### Winning the world series in X games

As a SABR member, I got my copy of "The Baseball Research Journal 35" in the mail this week. It's got a few sabermetric articles in it, and I've so far only had a chance to take a look at one of them. It's called "Relative Team Strength in the World Series," by Alexander E. Cassuto and Franklin Lowenthal.

(Disclosure: when articles for the book were being selected, I was asked to briefly give the editor my opinion on some of them, and so I am listed as a "peer reviewer and designated reader." Also, one of my own articles appears in the book.)

Cassuto and Lowenthal (C&L) start by figuring the chance that the World Series will go a certain number of games, assuming that the teams are exactly equal in strength:

A 4-game series has probability .125;

A 5-game series has probability .25;

A 6-game series has probability .3125;

A 7-game series has probability .3125.

Note that the 6- and 7-game chances are equal. That's because any series that goes at least six games will necessarily be 3-2 after the first five. Then, there's a 50% chance it'll go six (if the "3" team wins), and a 50% chance it'll go seven (if the "2" team wins).

C&L then compare the distribution of real-life World Series to the theoretical one:

4 games: 19 actual, 12.1250 predicted;

5 games: 21 actual, 24.2500 predicted;

6 games: 22 actual, 30.3125 predicted;

7 games: 35 actual, 30.3125 predicted.

It does seem like the theoretical model isn't very accurate, but that's probably because teams aren't really equal in strength. Using a Chi-Squared test, the authors found that the "teams are equal" hypothesis can't be rejected at the .05 level, but *can* be rejected at the .10 level.

The authors then turned to the question of just how unequal the teams would have to be to lead to the observed results. They find that if, in every world series, the stronger team overall had a 51.38% chance of winning every individual game, that would account for the results. That's a lot more balanced that I would have expected.

However, even with these results, there are many more game 7s than predicted. Why is that? The authors argue that

"The team that is behind must play to win games at all costs; thus it may change its rotation to start its best pitcher, use its best reliever for more innings than usual, etc., while the team that is ahead will formulate its strategy to win one more game, not necessarily the sixth game."

I wonder if there are more likely explanations, such as home field advantage. The team behind 2-3 is more likely to be at home for game 6. (The authors mention home field advantage, but not in this context.)

C&L also briefly discuss the economics of long and short series. They note that, assuming two equal teams, the average series length is 5.81 games. As the teams become more unequal, the expected length decreases. However, it decreases slowly: if one team is an expected .550 against the other, the expected games becomes 5.78, which is hardly a drop at all. A .600 team vs. a .400 team results in an average 5.7 game series, still a small drop.

In that context, the authors write,

"[Television] contracts that do not consider the probability of less than seven game series are likely to cause marginal economic losses to the networks and windfall gains to MLB."

Which is true, but I doubt that any networks are naive enough to believe that every series runs seven games. And I'm sure they have qualified staff on-hand to do some rough calculations before submitting a bid.

The part of the article I found most interesting was where the authors figure out the level of imbalance that maximizes the chance of the series going X games.

For a 4-game series, you want one team to have a 100% chance of winning each game. For a 7-game series, you want the teams as balanced as possible: 50% each, which minimizes the probability that one of the teams will win earlier.

What about a 5-game series? The authors calculate the optimal level is .789/.211. This makes intuitive sense; in a 5-game series, the winning team will finish with a record of .800, and .789 is close to that. It turns out that if you do have one team being .789, the series will go five games 33.33% of the time.

Which leaves a 6-game series. How should you imbalance the teams to get the best chance of a 6-gamer? The answer surprised me – you might want to think about it before reading on.

Ready?

The answer the authors give is 50%. The chance of a 4-2 series is maximized when both teams are of equal strength.

This can be proven mathematically, but is there a good intuitive explanation? The only one I can think of is that a six-game series must start out 3-2. The best chance of 3-2 is when the teams are equal. But then you still only have a 50% chance of the "3" team winning game 6. Suppose you make one team stronger. Then it will be less likely that the series will start 3-2, but more likely that the leading team will win game 6. Off the top of my head, it would seem possible that you could trade a small reduction in 3-2 chance for a larger increase in game-6 wins. But the math says no.

Can anyone come up with an easy intuitive explanation of why two equal teams maximizes the number of 6-game series?

Labels: baseball, competitive balance

## 12 Comments:

From where I am standing the math says the following thing:

1) It only becomes more likely that a series will last 6 games rather than 5 once one team has a probility of winning a game of .640 -- this means that most series will be at least 6 games

2) The break-even point between 6 and 7 game series is about 0.550

So obviously as the length of series increases, to have n-1 games (where n is max # of possible games) the two teams have to be as closly matched as possible.

Anyway, back to Phil's question. I don't think his explanation is too far wrong.

In order for a team to win in 6 it has to be 3-2 at some point. Per the math above to maximise the likelihood of a team going 3-2 then each team should be .500 -- imagine it was a 5 game series -- probability of either team winning should be 50% to have the maximize the chance of going to 5 games (i.e 3-2).

It follows that a 50% team has greater odds of going 4-2 (because this to get to this state it has to be at 3-2 first).

A team that is .510 is (very slightly) less likely to have opportunities to go 4-2 (although it is a very close thing).

All you really care about is how to get into that 3-2 state -- everything else is irrelevant.

You can find the math here

http://tinyurl.com/2lqmrw

That probably isn't quite the non-math explanation you were looking for!

Beamer

John,

Thanks! Did you that spreadsheet just for this?

Yes! I should have better things to do with my time really ...

___________

One reason why it work is because 0.5*0.5 >0.51*0.49

Phil:

When the authors say it takes a .5138 average stronger team to yield these results, which result do they mean? To produce 19.6% 4-game series, it would take an average .650 stronger team (if I'm doing the math right). But of course, that would then mean the number of 7-game series is far too high.

I think the high frequency of BOTH 7- and 4-game series is quite interesting. It's hard to believe the average gap is .650-.350, so perhaps teams "fold" once they are down 0-3?

And I think their go-for-broke theory about 7th games makes sense. Why are you skeptical? (I don't think home/away explains it: if 53% of 3-2 teams play away, they should still win .498).

Guy: .5138 is the percentage at which the Chi-squared value between observed and expected is minimized. They say it's 6.886, which still looks fairly significant (the original chi-squared was 7.338).

I'm skeptical because, well, does shifting starting pitchers really happen a lot (when down 3-2)?

Another explanation is that a team down 2-3 may have used its ace only once, while the other team used its own ace twice. The team down is then more likely using its ace for game 6.

A second explanation is luck. 35-22 (in game 6s) is a very large difference between the two teams. Doesn't seem to me like there's any explanation that could account for *that big* a difference.

The theory that teams down 3-2 put everything into game 6 seems pretty easily testable.... How do they usually fare in game 7? Does anyone have this data?

Also, I think we have to come up with a more sophisticated model here-- obviously your winning percentage changes each day, most notably with the starting pitchers.

All in all though, it seems like the prevalence of 4 game sweeps is very tough to explain. I'd guess a very lopsided World Series would be about 55/45 since you're talking about 2 really good teams here regardless and a .600 winning percentage is excellent.

Phil:

A higher % of elite starting pitchers for the 2-3 team in game 6 is plausible -- would be an enteresting research project. And of course, luck is likely a factor. However, I do think trailing teams may be quicker to go to their bullpen, and more willing to have their best relievers go longer, which may confer an edge.

I wonder if the sequence matters. Is there any difference between series that went 3-1 then 3-2, vs. 2-2 then 3-2. Small sample size of course.....

OK, fair enough about the bullpen. Would indeed make an interesting research project.

Something in this article that caught my attention was that the authors say the "probability that out of the 57 series that lasted more than five games, 35 or more would last the full 7 games was 0.0427". They base this on a normal approximation to a binomial distribution, which in this case would be a B(57,0.5) binomial, since each of these 57 series is equally likely to go 6 or 7 games long. However, when I tried to recreate these results, I get a value of .0559 instead.

Can anyone else verify the results here? Not sure what I am doing wrong.

Yeah, I don't see an easy or intuitive explanation for 50% maximizes the 6 game result. My only point to add here is that any deviation from 50% doesn't help you gain a w in game 6 for the "3" team in the 3-2 standings as much as you'd think, because there is a pretty good likelihood that the "wrong" team is holding 3 w's already... So the cost of reducing prob of 3-2 position is greater than the small gain of the 3 team being the team with 50+% game probability.

On a different note, a ran a quick and dirty monte carlo (200k trials each time) trying to see effects of HFA on the game 6 vs. game 7 distribution. 3 results:

1. If teams are .500/.500 on a neutral field and hfa is a linear 6% benefit to home team, the increased likelihood of 7 games vs. 6 games is +0.7% (due to your note that the game 6 home team is more likely to be down 2-3.

2. If teams are .600/.400 on neutral and the .600 has 4 home games (2-3-2) (using same hfa %), then the 7 games vs 6 game delta is -3.4%

3. If teams are .600/.400 but with the .400 team having 4 home games, the delta is again -3.3%!

I think the driver is that the favorite can override HFA effects.

Neat! Taking 2. and 3. together, the chance of the .400 team taking the series to 7 games barely changes depending which team is at home.

Probably different for hockey, where it's 2-2-1-1-1 instead of 2-3-2. I suppose 2-3-2 was set to deliberately make it as even as possible.

a small correction (after rerunning the #s. in case 3: the .400 team getting the 4 home games, the delta between 6 and 7 games is actually closer to -1.1%. But the principle is the same: when there is a clear favorite, the bias goes to more 6 game series than 7 game series regardless of who gets the 4 home games under general HFA assumptions...

I still have no clue why there are so many 4 game sweeps. (if there was ever anecdotal evidence for "momentum" this is as close as it gets...)

also, i used 6% hfa, when regular season is maybe 4%. i'm assuming that playoffs get an unlift over a typical reg season game (Oct in yankee stadium has a bigger effect than a tue night in June in Tampa Bay, for example...)

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