A little potpourri for you on this double-dip Saturday, as the A’s sit in a virtual tie with Tampa Bay for the #1 seed and suddenly comfortably ahead (7 games, 8 in the loss column, 9 if you factor in the tie-breaker) in the AL West.
You knew the 2020 playoff format was going to cause some funkiness and indeed here it is. Because much of the AL winningness resides in the AL Central, where the 3rd place team cannot be higher than the #7 seed, we have a scenario where quite likely the #7 seed is going to be a better team than the #5 seed.
As of September 12th:
1. Oakland A’s (28-15)
2. Tampa Bay Rays (29-16)
3. Chicago White Sox (28-16)
4. Minnesota Twins (28-18)
5. Toronto Blue Jays (24-20)
6. Houston Astros (22-23)
7. Cleveland Indians (26-19)
8. New York Yankees (24-21)
So currently the #5 seed would be the Toronto Blue Jays at 24-20, whereas the current #7 seed, the Cleveland Indians, are 26-19 — and just four days ago they were a robust 26-15. The teams may change (just a couple days ago the Twins were in 3rd place, and the Yankees are 1⁄2 game from catching Toronto), but the math has remained the same for weeks: the AL Central’s 3rd place team is better than the second place team on either coast. And a lot better than the #6 seed, which is currently a team under .500 — your Houston Asterisks.
What that would suggest, at least on paper, is that you would be much better off to be the #3 seed, matched up against a 22-23 team, than to be the #2 seed, matched up against a solid 26-19 club.
Where will the A’s land? That’s hard to say when the top teams haven’t even played the same number of games. The A’s have played 43, the Rays, with whom Oakland is virtually tied for the #1 seed, have played 45, and the White Sox, currently the #3 seed, have played 44 — while the Twins, just a game back of Chicago, have played 46.
And then there’s the question of a possible tie-breaker between, say, Oakland and Tampa Bay, where they can’t look at head-to-head matchups because the A’s have played the Rays and the Buccaneers the same number of times. This year’s tie-breaker rules are complicated, but I think I recall that the second level is a homerun contest with Sean Manaea throwing to Yandy Diaz.
Can The A’s Beat Good Teams?
There has much back and forth on AN around whether Oakland’s record reflects that they are as good as anyone in the league, or just that they have played a soft schedule. For more on this question, here is a spirited debate between the voices in my head:
Me: Sure the A’s are 28-15, but that’s mostly because they have played terrible teams. 3 games against the Padres represent the only games all year so far against teams more than a game over .500.
Myself: OK, but you can only play the schedule in front of you. If you only have the opportunity to play against mediocre and bad teams, a skeptical fan would say, “If you want to show you can beat good teams, then you should be annihilating the bad ones.” And that’s exactly what the A’s have done, playing .651 ball for the year.
Me: Right, but that doesn’t mean they can beat good teams when they have to face them. Going 7-1 against the Bottomfeeders doesn’t mean you can go 4-3 against the Dodgers or 3-2 against the Rays.
Myself: So then how would the A’s prove, right now, that they can beat the Rays and the Dodgers? What can they do other than dominate the teams they are playing? Which is what they have done. Only the Dodgers have dominated their competition more, and their schedule isn’t that hard either — they haven’t played the A’s yet, the rest of the AL West is under .500, and the only other good team in their division is the Padres. The A’s are really good. They can play with anyone.
In the comments, let us know do you agree with Me or with Myself? And what else would you add to the analysis?
And then enjoy game 1 of the double-header, featuring the major league debut of “Dalton Jefferies: Private Eye”. The A’s seriously need to sign pitchers with last names of Columbo, Rockford, and Quincy to round out the future rotation.