Here's what happens if the Mets, Cardinals, and Giants all tie for NL wild card
The three teams are 80-71 at the moment
By
Mike Axisa
•
5 min read
The 2016 season isn't over yet, but, if it were, there would be a three-team tie for the two wild-card spots in the National League. The Mets, Cardinals and Giants all came into Wednesday with identical 80-71 records.
It goes without saying the odds of those three teams finishing the regular season with identical records is tiny, though MLB is prepared for such a scenario. Here are the tiebreak rules:
Three-Club Tie for Two Wild Card Spots: After Clubs have been assigned their A, B and C designations, Club A would host Club B. The winner of the game would be declared one Wild Card winner. Club C would then host the loser of the game between Club A and Club B to determine the second Wild Card Club.
It's not that complicated, right? Two teams play a tiebreaker game and the winner is one wild-card team. The loser plays the third team and the winner of that game gets the other wild-card spot. In this scenario you want to be Club A because you get the home tiebreaker game and a loss would not automatically mean elimination.
The Club A, B and C designations are determined according to a long cookbook formula. First up is head-to-head records. Here they are:
| NYM | SFG | STL | |
| vs. NYM | -- | 3-4 | 3-3 |
| vs. SFG | 4-3 | -- | 4-3 |
| vs. STL | 3-3 | 3-4 | -- |
| Total | 7-6 | 6-8 | 7-6 |
These three teams do not have any games remaining with each other, so those head-to-head records are final. The Mets and Cardinals have identical combined head-to-head records while the Giants lag. That brings us to this tiebreak scenario:
If Club 1 and 2 have identical records against one another, but each has a better record against Club 3, then Clubs 1 and 2 would follow the two-Club tiebreak rules to break their tie to pick the first designation.
The Mets and Cardinals are 3-3 against each other and 4-3 against the Giants, so they're our Clubs 1 and 2. The Giants are Club 3.
OK, so now what are the two-club tiebreak rules? They're more straightforward:
1. Head-to-head winning percentage during the regular season.
2. Higher winning percentage in intradivision games.
3. Higher winning percentage in intraleague games.
4. Higher winning percentage in the last half of intraleague games.
5. Higher winning percentage in the last half plus one intraleague game, provided that such additional game was not between the two tied clubs. Continue to go back one intraleague game at a time until the tie has been broken.
Because the Mets and Cardinals have identical head-to-head records, we move on to the second set of criteria, their intradivision records. That means the Mets against the NL East and the Cardinals against the NL Central. Here are their records:
Mets vs. NL East: 33-32
Cardinals vs. NL Central: 36-30
The Cardinals have the edge here, though it's worth noting both teams have a bunch of intradivision games remaining, so those records will change. We're just taking a snapshot in time, so we'll stick with those intradivision records and keep moving forward.
Based on those intradivision records, the Cardinals are Club 1, the Mets are Club 2, and the Giants are Club 3. That does not mean the Cardinals are Club A for the tiebreaker scenario. It means the Cardinals get to pick whether they want to be Club A, B or C. Then the Mets get to pick. The Giants get whatever is left over.
Since Club A has the most favorable situation -- again, you get the home tiebreaker and a loss doesn't equal elimination -- it's safe to assume that the Cardinals would elect to be Club A. That leaves the Mets the choice of being Club B or Club C. Each has their pros and cons:
- Club B gets two chances to win one game to claim a wild-card spot, but both games are on the road. That means travel and unfriendly environments and all that.
- Club C plays one winner-take-all game for a wild-card spot. That game is at home and you get to play a tired team that just lost the first tiebreaker game and traveled from St. Louis.
There's a case to be made for being either Club B or Club C. I'm not sure there's a right answer. It comes down to each team's preference. The Mets have a slightly better record at home (41-35) than on the road (39-36), so that doesn't help us much. The travel schedule could be a factor as well:
- The Mets finish the regular season in Philadelphia, so selecting Club B means going from Philly to St. Louis. A loss in the tiebreaker game then means a trip from St. Louis to San Francisco.
- Selecting Club C means going from Philly back to New York and awaiting the loser of the first tiebreaker game.
What do you think is better, Club B or Club C? I lean Club B. Yeah, you have to play on the road, but it gives you two chances to win one game. Let's assume the Mets choose Club B. That makes the Giants our Club C.
So, after all of that, here are the tiebreaker games:
Game 1: Mets (Club B) at Cardinals (Club A), winner gets one wild card spot
Game 2: Loser of Game 1 at Giants (Club C), winner gets the other wild card spot
All of that only determines the two wild-card teams. So there's two tiebreaker games then the Wild Card game itself. Zoinks! Home-field advantage for the Wild Card game is determined through the five-step criteria cited above, so first head-to-head record, then intradivision record, so on and so forth.
Got all that? It's not as bad as it sounds. The only real decision here is whether Club 2 wants to be Club B or C for tiebreaker game purposes. The rest is all based on head-to-head records and things like that.
Keep in mind this only applies if the three teams finish with identical records and tie for the two wild-card spots. There are all sorts of separate tiebreak scenarios for two-team ties and four-team ties, division races, wild-card spots ... all of that. MLB has tiebreak scenarios in place. With any luck, we'll get to see some of them in action this year.
