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Top half of the inning: B1 - single but gets picked off by the pitcher. One out. B2 - Triple but gets picked off by the pitcher. Two outs. B3 - Triple. B4 grounds out. Three outs and no runs in. Bottom half of the same inning: B1 - single but gets picked off by the pitcher. One out. B2 - triple but on the next pitch tries to steal home and is thrown out by the pitcher. Two outs.B3 - single but is thrown out on the next pitch trying to steal second base. Three outs and no runs in. ________________________________________________________________ This is my first time doing this so I'm not really sure how this all works. I just stumbled across this question and noticed that the answer above is incorrect so apparantly anyone can answer anything whether or not they are correct. Here's the CORRECT answer: B1 - hits a triple but gets thrown out at the plate, out 1. B2 - also hits a triple and is thrown out at the plate, out 2. B3 - safetly hits a triple. So far we've got 3 triples, 2 outs, and a runner on third. B4 - hits an infield single and runner on third does not score. B5 - also hits an infield single advancing B4 to second yet runner on 3rd is still unable to score. So far bases are loaded, 2 outs, 3 triples and 2 singles hit leaving the remaining single. B6 then hits a grounder (or line drive) that strikes one of the base-runners. The base-runner is automatically ruled "out" and the batter is automatically credited with a hit.

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That is immpossible because if you hit a triple you have a guy on third. Then if you hit another triple then the runner scores because you force him/her to go off the base to home.

<not so fast> It is possible. This play actually happened at the collegiate level. In one inning there was two triples, two doubles, two singles, nobody scores.

1st batter: deep shot in the gap runner rounds third and tries to go home and is gunned down (ruled a triple)

2nd batter: exact same play, runner gunned down trying to go home (ruled a triple)

3rd batter: hits double in the gap lands on second:

4th batter: hits double in the gap, runner on second tries to go home but trips on third base and sprains ankle. Crawls back to third. Runners on 2nd and 3rd with two outs.

5th batter: single in the 3 4 gap, left fielder prevents the run from scoring from 3rd.

6th batter: bases loaded two outs, hits the ball towards the 4 5 gap and hits the runner going from 2nd to 3rd. Runner is ruled out (interference) batter gets scored a single.

Q: How can you get 3 singles 3 triples in a inning without scoring a run?

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If the first batter walks or gets on base, but the next three batters get out, that's three outs after four batters without scoring a run. Then, the next three batters could also get out without the runner advancing or scoring, resulting in seven batters coming to the plate without scoring a run.

K, po-7, k

The answer is six. Load the bases with 3 singles (3) single up the middle, runner thrown out at home 3 times in a row. (6) The answer is six, but not in the manner listed above, which are actually 3 fielders choices. The real answer is: 2 singles to start the inning. Runners at first and second. 2 more singles with the lead runner thrown out at home. First and second, two out. 1 more single to load the bases. Bases loaded, two out. The sixth batter hits a ball that hits one of the runners. The runner is the third out, and the batter is credited with a single.

on April 18th, 2009 the Cleveland Indians set a major league record scoring 14 runs in one inning.[2nd inning]

I don't know what the actual answer is, but I say 6 hits without scoring a run is theoretically possible: three hits to load the bases, two fielder's choice outs of the lead runners, then two more singles to load the bases again. Then the 6th hit would be a batted ball into a baserunner. The baserunner would be the 3rd out and the batter is credited with a hit. Or the 6th hit could be, although unlikely, an infield hit where there's an attempted putout at 1st, then the batter rounds 1st in attempt to go to 2nd and is immediately tagged out before the run crosses the plate.

Easy, if in multiple innings. They could get nine straight out. If you mean in one inning, i do not believe its possible. Even with bases loaded when the last out is made, it would still only be a total of six batters. Hope i helped

five unearned runs.

If a runner reaches on a walk (or base on balls) and scores that will figure into a pitchers ERA, only runners reaching or scoring on errors, or scoring after the 3rd out should have been made (due to earlier errors in the inning) do not count in the ERA

Milwaukee vs. Toronto (31 hits) august 28th, 1992.

The pitcher and the rest of the team.

in baseball an inning is split into top and bottom because of the way the scoring is (check it out if you ever go to a pro like game). the top of an inning is when the away team bats and the home team fields. it's vice-versa for the bottom.

The run will count if the runner on third crosses the plate before the runner on second gets taged out. All force place end the inning without any runs scoring, however, all tag plays end the inning at the time of the third out.