## Probability of successive wins in baseball

Dan Goldstein did an informal study asking people the following question:

When two baseball teams play each other on two consecutive days, what is the probability that the winner of the first game will be the winner of the second game?

Dan writes:

We asked two colleagues knowledgeable in baseball and the mathematics of forecasting. The answers came in between 65% and 70%.

The true answer [based on Dan’s analysis of a database of baseball games]: 51.3%, a little better than a coin toss.

I have to say, I’m surprised his colleagues gave such extreme guesses. I was guessing something like 50%, myself, based on the following very crude reasoning:

Suppose two unequal teams are playing, and the chance of team A beating team B is 55%. (This seems like a reasonable average of all matchups, which will include some more extreme disparities but also many more equal contests.) Then the chance of the same team winning both games is .55^2 + .45^2 = .505. Even .6^2 + .4^2 is only .52.

Dan and his commenters discuss other factors such as home-field advantage and the quality of the starting pitcher, but I think the above reasoning basically works.

What time is it when your team loses? Time for the coach to get fired.

Dan has another question that I think I have the answer to. He writes that he “has always wondered why teams are so eager to fire their coaches after they lose a few big games. Don’t they realize that their desired state of having won those same few big games would have been mostly due to luck?”

My guess is that these are situations where the management has already decided they want to fire the coach, and they’re just waiting for a convenient time to do it so as not to antagonize the fans.

1. Ken Williams says:

I think you have to consider the Monty Hall effect here, inasmuch as the wording is ambiguous.

Given your example probabilities, if the question is "the winner of game 1 is known, what's the probability of the same team winning game 2?", then the answer is either 55% or 45%.

If the question is "what's the probability that a single team wins both games?", then it's as you've calculated, 50.5%. Which is just a weighted average of the 2 outcomes in case 1.

Personally I think the starting-pitcher-effect is a pretty big deal, so much so that if it's known which team won the first game, then there's a fairly high probability that they had their "good pitcher" in that game (and/or the other team had their "bad pitcher"), taking him out of the rotation for the next game, and flipping the odds backward significantly.

Not sure how to quantify that with probabilities without looking through a lot of data though.

2. zbicyclist says:

My guess was .6, given home field advantage. But I'm in Chicago; what do I know?

"You know the law of averages says:
Anything will happen that can
But the last time the Cubs won a National League pennant
Was the year we dropped the bomb on Japan"

A Dying Cub Fan's Last Request
by Steve Goodman (1983, but still accurate)

3. Michael Lugo says:

I agree with your guess. It probably helps that I'd heard that doubleheaders are slightly more likely to be swept than split. Sean Forman of baseball-reference.com (a former math professor!) looks at the data as well. Over the past thirty years 48.83% of doubleheaders have been split; 48.85% of pairs of games played on back-to-back days have been split. (But I don't know if anybody's split this out based on how mismatched the teams are.)

I think starting-pitcher effects would lower the probability of the same team winning two consecutive games. If the Phillies beat the Qankees today (these are fake team names, chosen to start with P and Q), then it's likely that the Phillies used a better starting pitcher than the Qankees. Now assume that both teams choose their starting pitcher for tomorrow uniformly from among those that didn't start today (which is admittedly not the right model, but I'm assuming that rotations get scrambled enough that it's right for one game). Then the expectation of the skill of the Qankees' pitcher in tomorrow's game is higher than the expectation of the skill of a "generic" Qankees pitcher; for the Phillies substitute "lower" for "higher".

(As the real Phillies have Roy Halladay pitching in the first game of a doubleheader today, my interest here is not entirely academic.)

Home field advantage should work in the opposite direction. (If you won today, you're more likely to be at home than away, and that's where you'll be tomorrow.)

I have no idea, off the top of my head, how to estimate the magnitude of these effects. In particular it's hard to estimate how big the rotation effects would be without knowing a lot more about typical pitcher usage patterns than I have in my head. I'd have to do some calculation, and if I'm going to calculate I should do some real work instead of this.

4. doug says:

With independence, it's fairly difficult for the probability to be far from 1/2, since if your probability of one team in the pair beating the other is p (which has expected value 1/2 by exchangeability) and some standard deviation which is unlikely to be more than .2 which gives E[prob both games won by the same team] = E[p^2 + (1-p)^2] = 1 + 2[V(p) – (Ep)(1-Ep)] = .5 + 2 V(p)

5. afinetheorem says:

Isn't this just the standard "paradox" about convergence? Consider each team as having a density of quality on, say, 0-10, with one team stochastically dominating the other, and let the winner be the team that draws the highest number. Even given the dominance, the loser of the first game is, conditionally, more likely to have drawn a low number from his distribution.

Numerically, let A win 60% of the games. Then 40% of the time, A lost game 1, so will win 60% of game 2's. 60% of the time, B lost game 1, so will win 40% of game 2's. .4*.6+.6*.4=48 percent of the time, the loser wins game 2.

6. Albert Lin says:

What about for basketball? It seems that there's a lot more variability in baseball games. Basketball seems to come to "equilibrium" within a single game (and almost certainly within a 5-game series). By "equilibirum," I mean that the winner is truly the better team.

7. Steve Sailer says:

8. Andrew Gelman says:

Albert: For basketball it will still be close to .5, I think. As noted above, even .6^2 + .4^2 is only .52.

Steve: It's .513; see quote from Dan above.

9. Jonathan Falk says:

JC Bradbury, author of the baseball economist, had a blog entry on this very point this week:
http://www.sabernomics.com/sabernomics/index.php/

10. jonathan says:

I like to use this way of showing the reality, so people get the idea without needing to calculate the exact probability. Say a team loses 91 games. Bad team. Say a team wins 91 games, good team, maybe even a playoff team some years in some divisions. Difference is 20 wins over the entire 162. That means the good team wins 1 more game out of 8 games.

I find that's still too hard for people to understand, because they don't get that the example compares a 71 win team to a 91 win team. They seem to get it a little better if you take the difference between a 91 win team and a .500 team with 81 wins. That's 10 wins over a season, which means the team literally wins one more out of 16 games.

All this stuff means to me – in every sport – is that a few teams are really good and a few are really bad but most are coin tosses.

This mild stuff is, to me, really deep because it gets at the way people see how life works. They look at each game and can see why team A won or lost but they don't get that if you fix the mistake in this game then your team won't be perfect, that the team and the league will over time generate results that are mostly coin flips. We apply this approach to nearly everything: drunk driving accidents always have a cause but the system will generate a certain number of these accidents per mile driven (given the current system constraints, meaning mostly laws and enforcement); that we can find a reason why this long-tail risk happened in financial modeling or oil drilling in the Gulf but that lesson won't / can't protect us from the next long-tail risk.

11. Albert Lin says:

Did a quick count of this year's playoffs so far: 62% (31/50) of games are won by the team that won the previous match-up.

12. Ben Walter says:

From Calculation to Observation: Crack computer jockeys ought to be able to sift out the results of two-game series for as far back as they were accurately recorded – – and tell the rest of us all the empirical results.

13. aram says:

You don't want to average P(strong team wins) and get a number like .55, you want to average P(strong team wins)^2 + P(weak team wins)^2, which generally will give higher numbers. Also, the bias would increase if there were random factors that affected teams' performance on a scale longer than a few days (like injuries). On the other hand, it's hard to argue with the empirical numbers.

14. Andrew Gelman says:

Aram:

As I noted above, .6^2 + .4^2 is only .52, so the general principle applies. With some effort in your modeling, you could perhaps get this up to .55–I wouldn't believe such a number, but you could probably get there–but there's no way to get numbers like the 65%-70% given by Dan's colleagues.