How many hits can a MLBaseball team get in an inning without scoring a run?

Question

Please explain how they got the hits.

Answer 1

Six. I'll give an example.

The first two batters get hits, but are eliminated via pickoffs.

Hitters three, four and five get singles to load the bases.

The sixth batter hits a ball that hits one of the runners. He is given credit for a hit, but the runner is out, and the inning over.

Does that make sense?

Answer 2

It is 6.

In theory, a team could have 6 hits in an inning and not score a run. It has never actually happened.

Here's what could happen. First batter gets a single then gets thrown out trying to steal second. Second batter gets a single then gets thrown out trying to steal second. The next three batters all get singles to load the bases. So far we have 5 hits, no runs, 2 out. The 6th batter hits a ground ball which hits the runner on 3rd base in fair territory. The runner is ruled out, no run scores but the batter is credited with a base hit - the 6th and final hit of the inning.

Answer 3

Craig S is right. The only way you could get 6 is if the batted ball hits the runner for the third out. Batter gets credit for the hit, but no run scores on the dead ball 3rd out.

Everyone saying that a runner being forced at home or tagged out should know that this would only be a fielder's choice, and not a hit.

Answer 4

the number is six and here is a way it can happen first three hits go to the outfield the next 3 comes as a result of a runner being hit with a batted ball. The runner who was hit with the ball and the ball is dead so no runners can advance. The batter is credited with a hit. This is one way for it to happen, there are many other ways.

Answer 5

Six.

Just like the other 99 times this has been asked in the past two months.

First five are easy.

Aaron singles -- gets thrown out trying for second.
Banks singles -- then gets picked off.
Carew singles.
Doby singles.
Evers singles.

Five hits, bases loaded, two outs, no runs. That was the easy part.

Foxx slaps a hard grounder that strikes one of his baserunner teammates in fair territory (say, Doby) before a defensive player has had a chance to make a play on the ball. Doby is out, Foxx is credited a single, Evers advances to second because he is forced to do so by Foxx's single, Carew does not advance.

Three outs, three left on base, inning over, zero runs on six hits.

Answer 6

I think 5 because if there were three force outs at home after loading the bases then they wouldn't be considered "hits". If the bases are loaded as a result of 3 hits and then two bases are cleared for any number of reasons, then two more hits to fill the bases before three outs or a run scored would be five hits.

Answer 7

Some people are saying you can get six by loading the bases and then getting force outs at home. Force outs do not count as hits for the batter, but rather an 0-1 with a fielders choice. So loading the bases via hits and getting force outs at home will only count as three hits.

I think the correct answer is 5, and there are many ways of doing that as outlined by others above.

Answer 8

My guess is 5. 3 to load the bases. get two outs some how then fill those 2 bases. It seems like anything more would be a fielders choice, therefore not a hit. I will keep thinking about it, but for now, that is all I can come up with.

Answer 9

I think it would be 6.

1. Hits a single to left.
- Runner picked off at first.
2. Hits a single to left.
- Runner picked off at first.
3. Hits a single to left.
4. Hits a single to left.
5. Hits a single to left.
6. Hits a single to left, runner on third twists his ankle and falls down halfway between third and home. Tagged out*

*I'm not 100% sure that this would be a hit. If not, then I believe it's 5.

Answer 10

6,
hit
hit
hit
pick off (1)
hit
Pick off (2)
hit
hit but runner is hit with ball (3)