Since I last checked in on the playoff picture, the probability of pandemonium in both the American and National Leagues has been reduced significantly. If you're dreaming of a five-, six- or seven-way tie in the AL wild card race, this isn't your year. But if you're a connoisseur of chaos—if you’re rooting for Team Entropy, as I call it—you don't have to throw in the towel quite yet. What follow here is rundown of the scenarios that will produce extra baseball in the AL (I’ll tackle the NL separately today).
In the Junior Circuit's only remaining division race, it's the Red Sox (88–64) with a three-game lead over the Yankees (85–67). Boston's magic number to clinch is eight, but they own neither the tiebreaker (the Yankees won the season series 11–8) nor the easier remaining schedule. The Yankees don't have any remaining games against teams .500 or better as they sandwich a pair of three-game series against the 71–82 Blue Jays (in Toronto this weekend, and at home for the final one) around a makeup game against the Royals (75–77) and then three against the Rays (74–79), all at home.
The Red Sox, on the other hand, close with four games at home against the Astros (93–59) after playing three against the Reds (66–87) in Cincinnati and then three at home against the Blue Jays. But if Houston can't recapture the top seed from the red-hot Indians (96–57, including 27 of their last 28), they're unlikely to tip their hand against a potential Division Series opponent by sending out the varsity squad, lowering the degree of difficulty for Boston.
If the Sox and Yanks wind up tied—and somewhere, you can bet Rob Manfred is praying for this—they would play a one-game tiebreaker in the Bronx to determine the division champion, with the loser most likely winding up as the host of the wild card game.
AL Wild Card
When I wrote the first installment of this year's Team Entropy outlook on September 11 (covering play through September 10), the top eight AL teams in the wild card race were separated by seven games from top to bottom, with seven of them (all but the Yankees) clustered within 3 1/2 games. Somehow, the Yankees are the only team to have played better than .500 ball since then, going 8–2 to open up a 6 1/2-game lead over the Twins (5–5 during that span, 79–74 overall) for the first wild card spot (WC1). Their magic number to clinch a wild card berth is two, and to clinch WC1 is seven, since they hold the tiebreaker over Minnesota via their 4–2 head-to-head record, including this week's three-game sweep in the Bronx.
The Orioles spit the bit, having gone 3–8 since we last checked in. At 74–80 overall, they’re now 5 1/2 games behind the Twins, with an elimination number of four, while the Mariners (3–7, 74–79) and Rays (3–6, 74–79) are both five out with elimination numbers of five; the three teams have Baseball Prospectus Playoff Odds of 0.0%, 0.3% and 0.4%, respectively, so it's safe to disregard them for our purposes. We fans of Team Entropy are as disappointed in those teams as they are in themselves.
That still leaves the Royals (75–77, 3 1/2 out, with an elimination number of seven), Angels and Rangers (both 76–76, 2 1/2 out, with elimination numbers of eight) in the mix for WC2 (you can do the math yourself to see why WC1 is virtually impossible). The trio's BP odds are just 3.2%, 9.8% and 10.4%, respectively, but together that's a 23.4% chance, nearly one in four, of somebody crashing the party. Let's go to the big board showing the teams' head-to-head records:
There are no remaining head-to-head games among any of the four potential WC2 suitors, so these records are final. If any of these teams wind up tied, they will then be seeded based upon the following:
1. Head-to-head winning percentage among the tied teams 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.
If “only” two teams tie, then home-field advantage for a Game 163 play-in is straightforward. No pair of these teams split a season series evenly, so if it's, say, the Twins and Royals, the game would be played in Minnesota thanks to their 11–8 edge, and in fact, the Twins would host any of the other teams. If it's the Rangers and Royals, to Arlington they go due to Texas' 6–1 edge. And so on.
For three tied teams, the head-to-head records among them are combined in order to determine the pecking order. If it's the Twins, Angels and Rangers, that would shake out as Twins .643 (9–5), Rangers .538 (14–12), Angels .385 (10–16). If it's the three teams besides the Twins (call it the Minnesota Meltdown), they would go Rangers .654 (17–9), Royals .500 (7–7), Angels .346 (9–17), and so on.
All four of the possible permutations would shake out without invoking the intradivisional records or anything more complex. Once the three-team hierarchy is determined, the teams would draft Club A, Club B or Club C designations in the following scenario: Club B @ Club A, and then Club C @ Club A/B winner, with that winner claiming WC2. Realistically, first pick is an A/C decision, in that the team in question could either choose to take their chances by playing the first tiebreaker at home, or let the other two duke it out, and then become the road team against the winner.
If all four teams are tied, the same principle applies, and here we have a second knot to untangle. Both the Twins and Rangers went 20–13 (.606) against the other three teams in this muddle, with the Royals 15–18 (.455) and Angels 11–22 (.333) behind them. So it's onto the intradivisonal records, where the Twins currently have a .522 winning percentage within the AL Central (35–32) with nine games to play (six against the Tigers, three of which are at home, and three against the Indians in Cleveland), while the Rangers are at .485 (32–34) with 10 to play (three in Oakland, then three hosting the Astros and four hosting the A's).
If the two teams somehow wind up with identical intradivisonal records, the Twins would have the better intraleague record, since they had the worse interleague record (13-7 vs. 14-6). After the order is determined, the four teams would draft Club A, Club B, Club C or Club D designations in the following scenario: 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.
Easy peasy lemon squeezy, right? We've still got the NL to consider, which I'll do so in a separate installment.