Baseball records...

[rhu]

So I was wondering idly this morning... The best record that a baseball team could (theoretically) have and not be World Series Champions is 172-4. (Sweep all 162 regular season games, 3 games in the division pennant, 4 in the league championship, the first 3 World Series games, then lose the most heartbreaking four games in history.)

What's the worst record a team could have and still win the World Series? I don't know enough about how the season schedule is constructed to answer the key question of how many intra-division games there are --- they'd have to win exactly half of those plus one, right? That's assuming that every game is played and none are rained out; if we allow for canceled games, how few do they have to win to win their division?

[Edited to hide evidence of my lack of knowledge about baseball]

Comments

[michelel72.livejournal.com]

I can't answer the larger question, but I can contribute a bit.

The AL has 14 teams; the NL has 16. Each league has three divisions. Within each division, teams will usually play each other 18 or 19 times. Teams in other same-league divisions will usually be played 6-10 times. One team in the other league will be played six times; the remaining teams in one division of the other league will be played three times each.

So BOS might play TB and NYY 19 times, BAL and TOR 18 times, ATL six times, the teams of the NL West three times each, SEA 10 times, CWS four times, TEX six times, OAK six times, and most other AL teams three times.

But a team doesn't have to win its division to get to the World Series; it has to win its division or have the best record of the remaining teams. For a "worst winner", you'd probably have to be the best team in the worst division. A division's teams might theoretically have losing records against all other teams in the majors; in that case, two teams might be 10-152 (guessing) with the other two or three in the division being even worse, in which the tiebreaker would be their head-to-head matchups.

Tricky one. Good luck!

[rev69.livejournal.com]

If the three division's leaders in the AL (with 16 teams) went 162-0, and therer were no rainouts so that the total number of fgames won by the other 13 were 162 x 8 - 162 x 3 = 810, and this calculation would be more complicated because of interleague play. If there are 24 interleague games, and every one of the 13 teams lost them all, I'm guessing you can reduce the 810 total games won to (162-24)/162 * 810 = 690 games. 690 games / 13 teams means the average win record of those teams would be 690/13= 53.1-108.9. If all but one of those teams tie at 53-109, then the wild card recipient would be 54-108, a 33% record regular season (the exact opposite of the 1986 Mets 108-54 regular season record but still better than their 1962 debut at 40-120!)

with post season, all series would go the max. 3-2, 4-3, 4-3. Their record would improve to 64-117 (35.4% win record)

This is too much math on a Friday morning for this Mets fan engineer. :)