Monday, August 16, 2010

Equivalent Winning Streaks

Disclaimer: What follows is most certainly a freak show stat. It's a freak show stat that deals with something that isn't very important to begin with.

The longest winning streak in the majors in 2009 was eleven games; both Boston and Colorado put together ten-game skeins. Washington won eight consecutive games at one point. Which streak was the most impressive?

The obvious response is "What, are you nuts? Of course an eleven-game winning streak is more impressive than an eight-game winning streak. Why would you even ask such a question?" Truth be told, this is the response I would probably give if somebody asked me this question

So let's suppose that instead I asked you "Which streak was more likely?" Now that's a question that a sabermetrician or anyone else with an appreciation for probability can embrace. With some simple assumptions (independence of games and a constant team W% regardless of opponent or other factors), it's a very simple computation. Just take the team's W% and raise it to the nth power, where n is the number of games in the streak.

I'm going to assume that each team's 2009 W% is equal to their true, constant W%. Obviously I don't actually believe this, but in this case I'm trying to identify unusual streaks given the way the team actually played. If we believe a team is a .525 in true talent, but they actually played .575 baseball, then we are going to expect some longer winning streaks in retrospect. If we wanted to know the probability of a five-game winning streak in the next five games, then it would definitely be inappropriate to use .575 instead of .525.

That is not quite a satisfactory defense of treating the 2009 W% as the true W%, because if we treat all results as known, then we run into what you might recognize from probability class as an urn problem--for some reason they love to use colored balls in urns as examples in stats texts. If you start by taken a team's record (say 81-81) as known, then if you "draw" one game from that sample, there's a 81/162 chance it was a win. But if you draw a second game, the win probability drops to 80/161. Treating the games as independent is equivalent to replacing the balls that you draw, and it's easier, and if you assume that we know the true quality of the team (which is their actual record) but not their actual record then it's justifiable.

Back to the Red Sox and the Nationals. Boston had an eleven-game winning streak, and they were a .586 team, and thus the probability that they would win eleven in a row (over any particular eleven game segment, not for the season as a whole) was .586^11 = .28%. Washington won eight straight, but with a W% of just .364, there was only a .03% chance of an eight game win streak. Given their overall record, Washington's streak was the most unlikely in the majors in 2010.

If you are asking "Who cares?", you are not alone. This is quite admittedly a freak show stat.

There is a more elegant way to express those probabilities, on a scale which you probably find a lot easier to wrap your mind around than hundredths of a percent. We can express it as a winning percentage of equal length by an average (.500) team. Just ask the question "How many games would a .500 team have to win in a row for there to be a .03% chance of that streak occurring?" If you accept fractional games, then this is a pretty easy math problem:

.5^x = prob(streak)
where prob(streak) is for the observed streak, and is equal to W%^y, where y is the length of the streak (so for the Red Sox, W% = .586, y = 11, and prob(streak) = .586^11 = .0028)
so
log(prob)/log(.5) = x

x is the equivalent winning streak (i.e. equally likely) for a .500 team. For the Red Sox, this comes out to 8.5 games; for the Nationals, 11.7. Boston's winning streak, despite being tied for the longest in MLB by games, ranks only fifth by equivalent winning streak.

The shortest maximum win streak in 2009 was four games by Houston, but the Cardinals had the shortest equivalent winning streak, just 4.2 (a five game streak in reality). The Yankees longest winning streak was nine games, which given their .636 W% ranks seventh-to-last in equivalency (5.9 games).

One can of course also turn this around for losing streaks by simply using the complement of winning percentage (losing percentage, if you will, L/(W + L)) in place of W% in the formulas. The losing percentage of a .500 team is, of course, .500, so the constant does not change.

The longest losing streak in MLB was thirteen games by the Orioles, but the Rays' streak of eleven was the longest in equivalency (besting BAL 11.6 to 9.4). The shortest longest slides were four games for the Brewers and Angels; given the Brewers .494 W%, they have the shortest equivalent streak (3.9 games).

I've attached a table with the actual max win and loss streaks and the corresponding .500 equivalents for each team in 2009. Again, this has absolutely no analytical value and I'm not claiming that it does.



We could generalize this a little more by playing fast and loose with the normal approximation to the binomial distribution. Suppose a .550 team goes 6-14 over a twenty-game span. What is the equivalent performance by a .500 team (the number of games they would win which is equivalent to a .550 team winning less than or equal to < = 6)? I am not a stats professor, so use the approximation with care. There are some rules of thumb out there for when it is acceptable to use the approximation (n > = 20, np and n(1 - p) must both be > = 10, and others), and I can't guarantee you that what I'm doing here is kosher. But it's a freak show stat, remember, so I'll pretend from this point as if there is no issue at all.

Let n be the total number of games in question (here it is 20) and let p be the probability of winning a single game (.550). Then:

mean = np = 20*.55 = 11
std = sqrt(np(1 - p)) = sqrt(20*.55*(1 - .55)) = 2.225
z-score = (Wins - mean - .5)/std = (6 - 11 - .5)/2.225 = -2.47

The minus .5 is the continuity corrector, since we are converting between a discrete (binomial) and continuous (normal) distribution. Basically, we consider x wins to occur on the range from x - .5 to x + .5.

Now we need to convert the z-score to equivalent wins for the .500 team:

z-score = (EqWins - mean' - .5)/std'

mean' = .5n = .5*20 = 10
std' = sqrt(n*.5*(1 - .5)) = sqrt(n*.25) = .5*sqrt(n) = .5*sqrt(20) = 2.236

so

z-score*std' + .5 + mean' = EqWins
-2.47*2.236 + .5 + 10 = 4.98

So the .550 team going 6-14 over a 20 game stretch is *roughly* equivalent to a .500 team going 5-15.

1 comment:

  1. Does anyone think this might be the worst blog name of all time?

    ReplyDelete

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