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It could happen. I mean, it’s possible in the sense that it isn’t impossible. And is it ever fun to think about. For those of us who like to poke and prod at the edges of baseball just to see what will make it finally cry uncle, the idea of a massive tie for a playoff spot and the chaos that it would create is the forbidden fruit that we all hope to add to our lunch at some point. What if five teams ended the regular season tied for one playoff spot? Hooray Team Entropy! (Thanks, Jay Jaffe!)

MLB has had its share of memorable tiebreakers. The call of Bobby Thompson’s “Shot Heard ‘Round the World” ends with frantic cries of “the Giants won the pennant!” because it ended what was then a three-game tiebreaker series for the lone National League playoff spot for the 1951 season. (This was before divisional play and the idea of the “Championship Series.”) Somewhere out there, a lone Padres fan is still frantically searching for Matt Holliday’s home address and swears that he will bring a ball to finally tag him out, and the Padres-Rockies 2007 game will be continued 10 years after it ended.

When the second “Wild Card” was added to the MLB playoff format, the idea was that it would be fun to have that sort of one-game, win-or-go-home event every year, and it has largely succeeded. But there’s a bit of hidden math that came along with it that might eventually (and inadvertently) bring us even more of the chaos we crave. In most seasons, there a couple of good teams in the league, a couple of bad teams, and a bunch of teams that are scrunched together in the middle. In general, teams on the good side of that ledger are rewarded with a playoff spot (sorry, 1993 Giants). But the more playoff spots that baseball adds, the more likely it is that they will be reaching back into that giant cluster of mediocre teams.

It’s rare that two teams finish with 100 wins in a season, but two teams finishing with 86 wins? Yeah, that happens. What about three? But the dirty little secret about all this fantasizing about chaos is that … MLB has never actually had even a three-way tie situation. They have official written procedures for how it would handle a three-way and a four-way tie, but they’ve never had to use them. The four-way tie situation, however unlikely, at least has the aesthetically pleasing resolution that the four teams can play a single elimination tournament against each other. The three-way tie situation doesn’t feel as fair, because one team could theoretically “win” only having to have play one game.

Baseball is clearly aware of the unfairness of the three-way system. They do penalize the team that will only play one game by requiring that their game be played on the road, but even with home-field advantage taken away (the home team wins about 54 percent of all baseball games), they are playing that game after an off day on which their opponent had to play a win-or-go-home game. It hardly seems fair.

Still, the closest set of circumstances I could find that might have produced a three-way tie-breaker was in 1996. That year, the Yankees, Indians, and Rangers won their respective divisions, and the Orioles finished at 88-74 to snag the Wild Card spot. But had the second Wild Card policy been in effect that year, it would have found the Red Sox and White Sox sitting at 85-77. The Mariners finished the season at 85-76, having had a canceled game against the Angels that was never made up (because under the rules at the time, it was meaningless). Under the current rules, the Mariners would have made that game up, and if they had lost, would have entered a three-team playoff for that last Wild Card spot.

That’s the closest baseball has ever gotten to a three-way tie for a playoff spot. Eventually, it will probably happen in real life. And yet, we still long for the answer to the far more remote question of “what would MLB do if there were a five-way tie?” despite the fact that they’ve never gotten close to needing it. In 2012, Ben Lindbergh asked MLB’s senior vice president of scheduling and club relations Katy Feeney about what MLB would do. Her quote: “To be perfectly honest, considerations for tiebreakers do not go that far. It’s not there anywhere. It’s probably something that would have to be determined.”

To translate: ¯_(ツ)_/¯

So, as a public service to our friends at MLB, I’ve decided to help out. Let’s design a five-way tie-breaker that makes some sort of attempt at fairness.

Warning! Gory Mathematical Details Ahead!

Before we start, we need to talk about what our needs are. We need something that:

  1. Produces a winner.
  2. Takes a minimum number of days, because we can’t have one league still in tie-breaker land while the other is ready for its LCS.
  3. Is as close to fair as the laws of math will allow it to be.
  4. Has as few logistical (mostly travel) nightmares as possible.
  5. Looks great on TV.

As a spoiler, we’re not going to satisfy all of these. Like most things in life, we have to pick and prioritize which is the most important to us. But let’s give it a whirl.

***

Idea #1: Two days of double-header delights!

Let’s start with the idea that will never actually happen for a host of reasons, but would be so. much. fun. The twin problems of the single-elimination (loser goes home) model with five teams are that resolving it will probably take three days, and someone’s going to be in a position where they need to win three games, while some other lucky team will only need to win two. What if we tried something a little different?

First, we will take advantage of the fact that there are effectively four United States cities that have two MLB ballparks within a reasonably short distance of one another. New York has Yankee Stadium and Citi Field. Chicago has Wrigley Field and Guaranteed Rate Field. San Francisco has AT&T Park and Rickey Henderson Field across The Bay in Oakland. Los Angeles has Dodger Stadium and Angel Stadium of Anaheim. In all of these cases, if the two teams that lived in these respective stadia needed to play a day-night dual-stadium doubleheader against each other, they could. (The Yankees and Mets have done it a couple of times.)

Before the season begins, teams make a sacred pact that if there is a five-way tie, the tie will be settled in one of these four cities. Perhaps the designated city could rotate each year. Or be determined by the All-Star game. All five teams gather in the designated city. And then the murders begin. Did I say murders? Fun. And then the fun begins! Over two days, the five teams face off in a round-robin tournament, effectively consisting of two doubleheaders played on consecutive days. (Someone reading this in an MLB clubhouse just called me a name that I can’t repeat on a family-friendly website.) Everyone plays four games. Everyone plays everyone else. All teams get two games where they are the “home” team and two where they are the “away” team.

The schedule could go something like this:

Day 1, noon
B at A (Wrigley Field)
C at D (Guaranteed Rate Field)

Day 1, 4:00 pm
A at E (Wrigley Field)

Day 1, 8:00 pm
E at C (Wrigley Field)
D at B (Guaranteed Rate Field)

Day 2, noon
A at C (Guaranteed Rate Field)
E at D (Wrigley Field)

Day 2, 4:00 pm
C at B (Guaranteed Rate Field)

Day 2, 8:00 pm
B at E (Guaranteed Rate Field)
D at A (Wrigley Field)

Looking at the schedule for Day 1. Team A effectively experiences the day as a “traditional” doubleheader with a game at noon and a game at 4:00 (note that both games are at Wrigley), and Team E has a traditional doubleheader (again, both at Wrigley), with games at 4:00 and 8:00. The next day, Teams C and B have traditional doubleheaders (at Guaranteed Rate Park). Team D plays two fully split doubleheaders, but is rewarded by not having to do any in-day travel between parks.

Teams could bring their full, expanded September rosters. They’ll need them. No messy travel. No Team A playing in San Francisco on Monday and then Miami on Tuesday and then Los Angeles on Wednesday. On the downside, it’s possible that there won’t be a proper “home” crowd for any of the games, and certainly not for all of them. It also runs the risk of having the White Sox play a couple of games that are essentially a playoff game on the North Side.

Everything is over in two days. There’s no “but they only had to win two games!” whining. Although, there’s the distinct possibility that everyone could go 2-2. We could have a playoff spot being decided by tie-breakers like run differential within the tournament or “away runs” (the way that they do it in soccer) or maybe … a home run derby. (A modified version of this could even be used for a three-way tie. All three teams could go to a selected city, and essentially play a one-day, three-game round-robin. Same problems.)

This idea probably should mostly fail on the fact that it’s a liability to not produce a clear-cut winner. The nice thing about single-elimination tournaments is that eventually, in baseball, someone loses and can be summarily dismissed. But the reality is that it would fail because it’s nearly impossible to sell advance tickets for this. Fans wouldn’t know that such an event would be happening until the last day of the season, leaving about 48 hours to make travel plans. Even if there were five fan bases that might be interested in attending, along with some thrill-seeking locals and general baseball fans who just wanted to go see something cool, would it be enough to make the idea work economically?

***

Idea #2: Three plus two equals …

If a five-way tie were to actually happen, most likely MLB would melt together its procedures for a two-way tie with its procedures for three-way tie. Two of the teams would play each other. Three of the teams would engage in a three-way tiebreak. The two winners of those gauntlets would play a final game.

The schedule would probably look something like this:

Day 1
Team A at Team B

Day 2
Team D at Team E
Team C at Winner of A/B

Day 3
Winner of A/B/C plays Winner of D/E

Again, in the current three-team tiebreaker, MLB has some idea that it’s a bit unfair to teams A and B, because they have to win two games, while team C only has to win one. Team C is therefore relegated to the road for their confrontation with the winner of A and B. I think that same thought process raises an interesting question about how to assign home-field advantage for the final game on Day 3.

Team C plays their “semi-final” against a team that had to play a do-or-die game the day before, a game which they watched on TV. Teams D and E play their “semi-final” against each other on essentially equal footing. It’s reasonable to say that, knowing nothing else, if Team C makes it to the final, they had an easier path than did the winner of D vs. E. However, if A or B make it through, they clearly had a much tougher time than did the winner of D or E. I’d suggest that A or B would be given home-field priority against D/E, but that D/E would get home field against C.

If there ever were a real five-way tie, I’m guessing that this is what it would look like.

***

Idea #3: The Mulligan System

In a single-elimination system, the source of the unfairness is that two of the five teams are essentially playing off for the right to be in a four-team tournament, and have to play an extra game, despite the fact that everyone here is supposed to be an equal. Laying home-field advantage aside, the three teams who don’t have to play in—knowing nothing else about them—have a roughly 25 percent chance of emerging from the whole thing victorious, while the two short-straw teams check in with a 12.5 percent shot.

What if we could even up the odds a little bit, though? What if Team A played Team B for the right to enter the field of four, but came into the tournament with a mulligan? They were allowed to lose one game before being knocked out. It would look something like this:

Day 1
Team A at Team B

Day 2
Team D at Team E
Winner of A/B at Team C

Day 2a, if necessary
Winner of A/B at Team C

Day 3
Winner of D/E at Winner of A/B/C

Day 3a, if necessary
Winner of D/E at Winner of A/B

In this case, Teams A and B play on Day 1. The winner gets to play Team C, which had an off day and I’m going to give them home field (we’ll see why in a moment). If C wins the game, they don’t advance. In fact, they have to beat A/B twice, because A/B has a mulligan.

If A/B wins the first (or the second, if necessary) game, then they advance to play the winner of D and E. In theory, D/E should also have to beat A/B twice, although that could set up a five-day tie-breaker just to adjudicate one spot in the playoffs. In this case, we’re only going to grant A/B one mulligan, for practical purposes. The winner of A/B may lose a game to one of the teams it plays, but not both. Now, we get into some moving parts!

For a moment, let’s assume that all games are 50/50 bets. Team A and Team B would have a 50/50 shot of winning the first game against each other. If A/B has a 50/50 chance of winning against their next opponent in one game, they have 75 percent chance of winning one of two games. If they made it to the next round(?) they would go there with a 50 percent chance of only having one chance of winning and a 50 percent chance of having two chances.

Team A’s chances of emerging victorious are given, mathematically, as .5 * .75 * (.5 * .5 + .5 * .75) = 23.4 percent. Team B would have the same chances. Team C would have a one-quarter chance of surviving to the next round, because they have to beat A/B twice, again assuming that the games are 50/50 shots, and a 50 percent chance of winning the final round, leaving its odds at 12.5 percent. Teams D and E would split the remaining probability, leaving them each at 20.4 percent.

This is getting a little closer to even odds for everyone, though it’s not perfect. Team C gets the short end of things, but perhaps their situation is not all that bad. Their first game against A/B comes after an off day, while A/B has to play an elimination game the day before and A/B is probably using their second-best pitcher in the first game against Team C’s ace.

If Team C has to play a second game, A/B is playing their third all-hands-on-deck game in as many days vs. the second for Team C. Suddenly, Team C’s chances are moving up. We’d have a hard time pinning down exactly how much, but they’d probably be better than 50/50. We can certainly give them a bump up by letting them have home-field advantage.

This is where I broke out the Monte Carlo simulation. For those unfamiliar with the model, it essentially simulates a complex system the way that a series of dice rolls would. Since we’re working completely in the abstract, it will at least give us some idea of what we’re looking at. I first assumed—using the schedule shown above—that all teams were evenly matched in personnel, but that the standard home-field advantage (54 percent) applied. This gave us the following distribution of final winners:

Team

Final Wins

A

17.7%

B

26.8%

C

15.7%

D

21.5%

E

18.3%

That’s getting closer to equality for Team C, although we see that the home-field advantage in that initial A vs. B game is driving a lot of value. Perhaps that would have to be moved to a neutral site, for fairness reasons.

Or not. I made a minor tweak in the system, instead assuming that in Team C’s initial game, where they are fresh, at home, and can throw their ace against A/B’s second starter, they would have a 60 percent chance of winning just that first game (again, remembering that they have to win two). They maintain the home-field advantage in that second game, but I gave them no special advantage there. That one minor move made things look a little better.

Team

Final Wins

A

16.8%

B

25.0%

C

17.5%

D

22.0%

E

18.7%

How realistic is it that Team C would have that great of an advantage in that first game? We can’t account for exactly who the starters would be, but we do know that in the regular season from 2000-2016, there were 1,324 cases of a team playing on the road when they had played the day before against a home team that previously had an off day. The home team won 56.7 percent of those games. That’s not quite 60 percent, but again, we can’t really account for the starting pitcher effect either.

Team C is probably a little bit screwed, but I’d argue that this mulligan system at least distributes the probabilities of each team making it out alive—not entirely fairly—a little more fairly than a simple single-elimination tournament. The cost would be an extra day (this could take four days, rather than the three that a single-elimination format would take) and a lot of confusion for some fans.

***

Idea #4: Not Have Ties

Football figured this out. They don’t have tie-breaker games. They have a series of tiebreaks that start with head-to-head record, and go from there. One could imagine that run differential would be an easy next step in the event of a tied season series. We already have the drama of the winner-take-all Wild Card game, and someone will be in that game. Why not simply settle all other claims bloodlessly?

Oh, right. This is why baseball is better than football. Let’s Go Team Entropy!

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marshaja
9/07
I know it was for both wild card spots and not just one, but how would last year has been determined if all 3 teams that were chasing the NL Wild Cards had finished tied? Just do the normal 3 way tie breaker and that team advances to NLDS?
Deadheadbrewer
9/07
Or just have a mascot race. All five teams are at the park, on the field where the Wild Card game is about to be played, the mascots race, and the winning mascot's team heads for the dugout to bat in the top of the first. (boy, "mascot" is an odd word when one says it and types it too much . . . )
pizzacutter
9/07
The Freeze vs. Teddy Roosevelt.
dianagramr
9/07
Jay Jaffe approves of this article.
newsense
9/07
The additional argument for idea #4 is that you have the 5th thru 9th best teams out of 15 in the league vying for a single elimination Wild Card slot as the away team. It's a fight over a lottery ticket and none of these teams are more than minimally deserving of post-season play. In fact, rather than ruminate over what tie-breaker rules are fair, you could draw lots. Not much fun, but it's fair.
mitchiapet
9/07
The 5-way tie scenario is one where it seems likely MLB would use a hybrid between the tie-breaker and the play-in game. With the current alignment, it's a given that at least 2 teams in this scenario will be from the same division. With 19 games between teams in the same division, the head-to-head record would then be used to decide which team would represent the division in a 2-way or 3-way tie scenario. At that point, the current rules for those scenarios would take effect. The only time I can recall a tiebreaker being used was 2005, when both the Yankees and Red Sox both finished 95-67. They were both safely in the postseason, having finished 2 games ahead of Cleveland for the wild card. MLB decided to use head-to-head record as a tiebreaker rather than play a game to determine who won the division.
Robotey
9/07
This is a lot of fun, but could you also write a story--if you haven't already--about the three-team tiebreaker, perhaps comparing the equities and inequities of MLB's proposed format and alternatives that could be better? If I'm not mistaken, the current system doesn't sound perfect.