All right, so maybe it doesn't have the same impact as it would have years or decades ago when teams would have been fighting for a division title or, before 1969, a single league championship that would have gotten one team a berth in the World Series. All this one will give its winner is the right to play a one-and-done wild-card game.
Still, the five-team chase for the second wild-card spot in the American League is one that could wind up in a multiple-team tie -- something that's never happened in the history of baseball. And with more than two weeks remaining, there are far too many scenarios to go through to post about here.
First, even though there are five teams involved, they don't play all that many games against each other. The team that has the most games against any of the others involved is the Rays; seven of their 17 remaining games are against the Orioles (four) and Yankees (three). The Yankees have a three-game set with the Rays and the Indians face the Royals in a three-game series.
But there is a real chance that all five teams could wind up with an 88-74 record, and here's how.
Rays, 79-66, 17 games remaining. 3 at Twins, 3 vs. Rangers, 4 vs. Orioles, 3 at Yankees, 3 at Blue Jays. Let's have them take two of three at Minnesota, win two of three vs. Texas, split the four with Baltimore, lose two of three at New York and win two of three at Toronto for a 9-8 mark and 88 wins.
Yankees, 79-68, 15 games remaining. 3 at Red Sox, 3 at Blue Jays, 3 vs. Giants, 3 vs. Rays, 3 at Astros. For this schedule, say they win one of three at Boston, win two of three at Toronto, take two of three vs. Giants, take two of three vs. Rays and win two of three over the Astros. That would have the Yankees go 9-6 over this span, 88 total wins.
Indians, 78-68, 16 games remaining. 3 at White Sox, 3 at Royals, 4 vs. Astros, 2 vs. White Sox, 4 at Twins. To get to 88 wins: win two of three at Chicago, lose two of three at Kansas City, win three of four over Houston, sweep pair with White Sox, split four at Minnesota, a 10-6 overall record, and 88 victories.
Orioles, 77-69, 16 games remaining. 3 at Blue Jays, 3 at Red Sox, 4 at Rays, 3 vs. Red Sox, 3 vs. Blue Jays. This is the toughest one; to get 11 wins in those 16 games (and 88 overall) Baltimore will have to sweep at Toronto, win one of three at Boston, split with Tampa Bay, take two of three vs. Boston and sweep Toronto again.
Royals, 77-69, 16 games remaining. 3 at Tigers, 3 vs. Indians, 3 vs. Rangers, 3 at Mariners, 4 at White Sox. Kansas City also has a tough schedule, but they can do this by taking two of three at Detroit, winning two of three over Cleveland, taking two of three against Texas, winning two of three at Seattle, and winning three of four at Chicago. That might seem a tough road, but the Royals have won 13 of their last 18 and picked up 8½ games on the lead for this spot over the last three weeks. If they can make it to 88 wins, that would be their most since they won the World Series in 1985. That was also the last time Kansas City was in the playoffs; they've had just seven winning years since then. (And you think the Cubs have had a tough time.)
A five-team tie. It can be done at 88 wins that way, or perhaps even at 87 or 86 if you don't think the Royals or Orioles can go 11-5 in their final 16 games. MLB's tiebreaker procedures only include methods of breaking ties for up to four teams; I'm not sure what they'd do if five teams were involved.
In the absence of the Cubs being in the playoff hunt, this sort of thing is great fun to follow. Even if five teams don't tie, it's very possible that two, three or four teams might wind up with the same final record -- and that's not even counting the Rangers, who at 81-64 lead the rest of the pack by only two games.
Six teams tied? Now that really would make heads explode.