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I’m anxious to see how it all blooms together.” —Astros General Manager Ed Wade

Among the most enduring spring training clichés is the idea that every team has a shot. The standings are tied in March, with thirty teams sporting .500 records (undefined records, if you want to be a math nerd about it). This idea gets reinforced throughout the season, too, since the odds are always good that at least one unexpected team will end up in the playoffs. With the news that Adam Wainwright is sidelined for the season, these hopes live new life in the NL Central.

But of course we know it’s a big lie, a noble truth we tell our younger selves when the swallows come back to Capistrano. That’s why we have an elaborate system to calculate postseason odds, that’s why we predict winners and losers for the offseason, and that’s why the over/unders in Vegas aren’t all 81. But just how untrue is it that teams get a new chance and that there is always hope in next year?

Well, one way to look at it would be to compare last year’s actual playoff teams with our best estimate of 2011’s group of playoff teams.

 

Division

2010 Actual

2011 PS Odds

NL East

Phillies

Phillies

NL Central

Reds

Cardinals

NL West

Giants

Giants

NL Wild Card

Braves

Dodgers

AL East

Rays

Red Sox

AL Central

Twins

Twins

AL West

Rangers

Rangers

AL Wild Card

Yankees

Yankees

Note that I give the Dodgers the wild card because they are the team with the higher average win total in the Monte Carlo simulation. The Braves are very close and may in fact have an advantage for the NL wild card. Viewed this way, there doesn’t seem to be much hope—at least not for bottom feeders. Of last year’s eight playoff teams, five are projected to repeat the feat.

But there is another way of looking at this question: we can also ask how much better this year’s teams are expected to be than they were last year. We could do this in a couple of different ways. We might want to know the expected absolute improvement in the win column for each team. This is pretty straightforward. For example, here are the top gainers if we look at things that way:

 

Team

2010 Wins

2011 Projected Wins

Orioles

66

81

Pirates

57

71

Diamondbacks

65

74

Mariners

61

70

Brewers

77

85

This isn’t quite right, though, since all we’re really seeing here is the teams that were the lousiest in 2010. Put another way, the losers by this methodology are the Rays, the Twins, the Reds, the Padres, and the Blue Jays. That’s three playoff teams and two pleasant surprises. And does anyone really think the Pirates are going to be baseball’s second most improved team? There may be nowhere to go but up from 57 wins, but a Cinderella season is hard to envision. (Who’s the ace? Paul Maholm? James McDonald? Their rotation has literally been subsidized by the U.S. Department of Agriculture.)

Wouldn’t it be better if we could first correct for all the noise that makes bad teams likely candidates for a dead cat bounce? Of course it would be! That’s just what third-order winning percentage allows us to do. We’ll just take last year’s raw (i.e. unregressed) third-order wins from last year’s adjusted standings. Remember, third-order wins are calculated by taking aggregate batting and pitching statistics, adjusting for opponent strength, and then running them through Pythagenpat, thus eliminating most random fluctuations that aren’t included in individual batting and pitching lines (although some noise, like that introduced by the rate of hits on balls in play, remains). Comparing these more refined win percentages against the depth- chart-driven predictions yields a more interesting picture:

 

Team

2010 Third-Order Wins

2011 Projected Wins

Pirates

54.8

71

Orioles

70.2

81

Brewers

76

85

Cubs

73.3

80

Marlins

79.5

84

Argh, those pesky Pirates! It turns out that they were even worse than their record last year indicated. And sure, they have a young core of position players. The bullpen is better and more flexible than it has been in years past, and they have solid options at every position except perhaps shortstop (maybe they should try Ronny Cedeno as a starter?). So perhaps the Pirates have some hope.

What about the Orioles? They were a popular choice to improve last year, too, and look how that turned out. Aha, you say, but what about the Buck Showalter magic? Surely they won’t start the season 15-41 again! And what about potentially promising sophomore showings from Brian Matusz and Jake Arrieta? What about a new starter at more than half of the infield positions—Mark Reynolds at third, Derrek Lee at first, and J.J. Hardy at short? And what about the addition of Vladimir Guerrero?

The Brewers' hope is clear and manifest in their brand new one-two punch of Zack Greinke and Shaun Marcum. Perhaps more than any other team, the Brewers represent neither the hope of next year (the kind that sets in when teams fall out of contention) nor the reminiscing of last year (what was worth enjoying about the Brewers last year?), but the anticipation of this year. Greinke, Gallardo, Marcum; Braun, Fielder, . . . Betancourt?

It wouldn’t be hope if it were a sure thing.

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EnderCN
2/28
Reds are #3 in 2011 PS odds, why did you pick them as the playoff team in the NL Central?
TheRedsMan
2/28
Probably the same reason he has the Cardinals as the NL Central's 2010 playoff representatives.
kmbart
2/28
The Cardinals show as the 2010 division winner and I'm pretty sure the Reds were the champs last year, so it would appear that the columns were switched for this row.
willsharp
2/28
2010 Actual NL Central should be Reds, right?
pikapp383
2/28
The Cards won the NL Central last year? I must have missed all the playoff games on TV.
kmbart
2/28
Well, the Cards and the Reds had the same number of hits in the first game of the 2010 NLDS...
tbsmkdn
2/28
Sorry folks, Cards and Reds got switched around in the uploading. Should be fixed now. Thanks to all for the good catch.
chunkstyle
3/03
Well, if you use 1/infinity as your value for zero, then every team is has a winning percentage of 100%.... (1/infinity)/(1/infinity) = infinity/infinity = 1 Anyone care to out-nerd that?
matuszek
3/05
Who told you infinity divided by infinity was 1? It's undefined, same as zero divided by zero. Actually, what you have there is a nice little demonstration that if 0/0 is undefined (or NaN, not a number, in IEEE terms), then inf/inf must also be NaN, and vice versa.