirates have a 10-9 edge over the Cardinals in their completed season series.

Crazy win for the Yankees last night. Crazy American League wild-card race. While Joel Sherman wrote a column off the game, I looked ahead to what the Yankees’ left side of the infield might look like next season. It’s of course a hazy outlook, given the uncertainties surrounding both Derek Jeter and Alex Rodriguez.

September is the best baseball months for many reasons, and here’s one we’ll discuss today: Playoff scenarios. “What happens if the Yankees tie the Rays?” “What happens if the Yankees, Red Sox and Rays all tie?” “What happens if the Yankees and Indians are tied through six innings of a playoff game and then the world’s ugliest animal shows up and scares everyone out of the ballpark?”

Most of the time, none of this stuff actually happens. There never has been a three-way tie after 162 games. But it’s fun to discuss, isn’t it?

So for the Friday Five, let’s go over five possible (if unlikely) scenarios. We’ll include how the Yankees would factor into them, since we are a New York media outlet after all.

1. Two teams tie for the second wild card.

The two clubs would play a 163rd game to determine who makes the playoffs. Home-field advantage would be determined by the head-to-head record. The Yankees own the edge against Baltimore (10-9), Cleveland (6-1) and Kansas City (5-2), while they currently trail Tampa Bay, 9-7, with three games left.

2. Three teams tie for the second wild card.

In this scenario, you essentially construct standings among the three teams by using the games played within the trio. Let’s say the Rays, Yankees and Indians tie for the second wild card.

Here’s how it would look right now: 1. Rays (13-9) 2. Yankees (13-10) 3. Indians (3-10).

So the Rays, as the first seed, would choose their path – A, B or C – and the Yankees would select second. What would that path be?

First, Club A hosts Club B. The loser is eliminated. The winner hosts Club C to determine the wild-card team.

So would you rather be Club A, having to win two home games to reach the playoffs. Or would you rather be Club C, sitting out the first game and having to win one on the road? Club B is the least desired position.

3. Three teams tie for the two wild-card spots. Let’s say the slumping Rangers fall back into the pack alongside the Rays and Yankees.

Here’s how those standings would look: 1. Rangers (6-4) 2. Rays (10-9) 3. Yankees (10-13). The Rangers and Rays have four head-to-head games remaining and, again, the Yankees and Rays have three more.

Here, 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. So would you rather be Club A and give yourself room for error? Or would you rather take an extra day as Club C and create a win-or-go-home scenario?

4. Four teams tie for the second wild-card spot.

I won’t do the “five teams” scenario, even though the Orioles and Royals both sit at 77-69. Why won’t I? Because it’s not included in the scenarios I received from MLB. If the Rays, Yankees, Indians, Orioles and Royals finish with the same record, I think Bud Selig would do his famous shrug and they’d just cancel the postseason. (They’d probably come up with something, actually.)

Anyway, let’s include the Rays, Yankees, Indians and Royals in this scenario — don’t blame me, blame CoolStandings.com — and you can plug in the Orioles on your own if you’d like.

Here are the standings in head-to-head play: 1. Yankees (18-12) 2. Royals (15-15) 3. Rays (14-15) 4. Indians (12-17). There are three more Yankees-Rays and three more Indians-Royals games left.

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

So the Yankees would choose to be Club A, obviously. What would the Royals choose? Would they rather chance a first game on the road and then host the second game, as Club B? Or is it better to host the first game and go on the road for the second game, as Club C?

If there’s a four-way tie for two wild-card spots – the most likely combo would be: 1. Rays (14-11) 2. Yankees (16-14) 3. Rangers (7-9) 4. Indians (8-11) – then Club A hosts Club B, and Club C hosts Club D. The two winners are the wild cards. No contemplating there, obviously you choose the home game.

5. Three teams tie for the division title and also post the same record as a team outside the division.

What, you don’t think the Red Sox, Rays, Yankees and Rangers all will finish with the same record? Well, just in case, the Rangers would automatically become club D. Then the three AL East teams would go by these head-to-head standings: 1. Red Sox (22-13) 2. Rays (16-19) 3. Yankees (13-19), pending the Yankees’ three games remaining with each club.

Club A hosts Club B, and Club C hosts Club D (the Rangers). If the Rangers win, they automatically win a playoff spot, and the A-B game would determine the AL East champion. Then the loser of the A-B game would host Club C to determine the other wild card.

If the Rangers lost, then the winner of the A-B game would host Club C. The winner of that game would be the AL East champion, and the loser, having already won one game in this play-in “tournament,” would get a wild-card spot. The loser of the A-B game would host the Rangers for the right to the second wild card.

Got all that? There’ll be a test later.

–Have a great day.