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Interleague play by division


obobo55

AL is dominating overall but how does it breakdown by division?

 

AL East 34-27 (+7)

AL Cent 31-25 (+16)

AL West 30-24 (+6)

 

NL East 25-34 (-9)

NL Cent 30-35 (-5)

NL West 21-40 (-19)

 

Guess we know the culprit??? The west just stinks this year. The crew needs to cleanup against the Rockies, Giants, and Padres. I know we have never had much success going on our western road trips, but this year we need to come out of those at .500 or better. 8 of our 21 remaining games against the west come in the nest 2 weeks (4 @ AZ, 4 vs COL). Then out of the break, we play at SF for 3. We only have one other trip to SD and LA in mid-August. The Padres come here for 4 the first week of Sept, when they should be completely out of it and playing youngsters.

 

My hopes:

 

2-2 @AZ

3-1 COL

2-1 @SF

2-1 @SD

1-2 @LA

4-0 SD

 

14-6

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A four game sweep? Even if they stink and we are at home they should win at least one of them. If the Padres only have a 20% chance of winning a game, math says they have an 80% chance of winning one of those games, and a 40% chance of winning 2. (I think I did the 2nd one correctly, but I know the first one is right)

The poster previously known as Robin19, now @RFCoder

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I know that is a bit of a stretch, but it has been done before. If we are in position to make a run at either the division or wildcard at that point and SD is playing their youngsters...Why Not?

 

BTW--The scrubs just swept a 4 gamer against Colorado so it can be done.

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I know that is a bit of a stretch, but it has been done before. If we are in position to make a run at either the division or wildcard at that point and SD is playing their youngsters...Why Not?

I'm pretty sure these two teams were in reversed situations last year and we took 2 or 4 at home.

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Actuallly, if my memory of statistics is correct, they would have roughly a 24% chance of winning 2 games. You have to multiply the chance of winning by itself, then multiply that by the number of possibilities for winning 2 games.

 

Odds of winning 2 games = .2 * .2 = .04 (4%)

Winning 2 Game Possibilities: 1,2 - 1,3 - 1,4 - 2,3 - 2,4 - 3,4 = 6 possibilites

 

.04 * 6 = .24 = 24% chance of winning 2 games (if they have a 20% chance of winning each game individually).

 

 

Anyhow, I agree, a series sweep is probably wishful thinking there. I think 2 or 3 games is more realistic. I also believe that we can take 2 from the Dodgers, not just 1.

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