Three-team tie for two wild-card spots? The chaos would be great

Monday in Cleveland.

Tuesday in Texas.

Wednesday either back in Cleveland, or at Tampa Bay.

I like it. You root for your team in the American League wild-card race. I'm rooting for chaos.

It's not impossible. Heading into Wednesday night's games, the Rays lead the Indians by one game and the Rangers by two, with five games remaining and two wild-card spots up for grabs.

We need the Rays to finish 3-2 on the road against the Yankees (two games) and Blue Jays (three games). We need the Indians to finish 4-1 at home against the White Sox (one game) and on the road against the Twins (four games).

And we need the Rangers to go 5-0 at home against the Astros (one game) and Angels (four games).

The Rays finish 91-71. The Indians finish 91-71. The Rangers finish 91-71.

There are other combinations, but you get the idea. It's not impossible.

Here's what would happen:

The Rays and Indians would play Monday in Cleveland, because in the head-to-head games between the three teams, the Indians had the best record (7-5) and the Rays had the second-best record (7-6). The winner of that game would be in the playoffs as the first wild-card team, and would host Wednesday's play-in game.

The loser of Monday's Rays-Indians game would fly to Texas to play the Rangers on Tuesday (the Rangers were a combined 5-8 head-to-head against Cleveland and Tampa Bay). The winner of Tuesday's game would be the second wild-card team, and would be the road team Wednesday.

Cleveland to Texas to Cleveland, or Cleveland to Texas to Tampa.

Our travel budget would hate it. Baseball fans without a rooting interest would love it.

So would I.

A three-team tie for two spots. Let's do it.

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