Team Entropy Arrives: The American League Playoff Picture Is Wonderfully Cluttered
With three weeks left in the 2017 regular season, four of the six division races are essentially over. The Nationals have clinched the NL East, the Indians and Astros still lead their divisions by 13 games and the Dodgers by nine despite their 10-game losing streak. Don’t despair, because the possibility for late September drama and excitement still exists in the AL East, NL Central and both wild-card races. In fact, seven teams are within 3 1/2 games of the second AL wild card spot, producing possibilities for mayhem and confusion that MLB's schedulers have yet to officially untangle.
If you're a connoisseur of contemplating such scenarios, then Team Entropy is for you. Coined in late September 2011—a month that saw the Braves and Red Sox blow playoff spots at the expense of the Cardinals and Rays—with a nod towards the Second Law of Thermodynamics, which states that all systems tend toward disorder, Team Entropy describes the phenomenon of rooting for scenarios that produce end-of-season chaos. If you're a die-hard fan of a team trying to secure (or avoid blowing) a playoff spot, rooting for your squad of choice takes precedence. But if you've embraced the modern day's maximalist menu of options that allow one to watch scoreboards and view multiple games on multiple gadgets, you want MORE BASEBALL. Whether it's in the form of down-to-the-wire division and wild-card races, extra innings and tiebreaker scenarios, more baseball is always better. You want the MLB schedule-makers working around the clock in shifts trying to untangle once far-fetched scenarios. You want to go quad screen with MLB.tv and make sure your phone and tablet are charged as well.
The expanded Wild Card format introduced in 2012, with two teams meeting in a one-game playoff, was designed at least in part to capitalize on the frisson produced by the 2011 race. While it’s helped produce down-to-the-wire drama and confusion on the heels of that crazy year and a run of three straight Game 163 tiebreakers from 2007-09, there has been just one actual tiebreaker since the introduction of the new format (in 2013). Could this be Team Entropy’s year to throw a wrench in the works? Starting today and as necessary over the next few weeks, I'll do my best to untangle how such messes would be handled. For this, you'll want to keep in mind the Baseball Prospectus Playoff Odds and MLB's official tiebreaking scenarios (those are for 2016; we can expect an official update soon), as well as the general principle that MLB does its best to ensure that just about everything with respect to this stuff is decided on the field — nobody is going to be eliminated outright via a coin flip or a math problem; each tied team will get its turn at bat. As the current scenarios in both leagues are hairier than Chewbacca’s family, I’m focusing here on the AL and will follow with the NL in a separate column.
AL East
The Red Sox (81–62) lead the Yankees (77-65) by 3 1/2 games, and in turn the Yankees have a 3 1/2-game cushion for the top wild card seeding. The archrivals don't have any remaining head-to-head games scheduled, and the Yankees won the season series 11–8. Thus, if the two teams wind up tied, Yankee Stadium would play host to a tiebreaker game to determine which team wins the division and which has to go the wild card route. Assuming the two teams were to maintain their lead over the rest of the wild card scrum, the loser would host that game, to be played on October 3, but remember, kids—around here we never assume, because it makes an ass out of u and me.
It's possible—but hardly probable—that such a tiebreaker game would be a winner-take-all if at least two of the seven other wild card candidates leapfrog them. Suppose that the Red Sox channel their 2011 forebears by going 5–14 from here on out, finishing with 86 wins, and the Yankees putter home with an 9–11 record—not good, but good enough to tie. The AL East tiebreaker would become do-or-die if two of the following teams do at least this well: the Twins (74–69) go 13–6, the Angels (73–70) go 14–5, the Rangers or Royals (both 71–71) go 16-4, the Orioles or Mariners (both 71–72) going 16–3, or the Rays (71–73) go 16–2. I’ll leave you to do the mental mathematics if the target moves higher than 86 wins; obviously, that could eliminate the chances of certain teams making the jump.
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If the Red Sox, Yankees and one other team from that scrum wind up tied for a wild card spot, it's the pre-tiebreaker record of the loser of that game that will determine their place in the wild card pecking order. If the 86-win Yankees beat the 86-win Red Sox and just one of the aforementioned suitors finishes with 86 wins (subtract one win and add one loss to those hypothetical rest-of-season records), putting what we'll abbreviate as WC1 on the line, the host of the tiebreaker would be based upon the following:
1. Head-to-head winning percentage during the regular season.
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.
At the risk of spending 3,000 words with regards further scenarios that involve both the AL East and the wild card, we'll move on to the generalities of multi-team wild card ties.
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AL wild card
Whether there's a two-way tie for a wild card spot between the Sox/Yanks loser and one of those other seven teams, or between any two of them, the aforementioned pecking order applies for determining home-field advantage in a tiebreaker scenario. If the Yankees and Twins tie for WC1, the other teams are cooked, and the host for the wild card game would be based upon who has the upper hand in their season series; right now, the Twins hold a 2–1 advantage but the Yankees host them for a three-game series from September 18–20.
If the two teams wind up 3–3 against one another, then it's important that the Yankees are 34–26 against the rest of the AL East for a .567 winning percentage while the Twins are just 34–32 (.515). If somehow the pair wind up with the same overall record and a 3-3 split and the same intradivision records, then the Twins would have the upper hand when it comes to intraleague play. The Yankees have padded their overall record by going 15-5 in interleague play while the Twins are just 11–7 with two still to play.
If the Twins and Royals, say, wind up tied for the second wild card spot (WC2), the Twins' 11–8 season series advantage over the Royals would give them home field advantage for the tiebreaker game. If it's the Angels and Rangers tied, the Rangers, who lead the season series 10-6 with three to play, would be hosts. Here are the current head-to-head records of the nine teams still alive in this value pack of potential insanity:
Tm | BAL | BOS | NYY | TBR | KCR | MIN | LAA | SEA | TEX |
BAL | -- | 10-6 | 6-9 | 6-6 | 3-3 | 2-5 | 2-4 | 4-2 | 6-1 |
BOS | 6-10 | -- | 8-11 | 9-7 | 2-4 | 5-2 | 2-4 | 3-3 | 5-1 |
NYY | 9-6 | 11-8 | -- | 8-5 | 4-2 | 1-2 | 2-4 | 5-2 | 3-3 |
TBR | 6-6 | 7-9 | 5-8 | -- | 3-4 | 4-2 | 4-3 | 1-5 | 2-4 |
KCR | 3-3 | 4-2 | 2-4 | 4-3 | -- | 8-11 | 6-1 | 5-2 | 1-6 |
MIN | 5-2 | 2-5 | 2-1 | 2-4 | 11-8 | -- | 5-2 | 3-4 | 4-3 |
LAA | 4-2 | 2-4 | 4-2 | 3-4 | 1-6 | 2-5 | -- | 10-6 | 6-10 |
SEA | 2-4 | 3-3 | 5-2 | 5-1 | 2-5 | 4-3 | 6-10 | -- | 8-4 |
TEX | 1-6 | 1-5 | 3-3 | 4-2 | 6-1 | 3-4 | 10-6 | 4-8 | -- |
Note that all teams play 19 games within their divisions, so any pair that's short of that has games remaining (indicated in bold), and some of the pairings still have the winner at stake; for example, the Orioles have a 6–9 record against the Yankees with four games still to play in the Bronx. The other pairings outside the division are either six or seven games, and nearly all complete; the Twins and Yankees have a series in the Bronx as well, and the Royals have to come to New York for a make-up game.
Now, the fun really begins if three teams are tied for a spot—say, for WC2—which requires the determination of a pecking order. The first decision point is if they have identical head-to-head winning percentages within the trio, which probably can only happen if they're all either from separate divisions or the same one, since they will have had to play the same sum of games.
Suppose the Angels, Rangers and Mariners all finish with the same overall record AND with either 10–9 or 9–10 records against each other such that each one is 19–19 against the other two. Then, the pecking order would be determined by who has the highest intradivision winning percentage; right now, that would be the Angels (.516, 33–31), then Mariners (.483, 29–31), then Rangers .482 (27–29). In that order, the teams would choose whether they wanted to be Club A, Club B or Club C in the following scenario: Club B @ Club A on October 2, and then Club C @ Club A/B winner October 3, with that winner then WC2. Realistically, first pick is an A/C decision, in that the Angels could either choose to play the first tiebreaker at home, or let the Mariners and Rangers duke it out, and be the road team against the winner.
If the three teams tied for WC2 have unequal records against each other, then the pecking order is by winning percentage against the other two. For example, if it's the Orioles, Royals and Mariners who wind up tied, then the order for first pick in the A-B-C draft goes Royals .615 (8–5), Orioles .583 (7–5), Mariners .308 (4–9). If it's the Rays, Orioles and Angels, the order could hinge upon the outcome of the seven games remaining between the first two teams; right now it's Angels .538 (7–6), Rays .526 (10–9) and Orioles .444 (8–10). Obviously, there are a dizzying number of permutations among all these teams, so I can't lay them all out.
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But wait — there's more! If three teams are tied for WC1, it again comes down to head-to-head among those teams to determine who chooses where in the A-B-C draft. If it's the Yankees, Twins and Angels, the current pecking order is the Twins .667 (6–3), Angels .462 (6–7) and Yankees .333 (3–6), but of course there's that aforementioned three-game series in the Bronx; one win by the Twins guarantees them the first pick here.
What you have to know is that the format to untangle this is: Club B @ Club A (on October 2), with the winner claiming WC1, and the loser @ Club C on October 3 to decide WC2. The actual wild card game would presumably be pushed to October 4.
If four teams are tied … oh, baby. If it's four teams for WC1, then there's an A-B-C-D draft based upon the head-to-head records among the group. Throw the Yankees, Twins, Royals and Angels into the stew and the order (excluding the aforementioned Twins-Yankees series) is Twins .621 (18–11), Royals .500 (16–16), Yankees .467 (7–8) then Angels .350 (7–13). At best the Yankees can move up to second pick by winning either two or three of the games against the Twins; if it's two, then they hold the 4–2 season series advantage over the Royals. Once the four teams pick, then it's Club B @ Club A and Club D @ Club C, both on October 2. The two winners would be the wild card teams, with home field advantage then determined by head-to-head record between the two, then intradivision record, etc.
If the four teams are tied for WC2, the format is Club B @ Club A and Club D @ Club C (both on October 2), with the winners of the two games meeting at the site of the A/B winner on October 3 to determine the road team for the actual wild card game on October 4.
If there are five teams… officially, the league office hasn't figured that out, and it's a problem for another day. So too is the NL slate, which features a three-team NL race and the outside chance of one of those trailing teams winding up in the wild card mess. Stay tuned.
