Breaking Down a U.S.-Canada-Colombia Tiebreaker Nightmare

As of this writing, Mexico leads Canada 10-3 in the ninth inning of its final pool stage game. If that result holds, Mexico will win World Baseball Classic Pool C. (Update: That result held.) This evening, the U.S. will take on Colombia to determine who advances as the second-place team in the group. If Team USA wins, it’s simple: both the U.S. and Mexico will finish 3-1, with Mexico advancing as the no. 1 seed by virtue of its head-to-head win on Sunday. If Colombia wins, it will finish 2-2, as will the U.S. and Canada, with each team having a head-to-head win over one of the others. That throws things to the WBC’s wonderfully confusing runs-allowed-per-outs-recorded tiebreaker. Making things more complicated: Because the U.S. beat Canada by mercy rule, all three teams will finish the group stage having recorded a different number of outs. That makes the math all the more complicated.

Fortunately, I did it for you. There’s no guarantee the game will actually go nine full innings; in addition to the potential for extra innings, Colombia, the home team, could win in nine innings but only bat eight times, leaving the U.S. on 45 defensive outs in the relevant games rather than 48. But assuming the U.S. and Colombia each record 27 outs tonight, here’s who would advance given a particular score:

If Colombia beats the U.S. 7-4, those two teams would finish with the same run quotient. In that case, I am not 100% sure what would happen. Either the tiebreakers would continue down the line to earned run quotient in games among tied teams (so far there have been no unearned runs), then batting average, then drawing of lots; or Colombia would advance by virtue of a head-to-head tiebreaker against the U.S. The latter case would make more sense, but the posted WBC rules make reference to “identical winning percentage,” which might preclude resetting the tiebreaker to only include two teams. So, uh, let’s just hope it doesn’t come to that.