## 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.

## Monday, September 21, 2009

### I'll Probably Regret This, But What the Heck?

I have always vowed never to be the old fogey who doesn't know anything about new technology, and yells "get off my lawn", and talks about how he walked eight miles to school uphill...well, you get the idea, and I'm sure you've vowed the same thing. It's an easy thing to say when you're young or middle aged, and probably a lot harder to actually do when you're older.

In any event, I'm really not old enough to worry about it yet, but just in case, I decided to start a Twitter. It's not going to be very interesting, and so far I've just been using it to snark, which is not the nicest side of me or anyone else. But it's nice to have an outlet for quick thoughts and non-baseball things (although a lot of my comments--I will not give and in and call them tweets--will be about baseball). Any real sabermetric content will continue to be posted here, so even if for some reason you find this worth reading, you won't be missing anything by not reading the twitter feed.

## Friday, September 18, 2009

### Larry Corcoran's New Tombstone

There is a nice story from the New Jersey Star-Ledger about some firefighters who have raised money to erect a headstone for Larry Corcoran. Corcoran, for those of you who don't know much about nineteenth century and haven't read any of my 1876-1881 NL series, was an outstanding pitcher for the Chicago White Stockings (today's Cubs). He was the first pitcher to toss three no-hitters, the ace of Chicago's 1880-82 pennant winning teams, and IMO the most valuable pitcher in the NL in 1880.

This is certainly a nice gesture, and it's always nice to see overlooked nineteenth century ballplayers get their due. However, they may have wanted to run the design by someone before going ahead with it...

h/t: SavoyBG at Baseball Fever

## Tuesday, September 15, 2009

### Mundane Comments on the Playoff Structure

Believe it or not, this is a post filled with opinions. The underlying assumption is that if you are not interested in my opinion, you will not read, and so I’m not going to go around and apologize for giving my opinion in what follows.

The current MLB playoff system is bizarre from a certain standpoint, as the number of teams vary from division to division, even within the same league. It is apparent that teams from different divisions do not have equal likelihoods of making the playoffs (and therefore winning the pennant or the World Series) due to this structure, even if one assumes that all teams are equally likely to win each game.

In fact, if you grant me another assumption, that there is no home field advantage (I would add a balanced schedule, but since we have assumed that all teams are equal, we really don’t care--your opponents are .500 even if you play the Dodgers in all 162 games) in the playoffs, it is relatively simple to calculate the chances of a team from each division reaching certain goals.

Using the AL East as the example, there are five teams in the division; each team therefore has a 1/5 chance of winning the division title. 4/5 of the time, they will not win the division and thus will be in competition for the wildcard. There are fourteen teams in the American League, three of which will win division titles, so they are competing against eleven teams for the wildcard, and are expected to win with probability 1/11. The probability of making the playoffs is straightforward:

P(Playoffs|AL East) = (1/5) + (4/5)*(1/11) = 7/22

Since we have assumed equality amongst team and no home field advantage in the playoffs, the probability of winning the pennant is simply:

P(Pennant|AL East) = P(Playoffs|AL East)*(1/4) = (7/22)*(1/4) = 7/88

And the probability of winning the World Series is just as simple:

P(World Series|AL East) = P(Pennant|AL East)*(1/2) = (7/88)*(1/2) = 7/176

The probabilities for the other divisions can be figured similarly, and converting to four place decimals (overkill, I know, but I feel bad about killing all the nice fractions), this is what you get:

I realize that I am not telling you anything you couldn’t figure out for yourself.

One thing to note here is that the difference between playing in the AL and the NL is not too much of an issue as long as the divisions are of equal size. An AL East team can expect an extra playoff appearance every 89 years and another world title every 714 years compared to its NL East counterpart. This isn’t really worth worrying about IMO, and since the AL was at a similar disadvantage between 1977 and 1992, it only seems fair in a way.

The real issue lies in the divisions of different sizes, not the fact that the leagues are different sizes. An AL West team can expect to make the playoffs 32 times in a century, while an AL East or Central team will make the playoffs 27 times. Of course, there are so many other factors at play (after all, teams are not perfectly balanced as we have assumed) that you may not find this particularly troublesome.

The gap between the two extreme divisions, the AL West and the NL Central, is pretty extreme. Nine extra playoff trips and one extra world title per century seems inherently problematic for my money.

It would be nice if you could balance the two issues (more teams in the division making a division win less likely, more teams in the league making a wildcard less likely) by putting the larger division in the smaller league, but of course it is only because the league is larger that a division must be larger.

Assuming that expansion to thirty-two teams is inevitable (and I believe it is, although any guess on the timeframe would be me blowing smoke), and that the two league structure will endure (no need to start a civil war here), what should the divisions look like when it happens? As a side note, it certainly appears as if there will be no expansion prior to 2015, which will make the period from 1998-whenever the longest period of non-expansion in the expansion era (surpassing 1977-1993).

What I would suggest is two eight-team divisions in each league, with the wildcards being the two non-winners with the best record. In fact, I would like to see such a structure right now, with the difference being that the AL divisions would each have seven teams. Larger divisions are my preference as they increase the probability of the best teams making the playoffs.

However, the entirety of American pro sports history shows a trend towards smaller divisions over time. I did not check thoroughly, but I do not believe that any of the four major leagues have ever reduced the number of divisions. Certainly baseball has not, going from no division to two divisions in 1969 to three divisions in 1994. Football has gone from two to three to four, and hopefully will serve as a cautionary tale for baseball.

After all, in the NFL, it is relatively common for 8-8 teams to win one of the four-team divisions; in 2008, the Cardinals and the Chargers both pulled this off, and while it was unusual to have two such cases in one season, individual occurrences are not particularly uncommon.

For the NFL, the issue of having two good teams in one of the divisions while a mediocre team wins another division is blunted by the presence of two wildcards. But still, something just feels wrong about having the 11-5 Patriots on the outside looking in while the Chargers are in. Any divisional format will occasionally result in teams with better records being left out, but the tiny divisions increase the odds.

For baseball, unless the playoffs were expanded, you don’t have a lot of choices. You can have:

1) 4 4-team divisions, no wildcard
2) the NL model of 2 5-team divisions plus a 6-team division, 1 wildcard
3) 2 8-team divisions, two wildcards
4) one big league

While this is certainly a matter of personal taste, option four is the clear winner to me, while option three is the most attractive that would have a prayer of being implemented.

As an aside, I find the phenomenon of increasing numbers of divisions as time has marched on somewhat puzzling. Travel times and expenses should have made smaller divisions and unbalanced schedules most appealing early in the history of leagues, not now. The universality and as of yet unrelenting nature of the trend could be explained in a few ways. The most generous explanation is that fans like the excitement generated by more, smaller divisions, and the leagues are just giving their customers what they want. I find this preference odd, as an 8-8 team winning a four-team division does not seem any more exciting to me than a 9-7 team beating out another 9-7 team for a wildcard spot. But it may well be there the cachet that goes with winning a division as opposed to winning a wildcard justifies the system economically.

It may be so, but the reasoning behind it is completely foreign to me. In the small division race, you have to beat out just three other teams. In the wildcard race, you have to beat out thirteen other teams (admitted they are all teams with lesser records than at least one other team in the league and possibly as many as three). Still, I’m not impressed by being a division champion if all that entails is beating out three other teams.

## Monday, September 07, 2009

### Magic Number in a Slightly Different Form

As you know, the magic number is the combined total of wins and opponent losses necessary for a team to clinch a playoff spot. If we let g = number of games in season, then the magic number (M#) can be calculated as:

M# = g + 1 - W - oL

where "o" denotes opponent.

Of course, there's a big difference between having a M# of 5 with 10 games to play and having a M# of 5 with 20 games to play. We could look at what I will call Magic Percentage (M%)--the percentage of game outcomes that must go a team's way in order for them to clinch. In this case, game outcomes include both the team in question's games and the games of their opponents.

Suppose that the race between the Alphas and the Bravos is shaping up like this, with ten games left for each team:

Alphas....91-61....599
Bravos....89-63....586

The Alphas' M# is 9. Again, that means that a combination of nine Alpha wins and Bravo losses will clinch the division. Since each team has ten games left, there are twenty total game outcomes outstanding, and 45% of them (9/20) must go the Alphas' way. Therefore, their M% is 45%. Since I always feel compelled to write out a formula, here it is:

M% = (M#)/(2*g - W - L - oW - oL)

Suppose that the race tightens up a bit over the next five games, and with five to play the standings look like this:

Alphas....93-64...592
Bravos....92-65...586

The Alphas have trimmed their M# to five, but their M% has increased to 50, as they now need half of the game outcomes to go their way.

One might be tempted to say that the M% is better than the M#, because it puts the magic number in context...a M# of one with 20 outstanding game outcomes is a sure thing (M% = 5%); with one game left it's a nailbiter (100%). But the M% can be just as misleading--a M# of one with two games left is 50%, but a one game lead after the first day of the season will produce a similar percentage.

So the M% is not a magic bullet by any means. You still need to consider it in conjunction with the number of games remaining, just like you do with the plain old M#. All that I've done is express the number of favorable outcomes required as a rate rather than a total.

The M% only works in relation to one set of leader and pursuer. As I pointed out before when talking about variations on games behind, when multiple teams are in the race, the results of your team are more important than the results of either of the other teams considered independently, because your victories help your position against both of your opponents. So if the Alphas have a two game lead over both the Bravos and the Charlies, one can't look at their M% and state that if X% of game outcomes go their way, they will win. You would need two M%s, one with respect to each opponent.

Through games of Sunday, here are the standings for the top two teams in each division, along with the leader's magic number and magic percentage (the magic percentage for the pursuer is simply the complement of the magic percentage for the leader, and is thus omitted.  This is an embarrassing oversight on my part; they are not complements.  The magic percentage only gives you the percentage of favorable outcomes to win the division outright.  A tie scenario is not taken into account, and thus the M%s for two adversaries do not sum to one):

If you assume (falsely, of course, but it's usually not too large of a distortion at this stage of the season) that both teams have an equal number of games remaining, then one way to look at the M% is that it is half of the minimum W% needed to clinch the playoff spot even if the opponent goes undefeated--or half of the maximum W% your opponent can play, even if your team goes winless.

All of the above is just spitballing. If you are serious about estimating a team's playoffs chances, then playoff odds approach as implemented by Baseball Prospectus and others is the way to go. However, that involves estimating team quality and running thousands upon thousands of simulations. The freak show stat here is intended to be one you can figure out with nothing more than a calculator and your morning paper, err, the standings on Baseball-Reference, and a way to put a little twist on the familiar magic number.

## Tuesday, September 01, 2009

### Yahoo! Math: 1 = 3

Alternatively, why does Yahoo! hate the Mariners?