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Fast forward a couple decades as the baseball Yankees Core Four

reminisce at the 2040 Old Timer's Game...

Andy Pettite: Remember that game back in 2012 against the

Red Sox when I gave up a lead-off home run to Pedroia and

then retired the next 27 batters? The closest I ever came to

a perfect game.

Derek Jeter: Yeah I think so, it was a special game for me,

too. I led off the game and batted in every inning! I went

5-for-5 and raised my average 10 points in a single game.

Mariano Rivera: And I watched the whole game from the bull

pen, hoping to get another save. But Andy finished strong,

and Girardi never called me.

Jorge Posada: I don't recall the exact final score, but with Jeet

coming up nine times, I can't imagine there would have been

a save situation. [Leading by no more than three runs when

the closer enters the game.]

Dustin Pedroia, in town for the game, got into the discussion

later, at Pete's Tavern: Save? You have to have the lead to have a

save. In my first at-bat I got the winning hit. You guys never scored!

Is there a way to reconcile the memories of these 60-something

ball players?

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Posted · Report post

I don't understand baseball. Can you translate it into cricket? ;-)

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Posted (edited) · Report post

No

Jeter only got singles. and never scored? Og course, Pedroia would've been other team. Or, they're old and can't remember correctly, or, since they're at tavern, too drunk, or a little of both. And I'm not sure how pinch hitting works, so maybe Jeter pinch hitted every inning?

Edited by NickFleming
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Posted (edited) · Report post

They played a double header that day

Edited by Equality 72521
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Leadoff Hit, couple of walks or singles / outs (Final out at batter #6). Then, in the 2nd, a CS for the 3rd out with Jeter at the plate. Lather, Rinse, Repeat a few times. No one scores. Sahx win 1-0.

Thought.... (btw, very sneaky bonanova)

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No, since Jeter batted in every inning, there must have been at least 9 batters per inning. That means they must have scored runs in each inning where Jeter batted at least 6th. Pedroia saying they never scored contradicts this.

Edited by Drexlin
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Leadoff Hit, couple of walks or singles / outs (Final out at batter #6). Then, in the 2nd, a CS for the 3rd out with Jeter at the plate. Lather, Rinse, Repeat a few times. No one scores. Sahx win 1-0.

Thought.... (btw, very sneaky bonanova)

Does it count as an at-bat if you are at the plate when the last out is made?

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Does it count as an at-bat if you are at the plate when the last out is made?

No, not even a plate apperance. However, it's the only way I could make the clues fit. :-/ Otherwise, you'd be right. No way to send more than 6 to the plate in an inning w/o having a run score.

Edited by isdtyrant
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No, not even a plate apperance. However, it's the only way I could make the clues fit. :-/ Otherwise, you'd be right. No way to send more than 6 to the plate in an inning w/o having a run score.

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No, not even a plate apperance. However, it's the only way I could make the clues fit. :-/ Otherwise, you'd be right. No way to send more than 6 to the plate in an inning w/o having a run score.

Thought so. I guess we need Bonanova to clarify. Or maybe he left that ambiguous on purpose.

;)

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No way to send more than 6. If the 7th batter (of an inning) comes to the plate, at least one run has scored.

If the first three batters get on base so that there is a runner on First, Second, and Third; and the next three batters hit singles while the lead runner is thrown out at home. The 6th batter would hit into the last out at home plate.

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If the first three batters get on base so that there is a runner on First, Second, and Third; and the next three batters hit singles while the lead runner is thrown out at home. The 6th batter would hit into the last out at home plate.

So then you agree that a 7th batter is not possible without scoring a run. Glad we are on the same page.

;)

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Thought so. I guess we need Bonanova to clarify. Or maybe he left that ambiguous on purpose.

;)

Jeter getting CS at 2nd on an 0-2 pitch to A-Rod last night. Thereby giving him a new count to seek #600 with. I thought it was a smart play at the time. Didn't quite work out that way. ::mutter::

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Maybe some players werent there, and lineup was shorter. How many people does a team have to have?

No, not even a plate apperance. However, it's the only way I could make the clues fit. :-/ Otherwise, you'd be right. No way to send more than 6 to the plate in an inning w/o having a run score.

You could hit the single, and make it to first, but then person running to home gets out.

Edited by NickFleming
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Maybe some players werent there, and lineup was shorter. How many people does a team have to have?
You could hit the single, and make it to first, but then person running to home gets out.

A team MUST have 9 batters in the lineup.

That doesn't count as a hit, its a fielder's choice.

Edited by Drexlin
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Oh :( I don't know that much about baseball, but could sacrifice flies have anything to do with it?

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No, not even a plate appearance. However, it's the only way I could make the clues fit. :-/ Otherwise, you'd be right. No way to send more than 6 to the plate in an inning w/o having a run score.

If it wasn't for this, I thought I had the solution.

1=Jeter

1:1(hit)23456

2:7891(hit)23

3:4567891(came to plate then they realized runner missed base on way home)

4:1(hit)23456

5:7891(hit)23

6:4567891(came to plate then they realized runner missed base on way home)

7:1(hit)23456

8:7891(walk)23

9:4567891(came to plate then they realized runner missed base on way home)

Every inning had 3 base runners, and 3 outs for a total of 6 batters per inning. However, 7 batters came up in the 3,7 and 9 but they’re at bats were cut short when they figured out the runner who scored (or any runner I think) missed a plate. It’s pretty incredible they managed to miss a bag 3 times in one game!

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If Jeter had 9 normal plate appearances and went 5-for-5, then four times he would have walked, been hit by a pitch or sacrificed.

But these ways for a plate appearance not to become an at-bat don't fit the shutout scenario.

Something else must have happened.

isdtyrant is right; the something else was inspired by last night's game.

The whole scenario, however, goes back to a Martin Gardner [rip] classic in which

"Mighty Casey" batted in every inning, never striking out, while his team nevertheless lost.

Four times while he batted, a team mate made the third out - by being picked off base or caught stealing.

This happened in the even numbered innings; in the following odd innings, the at-bat was continued.

Only in this puzzle, the outcome each time was a classic Jeterian single to right center field. B))

Posada's comment was a red herring, allowed here because he spoke forgetting the game details.

By including Mo's words, I was able to pay homage to the entire Core Four.

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If Jeter had 9 normal plate appearances and went 5-for-5, then four times he would have walked, been hit by a pitch or sacrificed.

But these ways for a plate appearance not to become an at-bat don't fit the shutout scenario.

Something else must have happened.

isdtyrant is right; the something else was inspired by last night's game.

The whole scenario, however, goes back to a Martin Gardner [rip] classic in which

"Mighty Casey" batted in every inning, never striking out, while his team nevertheless lost.

Four times while he batted, a team mate made the third out - by being picked off base or caught stealing.

This happened in the even numbered innings; in the following odd innings, the at-bat was continued.

Only in this puzzle, the outcome each time was a classic Jeterian single to right center field. B))

Posada's comment was a red herring, allowed here because he spoke forgetting the game details.

By including Mo's words, I was able to pay homage to the entire Core Four.

bonanova, I'm not sure I follow your logic. Although, I'm willing to bet it is me and not your logic...

You state that "Four times while he batted, a team mate made the third out - by being picked off base or caught stealing. With this scenario, the most batters that could have been up before Jeter (without scoring a run) is 5: 3 are on base and 2 got out. If one of the runners is picked off you only record 5 at bats in that inning. That is not be enough movement through the line-up to allow Jeter to make it up to the plate again in all 9 innings. At least 6 at bats must count each inning in order to move through enough of the line-up. You can see this in my previous post showing the inning by inning breakdown.

Am I missing something?

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bonanova, I'm not sure I follow your logic. Although, I'm willing to bet it is me and not your logic...

You state that "Four times while he batted, a team mate made the third out - by being picked off base or caught stealing. With this scenario, the most batters that could have been up before Jeter (without scoring a run) is 5: 3 are on base and 2 got out. If one of the runners is picked off you only record 5 at bats in that inning. That is not be enough movement through the line-up to allow Jeter to make it up to the plate again in all 9 innings. At least 6 at bats must count each inning in order to move through enough of the line-up. You can see this in my previous post showing the inning by inning breakdown.

Am I missing something?

Read isdtyrnat's post #5. Jeter leads off, gets a hit. Then 2 more hits/walks and 3 outs in no particular order (except that an out must be last). That is 6 batters. Batter 7 comes up in the next inning, gets a hit/walk. Batter 8 flies out. Batter 9 stikes out. Jeter up again. Batter 7 picked off at first. Jeter will be first to bat the 3rd. Rinse. Repeat.

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Read isdtyrnat's post #5. Jeter leads off, gets a hit. Then 2 more hits/walks and 3 outs in no particular order (except that an out must be last). That is 6 batters. Batter 7 comes up in the next inning, gets a hit/walk. Batter 8 flies out. Batter 9 stikes out. Jeter up again. Batter 7 picked off at first. Jeter will be first to bat the 3rd. Rinse. Repeat.

Got it. Thanks.

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