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# Oakland A’s magic number (Friday): 4 for division, 2 for Wild Card

This could be a significant weekend for the A’s

Today is Friday, Sept. 18, and the Oakland A’s have 10 games left in their regular season. They currently hold a six-game advantage in the AL West division, over the Houston Astros, and that means it’s time for one of the best countdowns of the year — the Magic Number.

If you’re not familiar with the concept of the magic number, or have heard of it but never understood it, here’s a quick summary. It measures how many positive results a team must rack up in order to mathematically clinch whatever they’re chasing, using a combination of their own wins plus the losses of whoever is chasing them.

For an easy example, if you’re up by one game with one left to play, then your magic number is one. If you win your final game then you clinch, or if the runner-up loses theirs then you clinch. If you’re up by eight with eight games to play, then your number is still one. Take the number of games remaining, subtract the number of games in your lead, and add one because you have to finish ahead by one in the standings.

More precisely, the formula is: “Games remaining +1 - (Losses by second place team - losses by first place team)” which accounts for moments when the two teams haven’t played the same number of games. But that’s effectively the same as my simpler version above. The magic number can’t end in a half, so if a team is up by 3.5 games then just add imaginary wins to the chasing team until they’ve played the same number of contests.

But wait! There’s one more wrinkle to add. Normally, if two teams tie for their division then they play an extra tie-breaking game to decide it. But in 2020, with a tighter and more precarious schedule, there will be no such bonus games and all ties will be decided on paper.

The first criteria is head-to-head record, and that will be enough to settle this if needed. The A’s beat the Astros 7-3 in 10 games, so if they tie then Oakland automatically gets the division. Therefore, in order to win the West, the A’s only need to tie, not finish ahead by one. Take my simple version of the formula above, and remove the part where you add one at the end. For the AL West magic number, we only need to consider games remaining minus Oakland’s division lead.

The current standings, entering Friday:

### AL West division

1. Oakland, 31-19
2. Houston, 25-25, 6 Games Back

That’s a six-game lead with 10 left. The A’s only need to tie to clinch, so their magic number is FOUR.

### Wild Card

1. Oakland, 31-19
2. (Seeds 4 thru 8: Minnesota, New York, Cleveland, Toronto, Houston)
3. Seattle, 22-28, 9 Games Back

(Baltimore, Detroit, and Kansas City are within a game of Seattle, so one of them could replace the Mariners as the team “chasing” the A’s for this spot.)

The tiebreaker rules are the same here (head-to-head), but the A’s and Mariners haven’t finished their season series yet. It’s currently 4-2 with four left to play, and those are the final games of the season, so for now we can’t assume the A’s get the tiebreaker and must assume they have to finish ahead of Seattle by one to knock them out. Therefore, a nine-game lead, with 10 left to play, plus one, equals a magic number of TWO.

If the head-to-head record can’t decide a tie, then next is intradivision record, where the A’s are 24-12, the Astros are 17-16, and the Mariners are 16-17.

Next up for Oakland is the second half of the Bay Bridge Series, with three games at home against the San Francisco Giants. The A’s could absolutely clinch this weekend, with only minimal help from the Astros. Or, they could get swept by the Giants and still at least clinch a postseason berth via Wild Card, if Seattle loses too. Point is, get those rally towels ready to fly.