The New Playoff Probabilities in 2013

The Houston Astros are heading to the American League and baseball is headed toward balanced divisions (numerically at least) and 10 playoff teams. There are many fans in Houston upset at the move, but one aspect that they ought to consider being excessively happy about is the team’s improved playoff probabilities under the forthcoming new alignment.

Currently baseball is a mess of uneven odds. The squads in the AL West only have three foes to compete with for a division crown while the Astros and others in the NL Central have five each. Furthermore, American League teams only have ten others to battle with for the current single wild card spot. National Leaguers must outpace 12 others in the race for baseball’s second chance bracket. Between the years of 1995 (no playoffs in 1994 remember? Good thing the new CBA’s already done) and 2012, there were four separate probabilities for making baseball’s postseason depending on which division a team played in.

For the AL West and its four teams, each had a 31.8%* at the playoffs, assuming teams of equal strength. The AL Central and East division teams were next with a 27.2% chance. The NL West and East followed with a 26.2% and the NL Central teams bring up the rear at just 23.1%.

*Each team had a 25% chance at an automatic berth and then a 75% (since they can’t win both) chance at a wild card shot, which carried a 9% (1/11) probability. 0.25*1 + 0.75*1/11 = 31.8%. The formula comes from Bayes’ theorem.

With the switch to 15 teams per league, five teams per division and two floating wild card, every team in baseball will enter the season with identical odds. Again, assuming a theoretical model here in which teams are of equal strength. Going again to Bayes’ that probability is 0.2*1 + 0.8*2/12 or the same 33.3% chance you expect when five of 15, in equal set up, qualify.

Every team in baseball sees a boost in their chances to enjoy and reap the benefits of playoff baseball, but the NL Central teams, and their fans, have the biggest jump to celebrate.

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Matthew Carruth is a software engineer who has been fascinated with baseball statistics since age five. When not dissecting baseball, he is watching hockey or playing soccer.

24 Comments
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Flharfh
14 years ago

Is there a way to account for, say, the Blue Jays having to play 3 good (Rays, Yanks, Bosox) and 1 terrible (O’s) team, while say, the Brewers have to play 2 goodish (Cardinals, Reds) and 2 terrible (Cubs, Astros) teams?

Flharfh
14 years ago
Reply to  Flharfh

Correction: Three terrible teams (forgot Pit)

BirdStackMember since 2026
14 years ago
Reply to  Flharfh

no this article is clearly not about the skill of a team. Over decades a teams current ability is negligible.

exxroxMember since 2026
14 years ago
Reply to  BirdStack

unless we are talking about the Orioles or Pirates..oh wait, he was

TK
14 years ago
Reply to  BirdStack

Or you don’t understand what “decades” means.

Toffer Peak
14 years ago
Reply to  BirdStack

Since 1976 (2.5 decades plus free agency started) the Pirates have a .472 record and the Orioles have a more respectable .492 record. NY and BOS kill at .570 and .544 respectively. So yes, even when looking at decades their are unfair divisions though the effect is often exaggerated and as noted in the long run it’s even smaller.

BirdStackMember since 2026
14 years ago
Reply to  BirdStack

Way to take the O’s best decade and cut it in half for that fact. Since 1960 they have 3 WS rings to Boston’s 2. And the AL East, or any division, has changed since then. The same Phillies in perennial contention were the first team to 10,000 losses in any sport. Yes the yankees are the yankees but beyond that you can’t tell me you have any clue who the good teams are in a decade.

Toffer Peak
14 years ago
Reply to  BirdStack

My reasoning was simple. That is the year MLB free agency began and it has dramatically and irreversibly changed competitive balance (barring major changes to the CBA and/or US legal decisions).

Matt
14 years ago

Shouldn’t the formula be .2*1 + .8*(1/12) since there is only one wildcard team in each league? It shouldn’t make a major difference but it’s just for correctness’ sake.

pete2286Member since 2016
14 years ago
Reply to  Matt

The first sentence says 10 playoff teams. I think MLB is adding a second wild card team in each league.

Matt
14 years ago

Thanks. I wasn’t sure if that was going into effect in 2012 or not

placidity
14 years ago
Reply to  Matt

It’s not. The title says 2013.

BenS
14 years ago
Reply to  placidity

Actually, 2012 is still up in the air. MLB is hoping to enact the change then, but 2013 is definite.

sheath1976
14 years ago

This was a relly cool regular season with the Rays and Cardinals overcoming massive odds to unseat the Red Sox and Braves respectively on the last day of the season. If this new system had been in place much of the Incredible month of September never would have happened because all four aforementined teams would have secured playoff spots.

Logical Step
14 years ago
Reply to  sheath1976

So if 2014 September is boring because the gap between WC#2 and WC#3 is big, the only solution is to add a 3rd Wild Card! Obviously!

adohaj
14 years ago
Reply to  Logical Step

I say we just play a bunch of single elimination games until only 8 teams remain. Obviously NYY BOS and PHI will get a free pass to the final 8 they deserve it for spending all of that money.

CampBrice
14 years ago
Reply to  sheath1976

Yet the argument could be made that the Blue Jays, Indians, Angels at the trade deadline or even into the waiver trade period would be that much closer to the playoffs and thus might act according and try – making September even more exciting – I’m sure this is the stance that MLB is taking

ben
14 years ago

Not questioning your formula, but is it really Bayes’ theorem? Bayes is for when you have one conditional probability and want to find the opposite one. This seems like just a direct multiplication of probabilities that you already know. For the West Coast teams example:

P(division win) = 1/4
P(wild card win) = p(not division winner) * p(beat 11 other non-division winners) = (3/4) * (1/11)
P(playoffs) = p(division win) + p(wildcard) = .318

I think Bayes would be more useful for something like p(wild card | playoffs) or p(division | playoffs).

Hurtlocker
14 years ago

Houston going to the American league is just another team that will get beaten up by the Al East. What’s the odds that Houston will add 2-3 wins each to the Yankees, Boston and Tampa Bay? Pretty good I would say.

akas247
14 years ago
Reply to  Hurtlocker

Unless Houston has a bullpen day and throws six pitchers against the Yanks

Pete Munro
14 years ago
Reply to  akas247

hey, I got the win. that’s all that matters.

…right?

Brad Lidge
14 years ago
Reply to  akas247

“hey, I got the win”

no you didn’t.

Peterskull
14 years ago

To get the overall odds for the past 18 seasons I believe you need to use some sort of a weighted average approach by the odds of each season because you need to factor in the expansion teams in 1998. As I recall, for 3 years the NL West also had only 4 teams This would slightly increase the NL West overall odds over teams in the NL east, except for Arizona. Over the same timeframe the NL Central also only had 5 teams until the Brewers switched leagues. This would give the NL Central slightly better odds. And of course, for 3 seasons both leagues had 14 teams.

Robin
14 years ago

Go Red Sox