How is it possible to hit three singles and three doubles without scoring a run in one inning of baseball?

Question

For one team...so technically a half inning.

Answer 1

Double Plays?

Answer 2

Haven't seen this one in about a week.

I know the answer, but let me start at the back end -- what the finished inning looks like. Six hits means six baserunners; three outs removes three of them, and there were three left on base. All we need to do is work out how they got there.

The first five are easy:
Batter #1 -- doubles, gets thrown out trying for a triple.
#2 -- doubles, thrown out trying to steal third.
(The first two somehow make outs on the basepaths. The specific methods are not important, just that they do not score.)
#3 -- doubles.
#4 -- singles, runner to third.
#5 -- infield single, bases loaded.

That part was easy. Fitting in the sixth hit is where it gets tricky, because merely throwing out the runner at home is scored a fielder's choice, and that's not a hit.

#6 -- hard infield grounder which strikes a baserunner teammate in fair territory before a defensive player has had a chance to make a play. Plunked baserunner is out, batter is credited a single, no runners advance except as forced by the batter-runner. Third out, inning over, three left on base, six hits, no runs.

MLB Rule 5.09 (abridged):
The ball becomes dead ... when --
(f) A fair ball touches a runner or an umpire on fair territory before it touches an infielder including the pitcher, or touches an umpire before it has passed an infielder other than the pitcher; ... runners advance, if forced.

Answer 3

First batter hits a double and gets picked off 2nd base. Second batter hits double and gets picked off 2nd base also. Next three batters all hit infield singles so the bases are loaded with two outs. The sixth batter hits a ball that goes to the wall and tries to stretch it into a triple, but gets thrown out at third base before any of the three men on base cross home plate. The batter is therefore credited with a double and no runs score.

Answer 4

first batter hits a single, second batter hits a double, third batter hits a single,

4th batter hits a single. Runner is thrown out at home.

Person leading off third is tagged out.

5 th batter hits a double, runner thrown out at home.


well, I'm close. I got 3 singles and two doubles, and no runs.


Well, another answered said the MOST hits you can get in an inning an not score a run is 5, and I accomplished that!!

Answer 5

Rich V has a great answer. The problem with his analysis is that if the last guy hits a single and the guy on third is thrown out, it would be scored as a fielder's choice and not a single. So Based on Rich V's fantastic analysis. You would need to score at least one run.

Answer 6

Caught stealing is probably the easiest way. also a double then single could be the single hit toward third so lead runner can't advance.

Answer 7

That is not impossible but extremely unlikely. There are only 3 outs in a half inning...so let's say the leadoff hitter hits a double, then is out at third stealing. The next guy hits another double and is out at third stealing. Two outs, two doubles. Next guy up hits a double. Next guy up hits a single, runner stays at second. Next guy up hits a single, runners advance one base, bases loaded, two outs, three doubles and two singles have been hit.

The ONLY possibility now is that the next batter up hits a single, but the runner at third is really, extremely slow, and was thrown out at home. That is a nearly impossible scenario.

Answer 8

SIngle, pick off 1 out
double, pick off 2 out
double
double (men at 2-3)
single,(men at 1,2,3)
single (ball hits runner, runner is out)
3 OUTS

Answer 9

There are many different ways but the simplest is if every runner gets thrown out at the plate.

Answer 10

I guess you didn't realize that runners can always be thrown out at the plate.

Anything up to 6 batters is possible. 7 batters, however, is not possible.