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While brushing up on "third out, run counts" rules and "fourth outs" the following came to my mind:

With 1 out could a scenario arise where one runner comes in, run counts, and second runner comes in, run does not count?

I consider scenarios, think I've got the answer, and then....nothing. It seems possible but I can't figure when.

-1

Yeah, with a ”time play”.

Scenario: Bases loaded, one out. The batter hits a short-hop to the second baseman, who tags second base to get out #2, then he throws to home (because he gets confused, or thinks he can't turn the double play at first) but it's a wild throw and the run scores. The catcher gathers the ball and then throws back to second base because the batter is trying to get there and he throws him out for out #3 just before the runner originating from second base crosses home plate. This would be a 'time play' and the first run would count and the second would not.

1
  • (Note: out #3 above is NOT a force out, but a tag out of the BATTER trying to make it to second. Had the third out been a force out, it would not have been a "time play", and NO runs would have counted.)
    – ipso
    May 15 '15 at 3:26
-1

No this is not at all possible. There are basically two different situations that can occur to end an inning.

  1. A force out in which no runs that previously scored will count.
  2. A tag out where any runs that score prior to the out will count.

If I am interpreting your question to mean can two players cross the plate during play and only one run count, the answer is no.

Now obviously a player could cross home after a tag out as @ipso has noted. If you are talking about during the actual play, this has no application. It is not necessarily clear if you mean during play or simply crossing the plate, because many players could cross home and not count if we don't limit the situation to during play.

We can get into little things such as base runner interference which when simplified will basically boil down to the two above scenarios.

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