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Jake (The Snake)
10-17-2005, 11:40 PM
This is a situation I see frequently in the ninth inning of games. It happened again today in the NLCS:

Astros are up by two and there is a runner on first base in the ninth inning. Baserunner on first steals second and they just let him go.

Basically, I'm wondering why the catcher wouldn't just throw down and see what happens, even if it looks like he has no chance. Isn't it a freeroll?
Jack of Arcades
10-17-2005, 11:42 PM
Not if he throws it into centerfield.
pudley4
10-17-2005, 11:47 PM
[ QUOTE ]
Not if he throws it into centerfield.

[/ QUOTE ]

That won't matter.

What does matter is that one of the two middle infielders will have to cover 2nd, thus leaving a hole somewhere in the infield.
Jack of Arcades
10-17-2005, 11:49 PM
[ QUOTE ]
[ QUOTE ]
Not if he throws it into centerfield.

[/ QUOTE ]

That won't matter.

What does matter is that one of the two middle infielders will have to cover 2nd, thus leaving a hole somewhere in the infield.

[/ QUOTE ]

Ah, I missed the "up two" part. Yeah.
Jake (The Snake)
10-17-2005, 11:50 PM
Yeah that explains it, thanks.
Voltron87
10-17-2005, 11:50 PM
with 2 outs and a 2 run lead the middle infielders wont break to cover the base, so the catcher has no target. they wont break because if they do they will leave gaps open, increasing the chance of a hit.

also, usually the runner on 1st is not being held (same reason, when a 1B holds a runner on it leaves a big gap) so the runner will likely have a great lead so the chances of getting him arent very good at all.