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?