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The Myth Of The 9 Hattebergs

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When the A’s unveiled the shocking news that a lineup of 9 Scott Hattebergs would outscore the 1927 Yankees, cure leprosy, feed the world’s homeless, and likely make Scott Hatteberg rather wealthy if not a bit tired, this was a revolutionary concept. Hatteberg was not a household name, and yet due to his above-average ability not to make outs he became the poster boy for an idea that was truly ground-breaking. And not really true.

For the purposes of this post, a “Scott Hatteberg” is a .280 hitter with a .374 OBP and .433 slugging percentage, which is what Hatteberg produced in the 2002 season Moneyball references to claim a lineup of 9 Hattebergs would have outscored every team in the league that year.

It’s a cute concept, but here are the problems, as I see it, with the premise—and with the A’s offense, even today:

  • 9 Hattebergs will go station-to-station most of the time, meaning that in any inning in which you do not get an extra-base hit you need three successful at-bats, before three outs, in order to score. But wait, they can’t come in just any sequence, because while BB, BB, single will probably score a run, BB, single, BB won’t. You actually need three successful at-bats in which the last one is a hit, and if you do the math you’ll find that .280 hitters who make outs 62.6% of the time will get a lot of runners on base, and will leave a lot of runners on base, but will not score that many runs. Sound familiar?

Now make Hatteberg fast, with the same stats, and you’re in business, because through steals and first-to-third jaunts you significantly increase the chance of scoring with just two “non out” at-bats, which dramatically increases your chances of scoring (the odds of two BBs or hits in an inning are far more than the odds of three BBs or hits in an inning —approximately twice as good, if my computations are correct).

  • If your lineup is full of base-stealers, what happens when Mark Buehrle and his terrific pick-off move are on the mound? If your lineup is full of HR hitters, what happens when Brandon Webb and his power-sinker are on the mound?

A lineup of 9 Hattebergs will excel against “Hatteberg-friendly” pitchers (pitchers with spotty command, etc.), but will struggle against all “Hatteberg-unfriendly” pitchers, because a lineup of 9 Hattebergs has no Plan B to turn to against a Carlos Silva, who throws a ton of strikes, or a Derek Lowe, who doesn’t give up a lot of hits or walks but also cannot shut down the running game. This would lead to the team scoring a ton of runs occasionally but also getting shut down far too often. Sound familiar?

See, when your lineup includes a high-average hitter, a power hitter, a good contact hitter, a base-stealer, a patient hitter, and so on, you have a variety of weapons against a variety of pitchers, something a lineup of 9 clones can never provide.

This brings to me the 2007 Oakland A’s, whose lineup is eerily close to 9 Hattebergs. They have a Hatteberg protégé (Johnson), a hitter having a very Hatteberg-like season (Nick Swisher), a hitter who is Hatteberg with power (Cust), hitter who is Hatteberg with average (Stewart), a hitter who has regressed into a poor man’s Hatteberg (Chavez), a hitter (Ellis) whose average and power remind one of…Hatteberg…

Obviously, one of the 2007 A’s problems is that they aren’t hitting .280 as a team, aren’t reaching base at a .374 clip, and aren’t slugging .433. But I think that even if they were, they would find that without speed, without balance, without versatility, they would do a lot of what they’re doing a lot of now: getting guys on and getting guys over--but not getting them in.