2007-05-29 03:56:39 · 17 answers · asked by BP 2 in Sports ➔ Baseball
Six is correct, as given.
Regarding nupectbrad's answer, the DP would not leave two bases open. The question said no runs score; you've got three men on, one man up; two make outs, where are the other two? Still on base. So there's only ONE base open.
The only way to get more would be to loosely interpret the word "batter"; say Hank Aaron bats, hurts himself on the first pitch, and is replaced by Willie Mays, who hurts himself . . . .
If we count all of those as separate "batters", then I guess the maximum is 25, the number of players on the roster.
But within what I think is the spirit of the question, six remains the answer.
2007-05-29 04:31:03 · answer #1 · answered by pob14 4 · 1⤊ 0⤋
6
2007-05-29 05:05:04 · answer #2 · answered by J Dub 5 · 1⤊ 0⤋
I am only going to answer this because of an answer I read up a little bit. The correct answer is 6. The answer 7 that was posted by nupectbr somehow got a thumbs up from someone even though it is a physical impossiblilty. If the bases are loaded and someone hits into a non scoring double play, that means that one of the runners had to be forced out at home, which means that it is impossible for there to be two bases open
2007-05-29 04:54:43 · answer #3 · answered by AngusAssassin 2 · 1⤊ 0⤋
6
2007-05-29 04:00:44 · answer #4 · answered by llk51 4 · 0⤊ 1⤋
12. 6 in each half inning. 3 get on base from "hit by pitch" or errors. The bases are now loaded and the next three must make outs for there to be no runs scored. There is no place to put any more base runners. That is your 6 per half inning.
2007-05-29 04:10:07 · answer #5 · answered by Indian Fan 1 · 1⤊ 1⤋
I would say that if the first three batters reached base on errors, and the next batter (4) hit into a non scoring double play , leaving two bases open, and the next two batters (5&6) reached on two more errors, most likely infield errors, leaving room for one more batter (7)to either get the last out or knock in a run.
So...I would say 7
2007-05-29 04:09:26 · answer #6 · answered by nupectbrad 1 · 1⤊ 2⤋
Well if you loaded the bases on errors and then made a double play that would be four. Then you could re-load the bases with the next two on errors. With the seventh batter a run would either have to score or make an out, I think.
2007-05-29 04:28:01 · answer #7 · answered by GOB BLUTH 5 · 0⤊ 2⤋
Be the home team and be in the lead going into the 9th. Then the closer comes in, records the out, and they win the game. Game over, without any of those happening. However, if you are only referring to the game ending by the team being at bat, then a passed ball or balk would do it.
2016-04-01 02:45:01 · answer #8 · answered by Anonymous · 0⤊ 0⤋
6
3 batters can reach on either dropped-3rd strikes or errors and then the other three batters record outs
that makes 6
2007-05-29 04:00:12 · answer #9 · answered by TheSandMan 5 · 1⤊ 1⤋
Six. Three outs, three left on base, each hit by a pitch.
Well, strictly that is a half-inning, so maybe "12" is the answer.
2007-05-29 04:05:35 · answer #10 · answered by Chipmaker Authentic 7 · 1⤊ 1⤋