The return of Alex Rodriguez
has made the AL wild-card chase even wilder. (Kathy Willens/AP)
Two years ago, when the Red Sox and Braves were in the midst of coughing up playoff spots to the Rays and Cardinals, respectively, I coined the phrase "Team Entropy" to describe the phenomenon of rooting for scenarios that produced end-of-season mayhem.
If you're a die-hard fan of a team trying to secure (or avoid blowing) a playoff spot, rooting for that team generally takes precedence, but if you've embraced the modern day's maximalist menu of options, the sheer volume of meaningful action is what's important. You want MORE BASEBALL via down-to-the-wire division and wild-card races, extra innings, scoreboard watching and tiebreaker scenarios. You want the MLB schedule-makers to contemplate entering the Federal Witness Protection Program instead of untangling once far-fetched scenarios. You want absolute chaos. Reaching back for that old Second Law of Thermodynamics term that states that all systems tend towards disorder, you're rooting for Team Entropy.
Entering play on Thursday, the American League wild-card race features five teams separated by just two games in the race for the second spot. Meanwhile, the wild-card-leading Rangers have lost three in a row. All of which means that with a bit more than two weeks to go, it's time to rev up the Team Entropy bandwagon and envision what kind of chaos could ensue.
Here's what the wild-card standings look like coming into today's games:
|Team||W-L||PCT||GB||Run Dif||SOS||BP Odds|
SOS stands for strength of remaining schedule, each team's weighted average opponent winning percentage via the Baseball-Reference.com Expanded Standings. BP Odds are of course the Baseball Prospectus Playoff Odds, which account for run differential and remaining schedule and thus show the Indians as slight favorites over the Yankees despite their half-game deficit in the standings. Here I've used the total odds instead of just the wild-card ones to account for the possibility that any of these teams winds up winning their division; the Rangers, three games behind the A's in the AL West, have a 19.1 percent chance of doing so, while the Indians (0.3 percent) and Royals (0.1 percent) would apparently need several Tigers to be mauled by actual tigers in order to gain the upper hand in the AL Central.
Say what you will about the addition of the second wild-card, but with 16 to 18 games remaining for these six teams — the final 10 or 11 percent of the season — the possibility of a multi-team pileup at the end is very real. Better make sure your MLB.TV accounts are up to date on all of your gadgets so you can do more than simply watch the scoreboards, and brush up on those tie-breaking procedures, because things can get complicated.
In each of the past two years, MLB.com has published a page of over 2,500 words on how to untangle the various tie-breaking scenarios for two-, three- and four-team ties for the division and wild-card spots. Suffice it to say that I can't cover all of that here, but the general principles are outlined in the procedure to determine who gets home-field advantage in a two-team play-in scenario, which is based first on the higher head-to-head winning percentage between the pair, then on winning percentage within the division, then within the last half of intraleague games, then within "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."
For two-team ties for division or wild-card spots, the games are tentatively scheduled for Monday, Sept. 30, the day after the regular season ends. For three- or four-team scenarios, second-level play-ins would take place on Tuesday, Oct. 1.
In three- or four-team scenarios, the teams are designated A, B, C and D (if necessary) based first upon their records against one another, then within their respective divisions. If that narrows the tie down to two or three teams, that respective subset procedures are followed, but if not, the interleague half, half plus one, and so on scenario is introduced until the ties are broken, with the "winning" team allowed to decide which letter it wants to be.
From there — well, here's a taste of what the rules say for the relevant wild-card possibilities, followed by my own attempt to devise scenarios involving the teams in question:
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.
Four teams are within two games of the Rays, so it's not too tough to see how this could come about. Suppose Tampa Bay goes 9-9 the rest of the way to finish at 87-75. That would require two of the following four teams to hit exactly these marks with none of them surpassing them: Yankees 9-7, Indians 10-7, Orioles 10-7, Royals 10-6. Obviously, every additional win by the Rays raises the bar one win for each of the other teams; if Tampa Bay went 12-6, the Yankees would have to go 12-4, the Indians or Orioles 13-4 or the Royals 13-3 to get to a three-way tie at 90-72.
Keeping things simple, let's follow the Rays' 9-9 scenario and imagine that the Indians (who have the easiest schedule) and Royals hit their marks as well while the Yankees and Orioles fall short. Tampa Bay, Cleveland and Kansas City won't have identical records against one another, one variant of the three-team tiebreaker scenario. The Rays have gone 6-8 against the other two and have no more scheduled against either. The Indians have gone 11-11 against the other two, with three games still to be played in K.C., while the Royals have gone 13-10 with those three against Cleveland still to come. If the Royals win just one of those three games, they would wind up 14-12 and claim the Club C designation, which means they'd await the winner of Indians-Rays. Because Tampa Bay holds a 4-2 advantage in its head-to-head matchup, it would presumably choose to be Club A and play host to to the Indians, who by default become Club B.
Three-Club Tie for Two Wild Card Spots:
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 be declared one Wild Card winner. Club C would then host the loser of the game between Club A and Club B on Tuesday, Oct. 1 (tentatively) to determine the second Wild Card Club.
Suppose the Rangers continue to tank, going 8-9 while the Rays go 11-7, and this time it’s the Yankees coming out of the pack to go 11-5 while the Orioles, Indians and Royals all fare worse, thus leaving three teams tied for the two spots at 89-73. The Rays would be 10-9 against the other two teams in question not including the results of four home games against the Rangers and three road games against the Yankees. The Yankees would be 10-13 against the other two, not including the results of their three games hosting the Rays. The Rangers would be 6-4 against the other two, not including the results of their four-game set hosting the Rays. If that latter series were a split, the Rangers would be 8-6 and claim first choice in this scenario, and the Rays would be 12-11, claiming second choice; the Yankees would have to win all three against the Rays to claim second choice.
Even then, it's not so clear-cut where the advantage lies. The Rangers could choose to be Club A, in which case they'd have a home game on Sept. 30; if they won, they'd be a wild-card team. They could choose to be Club B, in which case they'd play on the road on Sept. 30; if they won, they'd be a wild-card team, but if they lost, they'd have another chance by going on the road to face Club C. Or they could choose to be Club C, in which case they'd play host on Oct. 1 to a team that had already played an elimination game.
But wait! If the Rays took three out of four from Texas and were swept by the Yankees, that would leave the Rangers 7-7, and the Rays and Yankees both 13-13 — still tied, based upon winning percentage. In that case, the next tiebreaker would come down to intradivision winning percentage. That scenario definitely favors Texas (currently 46-20, .696, with 10 home games and zero road games remaining) to emerge as the team with the first choice to be Club A, B or C.
The Yankees (35-31 now, plus the hypothetical three-game sweep) still have six other AL East games to account for against the Red Sox and Blue Jays, while the Rays (34-31 now, plus the sweep) would have seven with the Orioles and Blue Jays. If the two teams somehow still remained tied — and I don’t even know if it’s mathematically possible, I’m just spitballin’ — the
half-interleague half-intraleague breakdowns would come into play.
Updating to correct my mistake... with 20 interleague games apiece (three still to come for the Yankees against the Giants), that presumably would mean each team's record over their final 71 intraleague games. At this writing, the Yankees are 32-26 with 13 intraleague games still to play over that stretch, while the Rays are just 24-29 with 18 still to play. They'd be unlikely to catch the Yankees under such circumstances, which would mean the Yankees get next pick for A, B or C.
Four-Club Tie for One Wild Card Spot:
After Clubs have been assigned their A, B, C and D designations, Club A would host Club B and Club C would host Club D on Monday, Sept. 30 (tentatively). The winners of each of those games would then meet on Tuesday, Oct. 1 (tentatively), hosted by the winner of the game between Club A and Club B, to determine the Wild Card Club.
Let's again assume the Rays go 9-9 and say the Orioles (10-7), Indians (10-7) and Royals (10-6) create a Yankees-free zone, a scenario that would surely relieve Bud Selig. Tampa Bay would hold a 14-14 record against the other three teams with four still to play at home against the Orioles. Cleveland would be 15-14 with three still to play against the Royals. Kansas City would be 17-13 not including those three games, and Baltimore would be 12-17 not including that series against the Rays.
If those relative rankings held through those final series, the Royals would get first choice as to which Club they want to be, followed by the Indians, Rays and Orioles. In all likelihood -- putting aside any potential specific complications related to matchups, rosters, etc. -- Kansas City and Cleveland would choose to be Clubs A and C, respectively, thus giving them home games. The Rays, as Team B, would then get to choose their opponent. The Orioles would thus be Team D and get whatever's left and have to be happy that they're in the mix at all.
Four-Club Tie for Two Wild Card Spots:
After Clubs have been assigned their A, B, C and D designations, Club A would host Club B and Club C would host Club D on Monday, Sept. 30 (tentatively). The winners of each of those games would be declared the Wild Card Clubs.
Are you trying to kill me?
Let's go back to the Rangers going 8-9 while the Rays go 11-7 and the Yankees 11-5; this time we'll throw the Indians into the mix at 12-5 as well, producing four 89-73 teams. Within that group, the Rays would be 14-11 with four against the Rangers and three against the Yankees still to come. New York would be 16-14 with three against the Rays remaining. The Rangers would be 7-9 with four games left against Tampa Bay, while the Indians 8-11 with no further games against any of them. If the current rankings held (i.e., skipping over the hypothetical permutations of those remaining series), the Rays and Yankees would be Clubs A and C, thus giving them home-field advantage on Sept. 30, with the Rangers (as Club B) picking their poison before the Indians (Club D).
As for a five- or six-team scenario, we're pretty sure that allows Bud Selig to open a special, sealed envelope containing a document that declares total victory for the forces of competitive balance and thus allows him to render the current Collective Bargaining Agreement null and void. The resulting playoffs will be conducted in a "Battle Royal" format where multiple teams take the field at once and play until each either throws the opposition's manager into his own bullpen or successfully submerges Alex Rodriguez in a dunking booth.
Sure, it's a crazy system, but Bud's thought this through and is certain it's in the game's best interests.