20 replies to this topic

[isdtyrant]

Spoiler for Still Accurate

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

Spoiler for Yes there is

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.

[Drexlin]

Spoiler for Yes there is

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.

Spoiler for

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

[isdtyrant]

Spoiler for

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

Spoiler for I think this was inspired by....

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::

[NickFleming]

Spoiler for Another possible part of answer, idk

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

Spoiler for Answer:

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, 28 July 2010 - 02:33 PM.

[Drexlin]

Spoiler for Another possible part of answer, idk

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, 28 July 2010 - 02:38 PM.

[NickFleming]

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

[Louie]

Spoiler for Answer:

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.

Spoiler for Counts plate appearances as at bats

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!

[bonanova]

Spoiler for Guilty of being sneaky

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.

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.

[Louie]

Spoiler for Guilty of being sneaky

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.

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...

Spoiler for can't get to plate

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?

[Drexlin]

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

Spoiler for can't get to plate

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.