Are the Indians in the wild-card driver's seat?

Jason Miller

In case you missed it, Thursday night the Cleveland Indians opened their Wild Card Victory Tour with an extra-innings victory over the Houston Astros. Friday night, the Tour continues.

What in the hell am I talking about? The Indians have by far the easiest remaining schedule among the American League's wild-card contenders. For a while now, there's just been this vague prospect that they might snag at the last instant the second wild card ... but now it's distinctly possible they'll grab the first one, and thus play that dreaded sudden-death contest at home. Whether anybody actually shows up is a different matter, of course ...

Anyway, counting Friday night against the Astros, here are the Indians' remaining games:

Astros - 3
ChiSox - 2
Twins - 4

Adding those up and prorating their opponents' winning percentages this season, the Indians basically have nine games against one team with a .398 winning percentage. We would expect the Indians to win roughly six of those nine games. Okay, so they're not a lock to win at least six games. But they're likely to win six, and more likely to win seven than five.

So are they likely to grab one of those wild cards? It's very very close. According to Baseball Prospectus, it's almost exactly a 50/50 proposition. Because while it's likely the Indians will win more games than at least one of the teams ahead of them, there are also teams behind them that might finish in a rush. That's the problem with a tight six-team race; nobody's safe, whatever the apparent advantages.

But compared to just a few weeks ago, let alone a few months, you have to love the Indians' chances.

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