mas@drutx.UUCP (SchwarzMA) (04/21/85)
Question to my fellow Detriot Tiger fans: Will they repeat????? My answer: Of course, the AL East and the rest of the league for that manner has some catch up to do. GO TIGERS Mike Schwarz (formally of Detriot, 1964, now Denver) Take me out to the ballgame, buy me some peanuts and Cracker Jacks .......
roy@hpmtla.UUCP (roy) (04/22/85)
>I don't know if they'll repeat, though they will still be very good >this year. The question that I'm more interested in is, *which* team >from the AL East is going to win the world series this year. Before howling about the AL Easts invincibility, maybe you should A. Check their world series record VS NL east teams. B. Consider the fact that they only have to eliminate the AL West winner to get to the series. (K.C. HA HA HA). C. Put their regular season victory totals into proper perspective by realizing they play exactly one half their games against the AL West. If the Mets or Cubs played half their schedule againts the AL West they'd win 110!
planting@uwvax.UUCP (W. Harry Plantinga) (04/22/85)
> Question to my fellow Detriot Tiger fans: > > Will they repeat????? I don't know if they'll repeat, though they will still be very good this year. The question that I'm more interested in is, *which* team from the AL East is going to win the world series this year.
jlh@hou2e.UUCP (J.HEATWOLE) (05/03/85)
> If the Mets or Cubs played half their schedule againts the AL West > they'd win 110! If the Mets or Cubs played half their games against the AL East they'd be lucky to win 81. Jeff Heatwole ..!hou2e!jlh
david@fisher.UUCP (David Rubin) (05/03/85)
Regardless of the merit of the rest of Roy's argument, AL East teams play MORE than one-half of their games against AL West opponents. That's because (1) the AL doesn't believe in divisional play, and schedules the same number of games in an inter-divisional (e.g. Detroit-Texas) matchup as in an intra-divisional (e.g. Detroit-Milwaukee) one, and (2) Each AL Eastern team has seven AL West opponents, but only six from the East. David Rubin
pete@umcp-cs.UUCP (Pete Cottrell) (05/05/85)
>Regardless of the merit of the rest of Roy's argument, AL East teams >play MORE than one-half of their games against AL West opponents. >That's because > > (1) the AL doesn't believe in divisional play, and schedules > the same number of games in an inter-divisional (e.g. > Detroit-Texas) matchup as in an intra-divisional (e.g. > Detroit-Milwaukee) one It's not a question of believing in divisional play, it's just the way the figures break down; incidentally, there are more games scheduled in an intra-division rivalry than in an inter-division (13 to 12) but the totals favor the inter-division (84 to 78). I agree that there should be more intra- than inter-divisional, but then the numbers get messy; the number of games to be played with the other divsion would not be divisable by 7, so teams would end up playing more games with some teams than with others. I would rather keep things even. If the National League had 14 teams they would be in the same boat; 14 is just an awkward number. -- Call-Me: Pete Cottrell, Univ. of Md. Comp. Sci. Dept. UUCP: {seismo,allegra,brl-bmd}!umcp-cs!pete CSNet: pete@umcp-cs ARPA: pete@maryland
wildbill@ucbvax.ARPA (William J. Laubenheimer) (05/05/85)
> (1) the AL doesn't believe in divisional play, and schedules > the same number of games in an inter-divisional (e.g. > Detroit-Texas) matchup as in an intra-divisional (e.g. > Detroit-Milwaukee) one, and Not quite. You get 13 games with each division opponent and 12 with each non-division opponent. > (2) Each AL Eastern team has seven AL West opponents, but only > six from the East. > > David Rubin Well, in the NL, you have 6 opponents in the other division and only 5 in your own - so what? Actually, the reason this happens is that Major League Baseball has decided that the fairest way to build a schedule is to have each team play a certain fixed number of games against each team in its own division and a different (and smaller) fixed number of games against each team in the other division in its league, and that 162 games should be played. If we use "d" to represent the number of division games and "n" to represent the number of non-division games, we get a Diophantine equation (equation with integer coefficients, only integer solutions desired) for each league: (NL) 5d+6n=162 (AL) 6d+7n=162 The solutions for the NL equation include the current d=18, n=12, which results in a nice schedule where you see division rivals more often, but still have enough competition with non-division opponents. Unfortunately, the only solution anywhere near the "reasonable" range for the AL equation is d=13, n=12. The next solution is d=20, n=6, which begins to look a lot like the current NBA schedule where you only play two games against each team from the other conference. Interestingly enough, if you try a 154-game schedule (the way it used to be) in the AL, the most reasonable solution is d=14, n=10, which also looks pretty good. This has the advantage that you wouldn't need yet another set of asterisks to put in the record book, and you can get a "perfectly balanced" schedule with equal numbers of home and away games against each team. But would the owners ever agree to anything that would reduce their beloved revenue? Naaaah... Bill Laubenheimer ----------------------------------------UC-Berkeley Computer Science ...Killjoy went that-a-way---> ucbvax!wildbill
david@fisher.UUCP (David Rubin) (05/07/85)
>> (1) the AL doesn't believe in divisional play, and schedules >> the same number of games in an inter-divisional (e.g. >> Detroit-Texas) matchup as in an intra-divisional (e.g. >> Detroit-Milwaukee) one, and >Not quite. You get 13 games with each division opponent and 12 with each >non-division opponent. Well, PRACTICALLY the same. >> (2) Each AL Eastern team has seven AL West opponents, but only >> six from the East. >> >> David Rubin > Well, in the NL, you have 6 opponents in the other division and only 5 in > your own - so what? I was permitting the reader to draw his own conclusion, but to clarify my point, I will do the multiplication: NL AL Intradivisional 5*18=90 6*13=78 Interdivisional 6*12=72 7*12=84 >Actually, the reason this happens is that Major League Baseball has decided >that the fairest way to build a schedule is to have each team play a certain >fixed number of games against each team in its own division and a different >(and smaller) fixed number of games against each team in the other division >in its league, and that 162 games should be played. If we use "d" to >represent the number of division games and "n" to represent the number >of non-division games, we get a Diophantine equation (equation with integer >coefficients, only integer solutions desired) for each league: > >(NL) 5d+6n=162 >(AL) 6d+7n=162 > >The solutions for the NL equation include the current d=18, n=12, which >results in a nice schedule where you see division rivals more often, but >still have enough competition with non-division opponents. Unfortunately, >the only solution anywhere near the "reasonable" range for the AL equation >is d=13, n=12. The next solution is d=20, n=6, which begins to look a lot >like the current NBA schedule where you only play two games against each >team from the other conference. > >Interestingly enough, if you try a 154-game schedule (the way it used to be) >in the AL, the most reasonable solution is d=14, n=10, which also looks pretty >good. This has the advantage that you wouldn't need yet another set of >asterisks to put in the record book, and you can get a "perfectly balanced" >schedule with equal numbers of home and away games against each team. >But would the owners ever agree to anything that would reduce their beloved >revenue? Naaaah... > > Bill Laubenheimer >----------------------------------------UC-Berkeley Computer Science > ...Killjoy went that-a-way---> ucbvax!wildbill Three comments to your last few points: (1) The solution d=13, n=12 is not really a solution, as you've dropped the restriction that d and n should be even. (2) Given that the "solution" must be flawed, it would (in my opinion, which every right-thinking individual shares :-)) be better to have intra- or inter- divisional opponents play an equal number of games in the season (the deviance limited to 2) so as to have truly divisiosal play (else why even have divisions?), and (3) 14 teams is a lousy number to have in a 162-game schedule, but as more baseball is a positive good, perhaps we should take this as another rationale for expansion to 16 teams per league (14 games for intra-divisional rivals, 8 games for inter-divisional ones) rather than an argument for eliminating 8 games from the schedule. The real reason that the AL has "divisional" play as they have it now is that the owners of AL West teams are extremely reluctant to forego any home dates with the league's best road attractions -- teams such as the Yankees and Red Sox. David Rubin {allegra|astrovax|princeton}!fisher!david