clock menu more-arrow no yes mobile

Filed under:

Predicting Possible Playoff Probabilities

Gary A. Vasquez-USA TODAY Sports

One of Billy Beane's most memorable quotes in Moneyball was that his [stuff] doesn't work in the playoffs. Baseball is, more than any other, a sport that can be understood through the lens of statistics. The stats that we have aren't perfect, but they often give us a pretty good understanding of what we're seeing on the field. But for those stats to work, you need a large sample. Luckily for us stats nerds, 162 games is a nice enough sample to tell us definitively that the Oakland A's are better than the Houston Astros (side note- this is why I have so many problems with year-to-year predictive stats in the NFL. 16 games isn't nearly enough to glean anything meaningful). But unluckily for us, the playoffs aren't 162 games. Starting Friday night, we square off against the Tigers for 5 games, not 162. What are the chances that the team that's not better wins?

The answer: pretty good. Let's do a thought experiment. Let's say that instead of playing the Tigers in this first-round series, the A's were playing the Astros. To make things simple, let's assume that the A's would beat the Astros in any given game 78.9% of the time (the A's had a .789 win percentage against the Astros this year). I used a binomial distribution to calculate the odds of the A's winning at least 3 games out of the 5 (if you want to learn more about the methods behind the binomial distribution, check out this website, which will provide a simple calculator if you want to do your own calculations, as well as a simple explanation of how it works).

If the A's played the Astros in a 5 game series, the odds of them winning 3 or more games (and winning the series) would be 93.3%. Pretty good!

But let's take it the next step. What if they played a team with the same talent level as the Astros in the ALCS? Again, I used a binomial distribution to figure out the odds of winning at least 4 games in 7 tries. The odds of beating the Astros in a 7 game series are actually even higher at 96.0%.

But let's say the A's played 3 consecutive series (one 5-game series, the ALDS, and 2 7-game series, the ALCS and the World Series) against teams with the talent level of the Astros. Their odds of winning the World Series would be (93.3% x 96.0% x 96.0%), or 86%. That sounds pretty good, but when you think about it, that's terrifying. Even if you have a talent gap as big as the one bridging the A's and the Astros, you still have a 14% chance of being eliminated.

That thought becomes much, much more terrifying when you consider that the teams the A's will actually be facing are much, much better than the Astros.

The A's win percentages against fellow AL playoff teams this year are as follows:

Tigers: .571

Red Sox: .500

Rays: .500

Indians: .286

If we assume that these win percentages are the A's true talent level against these teams (which I realize is an outrageous assumption considering it's all of a 6 game sample against each team, but this is my thought experiment, darn it, and I'll make the rules), then the A's have a slight edge against the Tigers and each game is basically a coin flip with the Rays or the Sox. So assuming the A's get the easiest path to the World Series, what's their probability of making it based on these percentages?

First, the ALDS. Assuming the A's have a 57.1% chance of winning each game against the Tigers, we can use the binomial distribution again to figure out their chances of winning 3 out of 5. In this case, it's obviously much lower than the theoretical 5-game series against the Astros. The chances of the A's winning at least 3 out of 5 against the Tigers is 63.1% based on these assumptions. Pretty good, but also a substantial chance of losing.

Then, if the A's were to make the ALCS against either the Rays or Sox, each game is basically a coin flip. So assuming a 50% chance to win each game, what are the chances of winning at least 4 out of 7? According to my calculations of the binomial distribution it's... exactly a 50% shot.

So even giving the generous assumption that the A's would play .571 baseball against the Tigers, the chances of even making the World Series are pretty slim. The chances would be 63.1% (the probability of winning the ALDS) multiplied by 50% (the probability of winning the ALCS), or 31.6% of even making the World Series. Of course, the chances of winning the World Series, assuming again .500 baseball against the NL representative, would be about half of 31.6%, or 15.8%.

The actual chances of winning the World Series in any given year once you've made the divisional round, even if you're the best team in the whole pack, is going to be no more than 20%. The small sample size of the playoffs means that there's no time for chance to regress to the mean. It's insane to think that the best team typically has less than a 1 in 5 shot of winning the whole thing, but that's what the numbers say. In other words, the playoffs are worse than a crapshoot.

Each year that you make the playoffs, you get a lottery ticket: there's just not enough time for the best team to win out. Here's to hoping this is the year that the A's lottery ticket is the one that hits the jackpot.