Tiebreaker Follies

There’s a problem with the WBC tiebreaker system—and it’s not that Canada was eliminated despite finishing with the same record as the US and beating them.

It’s this:

Last night, Korea was guaranteed to advance unless they lost while allowing 8 runs. Had they not scored in the eighth inning and the game had been scoreless going into the bottom of the ninth, it would have been in their best interests to issue for consecutive intentional walks, losing 1-0 and advancing to round two.

Tonight, the US faces a similar situation. With a win, of course, they advance without the need of any tiebreakers. However, if they lose 1-0, they also advance.

If they lose 2-1 or 2-0 with fewer than two outs in the ninth, they’re eliminated and Japan moves on. If they lose 3-0, 3-1 or 3-2 with fewer than one out in the twelfth they’re eliminated. If they lose while giving up 4 runs at any point, they’re out.

Mexico has only one chance to advance: to win 3-0 or 4-0 in 13 or 14 innings. If they give up a run, they’re out. If they win in fewer than 13 innings, by any score, they’re out. If they win 2-0 in 14 innings, they’re out.

If the score is 0-0 in the bottom of the ninth inning you have the situation where it is in the best interests of both teams to not try to win. If the US tries to get Mexico out, they risk giving up a 2-run homer with less than two outs, and being eliminated. If it goes to extra innings, there exists the possibility of giving up a 3-run homer or grand slam in extra innings that would eliminate them.

For Mexico, even a 4-0 win in nine innings eliminates them. They need to keep the game scoreless until the bottom of the 13th, and then try for a 3-run homer or grand slam.

So at 0-0 in the ninth, the US needs to intentionally give up a run. Four intentional walks? Well, Mexico should swing at the pitches and strike out. Hit the batters? Well, Mexico can swing at those two and if they get hit, they’re strikes. They can get four automatic balls by refusing to pitch to the first batter and then balk him home, or throw the ball into the stands, but the runner could abandon the basepaths and be called out. But either way, it’s in the interests of the USA to give up 1 run so they don’t give up 2, and that’s from the ninth inning on.

For Mexico, the farce starts earlier: at the first pitch. They’ll be trying to shut down the US, but on offense they need to not score—the first run they score before the 13th inning eliminates them from the tournament.

Yes you read that right, if they score, they’re eliminated.

If it gets to the 13th scoreless, it’s not enough to score, they need to hit a three run homer. So if the bases are loaded and the batter gets a base hit, a Mexican runner has to pass another runner on the basepaths so they won’t score fewer than three runs.

I’m sure that both teams will try to win—but neither should. While the US just needs to win to get in, the first run they score will eliminate Mexico which will make them try to win which could eliminate the US. So the US should try to not score, either. Mexico should be trying to keep the game scoreless until the 13th, when they should not score on anything but a 3-run homer or Grand Slam.

But it sure would be entertaining to watch it unfold.

Thanks to DKDC at BTF for coming up with some of the scenarios.