May 02, 2006

Implicature Watch

An article about the Yankees-Red Sox game contains the following statement:

The rule of thumb if you don't have a calculator, which I learned many years ago from a wonderful gentleman named Bob Fishel, a Yankees public relations director: If the records are over .500 and one team has one more victory and one more loss, that team is almost always behind the other team in the standings. If the records are under .500, the team with one more victory and one more loss would generally be ahead.

I quibble with the words in boldface. I was under the impression that, if 1 ≥ x/y > .500, x/y > (x+1)/(y+2), necessarily. And if 0 < x/y < .500, x/y < (x+1)/(y+2), necessarily. Unless we allow an infinite number of games, but last I checked even baseball does not have an infinitely long season.

The first paragraph of this article does have a correct implicature, however:

Larry Lucchino, the chief executive, graduated from Taylor Allderdice High School in Pittsburgh, Princeton, and Yale law school, so he has to be a pretty intelligent guy, too.

Well, I don't know about those other two schools, but the first implicature is right.
[UPDATE: Or maybe not. Stupid algebra error fixed.]

Posted by Matt Weiner at May 2, 2006 12:11 PM
Comments

It's "almost" because the teams could be in different divisions. So if the Yankees are 13-11 and leading their division, and Atlanta is 12-10 but the Mets are 15-7, Atlanta has a better winning percentage than the Yankees but is not ahead of them in the standings. (After all, if the season ended right away, the Yankees (boo) would make the playoffs and Atlanta (boo) wouldn't, unless they got a wildcard.)

It is easy to use this theorem to prove that the NFL wild-card system is undecidable (in the Turing halting-problem sense), but the proof is too short to fit in this margin.

Another way of looking at your algebra is that the team that has played more games has the same record plus a bit of .500 mixed in - in the example above, it's as if the Yankees went 12-10 and then added a 1-1, which brings their record down a bit. The reverse holds for teams under .500.

Posted by: Ben at May 2, 2006 06:49 PM

It's "almost" because the teams could be in different divisions.

That can't be what's meant, because then 5/6 of the time neither team would be ahead of the other in the standings. They'd be in different columns. (I'm cheating, because the original article refers to the "Games Behind" column, which only applies to teams in the same division or to the wild card race.)

I mean, we don't say that the Pirates are ahead of the Marlins in the standings -- we just try to look away. Actually, AOTW if they were in the same division the Pirates would be behind in the games-behind column but ahead in the winning-percentage column; they've played 5 extra games (for some reason), they've won two more and lost three more, but since both teams are below .300 five extra games of .400 ball improves the winning percentage.

Oh my. How 'bout them SUPER BOWL CHAMPION Steelers? NFL Tiebreakers, of course.

Posted by: Matt Weiner at May 2, 2006 07:42 PM

It's only 5/6 of the time if you assume that teams to be compared are drawn uniformly randomly from all teams. Which clearly they aren't, since most comparisons are in the context of playoff races. How often does one talk about how many games Boston is ahead of Atlanta? Almost never unless it's a question of playoff home-field advantage. See, even talking about baseball (BOR-ing) is really a question of prior probabilities.

Posted by: Ben at May 2, 2006 09:19 PM

In your formulae, shouldn't (y+2) be (y+1)?

Posted by: My Alter Ego at May 3, 2006 12:29 PM

No, that was the stupid algebra error; the numerator is wins and the denominator is total games played, and the other team has played two more games and won one.

(Ben: lalalala I can't hear you.)

Posted by: Matt Weiner at May 3, 2006 12:48 PM

My sources tell me that the sportswriter, Murray Chass, also went to Allderdice.

Posted by: Matt Weiner at May 3, 2006 12:51 PM

There's still a false implicature, because it's said to be "the rule of thumb if you don't have a calculator" not "the rule of thumb if you don't know whether the two teams are in the same division".

Posted by: Allan Hazlett at May 3, 2006 01:50 PM

Is it a rule of thumb if you don't have a calculator, but no longer a rule of thumb if you do have a calculator?

What if you don't use the calculator, but are just carrying it in case you're accosted by a gang of bloodthirsty baseball statisticians?

Posted by: Ben at May 3, 2006 11:45 PM

"If you don't have a calculator" could be a biscuit conditional. You know how I love biscuit conditionals.

Posted by: Matt Weiner at May 4, 2006 10:04 PM

I thought it was "Allerdice". Doesn't matter we pwned them anyway.

Posted by: Cala at May 4, 2006 11:07 PM

At what, missy?

Posted by: Matt Weiner at May 5, 2006 08:19 PM

MATH TEAM SMACKDOWN!!!

Posted by: Ben at May 7, 2006 11:19 PM