Suppose you’ve got runners on second and third, and two outs. The batter hits to the outfield; ball drops without being caught. The runners from second and third make it safely to home, scoring two runs. The batter rounds first and runs to second, where he is thrown out, ending the inning.
Do those two runs count?
It was my understanding that the play had to end with the batter safely on base, or the inning was over and the runs didn’t count.
I ask because I was just watching the Braves-Pirates game and the above situation occurred in the top of the first, except the runner barely made it safely to second. It seemed strange that he would have run for second instead of stopping at first, because he was effectively risking two runs for the sake of one base.
Is the rule that the batter has to make it safely to first for the runs to count, but after that he can be thrown out and the runs still stay? Do the runs have to be scored before the batter is out (which seems like an unlikely rule to me for some reason)?
(Wasn’t sure if this should go in GQ, as it’s a factual question. Mods, please move if appropriate.)
The batter has to make it to FIRST BASE safely, and every forced runner must make it to their next base safely.
If the batter is retired at first, or any other runner is made out by a force play (for instance, if the bases were loaded, and the third out was made by a force out at second base) then no run scores.
However, in the scenario you describe, the batter made it to first and was made out by a tag, not a force. Any run that scores before the batter was tagged out counts.
Suppose there is a force play at second, but the throw to the second baseman is off target and he has to leave the bag to catch it, then tags the runner for the third out?
OK, my inner lawyer got the best of me and I went and looked up the rules. Rule 409(a):
So part (2) of that rule introduces the question of whether a forced runner who is tagged out, rather than thrown out, is considered to have been “forced out.”
It appears that he is. In the definitions section, Rule 2.00, a “force play” is defined as:
In other words, the batter hits the ball and becomes a runner, and because there are no empty bases between them, the runner is forced to head for the next base. That’s not a perfectly precise answer, but it suggests to me that any forced runner who does not reach base safely is deemed to have been “forced out.”
There’s another reference to the force play in the rulebook (which I just found but now cannot remember the rule number of) that says that a player being forced, who is tagged BEFORE reaching the base, is forced out.
So if the runner is tagged before touching second, it’s a force play, no runs. But if the runner is tagged AFTER touching second - if he overslides the base, for instance - it’s not a force play.
Missed the edit window. My last post should have said, “…because there are no empty bases between them, the runner who was already on base at the start of the play is forced to head for the next base.”
It all depends on what the runner had accomplished. If he had not reached the base safely before being tagged out, then it is considered a force play and the runs do not count.
If the runner had safely reached the base, and then had been tagged out, after leaving the base (intentional or not), the force is not in effect and any runs that had previously scored would count (and the hitter would be credited with reaching base safely).
You mean a runner who was on first at the beginning of the play, right? That makes sense, because it’s analogous to the scenario in the OP: The “batter” becomes a “runner” after he hits the ball, but once he tags first, he’s no longer a forced runner.
I hate to be the ass that has to split hairs, but the “batter” actually becomes a “batter-runner.” He isn’t a “runner” until the play on which he became a runner
ends.
In fact, I’m going to split one as well, and suggest that you meant to say, “He isn’t a ‘runner’ until the play on which he became a ‘batter-runner’ ends.”
If the runner on first advances to second and touches the base, the force is off, obviously. But if the runner then deliberately retreats back to first, the force is reinstated.
For instance: The batter hits a high fly to right. The runner on first took off with the pitch and is already at second before the ball comes down. Realizing it is a fly ball and he’s in jeopardy, he leaves second to run back to first… but halfway back, the ball comes down and is dropped by the outfielder. So the runner now has to get back to second, and he turns to run back. *The force is back on *(Rule 7.08 e) and the runner can be forced out at second.
So it is in fact possible for a force situation to exist, not exist, and then exist again, all on the same runner on the same base on the same play.
I think there is something that has been missed in answering the OP - the runs count because they scored before the final out was made and the final out was not a force out.
This is no so uncommon - another situation:
runners second and third, 1 out.
fly out to left field, runner on third tags, runner on second doesn’t pay attention and strays from the base.
lf throws to ss, rundown on runner at second ensues.
tagging runner crosses plate, other runner is tagged out.
run counts only if runner scored before runner was tagged out
this could happen in a double steal situation too with runners on, say, 1st and third, 2 outs. If runner on first is in run down long enough for the runner to cross the plat, it counts as a run, even if the 3rd out is then recorded.
this could happen also on a botched squeeze bunt if the runner crosses the plate before the pitch strikes out the batter.
That would only be true if the runner touched 2nd on the way back to first, right?
Also, depending on where the outfielder is, the infield fly rule could be in effect too and the out would be recorded when the outfielder touched the ball, even if he subsequently drops it. there would be no force, but the runner needs to get back to first before the ball or it would be a double play. and if he gets there before eh ball but hasn’t touched 2nd on the way back, the ball will have a lot longer time to get there as the runner goes to 2nd and back again in an effort to beat the ball to 1st.
It could happen that way or the runner could simply touch second, stand there, and go back. According to 7.08e, if he retreats to first for any reason the force is reinstated. I think it’s a weird rule, but there it is.
Not exactly; on an infield fly, the batter is out the instant the umpire calls the infield fly. Rule 6.05(e). (The call is reversed if the ball lands foul, though.) The act of hitting an infield fly is what makes you out. Nobody has to touch the ball.
There is no force play on an infield fly - there can’t be - so it’s got nothing to do with force plays. However, you’re of course correct in that if the runner has advanced past second, he must retouch it if he needs to go back to first, in all cases.
But he can’t be said to be retreating to first until and unless he touched second n the way. Otherwise a rundown between 2nd and third would trigger the situation.
If the runner runs towards first base, he is returning to first base.
You don’t have to advance past second for this rule to kick in, is all I’m saying. Think about it; you run up to second, step on it, then run back to first. You have advanced to second, then retreated to first, but you never ran past it.
Yeah but think about it - you head towards third and retreat due to a run down, or just “going halfway” - are you saying that the runner is heading back to first according to the rules? I’d posit he is heading to second, not first. Once he passes 2nd, then he can be said to be going back to first.
Well, okay, sure. I don’t see what that has to do with the rule that a force play can be reinstated. The runner is returning to first if, in fact, he’s returning to FIRST. If he’s going back to second, he’s going back to second. It’s usually not hard to tell the difference.
Now, let’s suppose the runner rounds second towards third, and then he want sto go back to first and, in a moment of confusion, heads DIRECTLY back to first, forgetting to retouch second base. If the umpire judges that the runner is retreating to first, he’s returning to first. The force play is back on if the ball drops. The runner is in jeopardy to be put out two ways; he could be forced out at second, or if they don’t get it there before eh turns around again and gets to seocn base, they can appeal at second, and he should be called out for failing to retouch the base he’s standing on.
Say runner rounds second, throw comes in to second, runner slides in safely, 2b catches ball. runner slides past the base towards first, coming off the base, whilst doing so 2b holds onto the ball with foot on bag.
By definition of a force and “returning to first”, he would be out.
I’d maintain 2b needs to tag the runner, and the runner still has opportunity to scramble back safely without being tagged.
Variation: If this happens while batter has safely reached first, runner can’t return to first, as batter can’t go back to home. Of course if both end up at first, the lead runner is out, but even on the way, can he be said to be “returning to first” when he is not eligible to go there?