## Tuesday, September 22, 2009

### More Mundane Comments on the Playoff Structure

In the previous post I briefly mentioned my dislike of the five-game series format currently used in the Division Series and formerly used in the LCS. But what is the real difference between a five and seven game series? If we make some simple assumptions about team quality, how often will the better team win a series of X length? Common sense tells us that the longer the series, the more likely the better team will win, but let's attempt to quantify that. (Actually, it's not attempting, since given the assumptions that I will make, the answers are simple probability, and it's also tough to classify it as an "attempt" since many people have done it before).

* each game result is independent of the other games in the series (this assumption is likely weaker for the post-season than for the regular season, as the series status has a great influence on how the manager approaches the game, particularly with regards to pitcher usage).

* there is no home field advantage

* the probability of a win for the teams is the same from game-to-game--we are not making any allowances for the aforementioned home field advantage, the identity of the starting pitcher, etc.

With these assumptions in place, we can use the binomial and geometric distributions and the principles behind them to crudely model series of X length. Throughout the rest of the piece, I will refer to "better" or "correct" outcomes. Please understand that I am using these terms in conjunction with the stated assumptions--we know the precise probability of each team winning, and therefore we absolutely know which team is better and ideally will win the series. Obviously, in real life situations we do not know with certainty which team is better. Which is the point--if a playoff format does a poor job of rewarding the better team when we are certain about its identity, it will be even less efficient at that task when we don't know which team is better.

First, let's look at the probability of a team winning the series, given that it is of X length. This can be done with the binomial distribution. For example, for a seven-game series, we simply add up the probability that a given team will win all seven games (even though they will not all be played), six out of seven, five out of seven, and four out of seven. This is the probability that they will win the series.

I will present the probabilities for each interval of .01 in W% between .51 and .65. I have limited the range because realistically in playoff series we will rarely see matchups in which one team is a heavy favorite over the other. The most unbalanced realistic playoff matchup would pit a .700 team against a .500 team, with an expected W% of .700. And that is assuming that the team's sample W%s are their true talent W%s, which would be unlikely for a .700 team. Again, these W%s are the expectations for a single game between the two teams.

I figured the probabilities for series ranging in length from one to fifteen games. I went up to fifteen games because fifteen games was the actual length of the World's Series in 1887, even if the series was not treated with the full championship reverence of today's World Series:

I bolded the 53% line because I'm going to use it as the "average" playoff series--I realize this table is tough to read with fifteen different scenarios. The explanation for why I chose that particular W% is explained below--it's not profound by any stretch (*).

One takeaway from this chart is how silly it is when folks talk about locks to win a playoff series. Even in a situation in which one team has a 65% chance to win each game (which is a big mismatch in the playoffs--a .500 team against a 105 win team or a 90 win team against a 112 win team using Log5), that team only has an 80% shot at winning a seven-game series. Even if you more than double the series length to fifteen games, there's still an 11.3% chance of an upset.

When sabermetrically-inclined people say that the playoffs are a crapshoot, this is the kind of thing they're generally talking about. It's not that you have no way of knowing which team is better or estimating the degree to which they are, it's just that even in a case where you have clear superiority, the short length of the series makes an upset quite feasible.

It was quite amusing during the Roy Halladay sweepstakes to hear commentators talk about how the Phillies were a lock to win the pennant if they got Halladay. Just like it was amusing to read about how the Cubs were going to march right through the weak NL to the pennant last year, or how the Tigers were going to trounce the Cardinals in the World Series. I wish I knew one-twentieth as much about baseball as those folks think they know.

Let's express that table in a more useful form by showing the marginal probabilities for each extension of series length. For example, the team that wins 51% of their games will win a one-game playoff 51% of the time. Expanding to a three game playoff will lead to them winning 51.5% of the time, an increase of .5%. If we expand to a five game playoff, they will win 51.9% of the time, an additional increase of .4%. This will enable us to see the benefit to lengthening series in terms of ensuring the better team wins:

As you can see, the added benefit starts diminishing quickly and for the normal range, essentially levels out after you make the move to seven games. Of course, these are the marginal outcomes, so longer series are still "better"...but less so with every additional pair of games.

Since the marginal benefit levels off after lengthening to seven games (for the nearly even matchups at least--the more lopsided matchups continue to show significant increases), it seems like as good of a point as any at which to stop.

Of course, I have approached this solely from the perspective of encouraging correct outcomes. This is not the goal of a league--if it was, there would be no need for any kind of playoffs at all. The league is going to act in a way so as to maximize its profits. Which is well and good, but I am examining this from the personal perspective of what I'd like to see and/or what will produce the best outcomes.

There is one thing that overlaps between my perspective and the economic interests of the owners, and that is the desire for a competitive series. Close series encourage higher ratings, and longer series means more ticket revenue. For a fan, there's nothing more exciting that a decisive game for the world championship after a hard fought series. While I have a strong preference for better outcomes, I can't completely suppress the desire for a winner-take-all finale.

So, given the underlying assumptions of this post, let's look at the probabilities of a decisive game, given a series of X length. This is done with the geometric distribution, and I have included the formula (**) because I think many fewer people are familiar with it than the binomial distribution--just speaking for myself, I know the binomial function by heart but have to look up the geometric function just to be safe:

A five-game series with a fairly normal matchup will produce a game five about 37% of the time; a seven-game series about 30% of the time. So for a roughly 1% increase in the likelihood of the better team winning, you give up decisive games in 7% of your series.

The next step, moving from a seven-game series to a nine-game series, would result in roughly the same increase in the likelihood of the better team winning while sacrificing another 4% of series without a grand finale.

All told, it shouldn't be too surprising that the probabilities here can be read to suggest that MLB has correctly identified the series lengths that provide the best combination of practicality, uncertainty of outcome, and producing desired outcomes. The extra benefit in terms of desired outcomes from expanding to longer series is relatively small, and is offset by a larger percentage drop in the expected proportion of series with decisive games.

Finally, let's take a look at the potential value of home field advantage in a five-game series. I previously looked at World Series HFA (i.e. seven-game series), and the same principles will apply here. I have not looked at the empirical data in this case and will only be discussing theoretical results.

First, we can use the geometric distribution to calculate the percentage of series that are expected to go X games, assuming that each game is a 50/50 proposition (in other words, not considering HFA):

Just as is the case for a seven-game series, the probability of a full-length series and one short of it are equal. This makes logical sense, of course; in order to create this situation the first three games must have produced a 2-1 series. There is a 50% chance that the team that has already won two wins, ending the series and a 50% chance that the team behind forces a decisive game.

Unlike a seven-game series, it is impossible for the team with on-paper home field advantage to play more road games than home games, as the format is 2-2-1 (Obviously, I'm talking about the current format; I'm aware that it was sometimes different in the past). Theoretically, on-paper home field advantage results in a true home field advantage 62.5% time, and the other 37.5% of the time there is no HFA for either team.

In order to add HFA into the mix, we need to identify all the possible series sequences, which I will not reproduce here. Suffice it to say that from the perspective of the winning team, there is one series sequence that produces a three-game series (WWW), three that produce a four-game series (WWLW, WLWW, and LWWW), and six that produce a five-game series.

I will assume a home field W% of .573, which is the empirical World Series statistic. I believe that the "true" parameter is likely lower, for reasons discussed in the earlier post, but I'll use the sample statistic for the sake of discussion. Retaining the assumptions of evenly matched teams and independent game outcomes, the probability of the team with on-paper HFA winning a five-game series is 52.66%, compared to a 52.31% chance in a seven-game series. So HFA is theoretically more important in a shorter series (no surprise, but we've estimated the degree).

It should also be noted that we would expect the empirical home field advantage in a five-game series to be even stronger because in those series, the on-paper advantage usually goes to the team with the better record. The same applies to LCS games, but not to the World Series as on-paper home field advantage is chosen without regard to the specific teams competing.

That's it, except for the asterisked digressions.

(*) There is no particularly compelling reason to use 53% as a default W%; I just wanted a line that you could focus on that was reasonably telling, because the whole table is a bit much.

Anyway, I chose 53% because in the World Series (for 1923-2008 with a few years excluded), empirically the mean W% of the team with the better record has been .635, and the team with the lesser record has a mean of .594. Regressing 30% to .500, this results in .595 and .566. Log5 tells us that a .595 team should beat a .566 team 53% of the time. And there you are.

(**) The geometric distribution gives the percentage of time a certain number of failures (x) occur before a certain number of successes (r) occur for a binomial process. In the case of a baseball series, r is the number of wins for the victor in a series (3 for a five-game series, 4 for a seven-game series, etc.), x is the number of wins for the series loser (in a five-game series with a decisive games, x = 2; for a seven-game series with a decisive game, x = 3). We also need to know the probability of a success (P), and calculate the number of combinations using the combination function C(x + r - 1, x).

To find the probability of a decisive game for the series as a whole (with either team winning), we need to do a calculation for each team, which is why the results are summed below--one for the winner and one for the loser. Let G be the number of games in a full length series, W the number of wins for the winning team in such a series, L the number of losses for the losing team in such a series, and P the probability of one of the team winning an individual game. Then the probability of a decisive game is:

C(G - 1, L)*P^W*(1 - P)^L + C(G - 1, L)*P^L*(1 - P)^W

For example, the probability of a seventh game in a series (G = 7) in which one team has a 55% chance of winning each game (P = .55) is (W is 4 and L is 3, of course):

C(6, 3)*(1-.55)^3*.55^4 + C(6,3)*.55^3*(1-.55)^3 = 30%

One other thing to note is the expected number of games in a series. This is found by taking summing G*P(G) for all possible series outcomes. So in a five game series, the expected number of games is:

3*P(3 games) + 4*P(4 games) + 5*P(5 games)

Reverting to the assumption that each game is a 50/50 proposition, the expected number of games in a five-game series is 4.125. The expected number of games in a seven-game series is 5.8125. You can see that there are diminishing returns going on; despite lengthening the possible length of the series by two games, our expectation is that the actual number of games will only increase by 1.6875 games.

The probability of a decisive game hints at this as well, but this is another way you could attempt to quantify the real observed benefit of lengthening a series.