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How it works: Your guide to MLB playoff tiebreakers

By Dayn Perry | Baseball Writer

If these three teams wind up in a tie atop the NL Central, things will get complicated. (USATSI)
If these three times wind up in a tie atop the NL Central, things will get complicated. (USATSI)

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While a number of division races are settled or soon to be settled, there's still a chance that the 10-team playoff field will remain undecided when the regular season ends on Sunday, Sept. 29. For instance, at present, the NL Central crown is very much in play, and six teams are within 3 1/2 games of the two AL wild-card berths.

So what happens if tiebreakers are needed to sort through the mess? Let's take a walking tour, in English as plain as the topic allows, of how the various scenarios will work ...

Two Teams: What if the Cardinals and Pirates wind up tied for the NL Central title?

Division winners, of course, get to skip the one-game wild-card round, so winning the flag matters quite a bit under the current playoff structure. If two teams end the regular season in a tie atop the divisional standings, then they'd play a one-game tiebreaker, likely on Sept. 30.

As for home-field advantage in that game, the team with the best head-to-head winning percentage would host. In this example, the Pirates would play at home, since they went 10-9 against the Cardinals this season. In the unlikely event that the two teams played .500 ball against one another, then you'd go to intra-divisional winning percentage (i.e., the best record against teams from within the division) and, failing that, to intra-divisional winning percentage in the second half to determine home field.

In the case of the Cardinals and Pirates, the loser would drop into a wild-card slot. In a division in which both tied teams are not in wild-card position, then the loser of the tiebreaker game would see its season end.

Three Teams: What if the Cardinals, Pirates and Reds wind up tied for the NL Central title?

In this most unusual instance, the three teams would be assigned "A," "B" and "C" designations. The team with the best intra-divisional record gets to choose its designation from the three listed above. The team with the next best intra-divisional record picks from the remaining two, and the team that holds up the rear in terms of 2013 record against teams from within its division gets the letter that's left over.

If any of those records are the same, then you go to intra-divisional records in the second half. If those records are the same, then we go to the results of each team's most recent intra-league but not intra-divisional game. For example, in the case of the Cardinals their most recent intra-league -- i.e., against an NL opponent -- but not intra-divisional -- i.e., against a team from outside the NL Central -- game would be the one they'll play against the Nationals on Sept. 25. That's the game that would count in the (highly) unlikely event that a third tiebreaker is needed for slotting purposes.

OK, so then we go to a mini-tournament format among Teams A, B and C. In the event that the divisional runners-up will each claim a wild-card berth, which will almost certainly be the case with the NL Central this year, Team A would host Team B on Sept. 30. The loser of that game would get a wild-card berth. Then the winner of that first game would host Team C the next day (Oct. 1), with the winner of that game claiming the division title and the loser going into the wild-card round against, as it turns out, the loser of the first game. Again, the team with the best intra-divisional record gets to choose which route it wants to attempt, though it's safe to say the Team B path isn't getting chosen first.

In the event that only one wild-card berth is in play, then the loser of the second game would earn it. The loser of the first game would go home. If no wild-card berths are at stake, then the format proceeds the same way, the key difference being that there's no consolation prize for the losers.

And what if the runner-up in a divisional tiebreaker then falls into a tie for one wild-card spot with a team from outside the division? In that case, you'd have a Team D -- the team from outside the division. They'd visit Team C and, if they won, claim a wild-card berth, and the winner of the usual Team A-Team B game would win the division. If Team C won, then they'd advance to play for the division title with the wild-card berth as a fallback. Team D and the loser of the Team A-Team B game would call it a year.

So ... Three teams atop a division, tie with team from outside the division for both wild-card berths. In that case, you'd again have Teams A, B, C and D assigned as described above. Team D still gets a wild-card bid if it beats Team C, and the winner of Team A-Team B wins the division while the loser claims the remaining wild card. If Team D loses to Team C, then Team C would play the winner of the Team A-Team B for the division title. The loser of the divisional "final" would get a wild card, and the other wild card would go to the winner of Team D versus the loser of the Team A-Team B game.

Perhaps a more specific example is in order. While what follows isn't possible in 2013, let's say the Yankees, Red Sox and Orioles tie for the AL East title at 95-67. Over in the AL Central, the Indians finish behind the Tigers but in wild-card position with the same record, 95-67. So, both both wild cards and the AL East title are on the line in this imaginary tiebreaker ...

Team A: Yankees
Team B: Red Sox
Team C: Orioles
Team D: Indians

The Yankees beat the Red Sox at home, and the Orioles do the same to the Indians. The next day, the Orioles travel to the Bronx and beat the Yankees, thus claiming the AL East title and a bye into the ALDS. The Yankees claim one wild card. Meantime, the Indians top the Red Sox at Fenway and claim the remaining wild-card bid.

Four Teams: Yeah, it could happen

Under these semi-ridiculous circumstances, you'd in essence have a four-team tournament for the flag. Team B at Team A, Team D at Team C; winner of the first game hosts the winner of the second game the next day with the division title at stake. "Draft order" for team designations is assigned in the manner noted above. If one wild card is at stake in a four-team divisional tie, then the runner-up gets it. If two are at stake, then the runner-up gets one, and the two losers of the Team A-Team B and Team C-Team D games play for the second bid.

Wild Card: What about the various AL scenarios, which one might punnily characterize as being "wild"?

As noted above, it remains a crowded scrum for the two AL wild-card bids, what with the Rangers, Rays, Indians, Orioles, Royals and Yankees all having designs on a berth. The MLB rules for such affairs -- at least the ones publicly disclosed -- provide for, at most, four-team tiebreaker scenarios. So we'll cover those here ...

- Two teams tie for one wild card: Very straightforward. One-game playoff on Sept. 30 with home-field advantage determined by the methods noted above in the Cardinals-Pirates section.

- Three teams tie for one wild card: Teams A, B and C are assigned as noted above and they play as noted above -- i.e. B at A, followed by the winner of A-B at C with the bid at stake.

- Three teams tie for both wild cards: Game 1: Team B @ Team A, winner gets wild card; Game 2: Loser of Game 1 @ Team C, winner gets remaining wild card.

- Four teams tie for one wild card: Tournament format. B @ A, D @ C, winners of those two games play for wild-card berth.

- Four teams tie for both wild cards: Game 1: Team B @ Team A, winner gets wild card; Game 2: Team D @ Team C, winner gets remaining wild card.

All of that made perfect sense, right? Good. Now consider yourself prepped for the forthcoming chaos.

(Wink of CBS eye: Paul Hagen, MLB.com)

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