No.
But this rule does not come into effect until AFTER the force out at first base occurs.
In other words, until that force out at first happens, the runner who was on first must advance. But AFTER the force out occurs, the runner who was on first need not do so.
If a first baseman steps on first and then immediately fires to second base, the person covering second base must tag out the runner coming towards second base.
I've seen twenty-year veterans of MLB forget this -- they take the throw from the first baseman after a force out at first base, step on second base, and then walk away without tagging the runner.
Absolutely not -- a runner can (and invariable does) leave the base even before the pitcher throws the ball to the batter! That's called leading off.
A batter can attempt to run to the next base without the batter hitting it -- ie, stealing a base.
PERHAPS what you're asking about is what happens if the runner leaves the base before the ball is hit AND the batter hits the ball AND the ball is caught by a fielder before it hits the ground AND the ball is then thrown to the base where the runner was AND the ball is held there before the runner returns to that base. If ALL of those things happen, then the runner is out.
3 base runners on 3 bases
Yes. If the first baseman tags out the batter on his way to first, the three existing runners can return to their original bases; none can be "forced out."
No. When a batter is hit by a pitch, the ball is dead and no runners may advance. However, if the bases were loaded, then all runners are forced to advance and the runner from third would score.
It's an all girl game. Nice puzzle.
Ten - First batter reaches steals two bases tagged out at plate. Second batter does the same. Third batter steals two bases and fourth batter steals one. Ten is the answer but alternatively, first batter gets on and steals two bases (2). Next batter reaches first and steals second (3). Third batter reaches first and now with three runners on each base a triple steal could be executed with the 3rd base runner being tagged out (5). Repeat last sentence, (7). Then the fifth batter gets walked, and a triple steal is attempted. The runner from third gets caught in a rundown. The runner from second steals third (8). The runner from first steals second and third (10). Then one of the three runners now between third and home gets tagged out before the lead runner can score.
The lead runners could all have been passed by the batter before any of them touched the plate after the batter hit a home run. The three lead runners would all be called out.
Only the batter needs to touch first base to complete the walk.
Runners can attempt to advance on a fly out, provided that they tag up (touch the bade they are currently on after the ball is caught).
If it is a batted ball, the batter and all runners are awarded 3 bases. If it is a thrown ball, the batter and all runners are awarded 2 bases from whatever base they had occupied when the ball was thrown.
Yes, in Major League Baseball, all players need to advance to their next base with the bases loaded and the batter being hit regardless of the runner that is on third base being the winning run and therefore being the ending of the current game that's being played.
From major league rulebook:7.03(a) Two runners may not occupy a base, but if, while the ball is alive, two runners are touching a base, the following runner shall be out when tagged and the preceding runner is entitled to the base, unless Rule 7.03(b) applies.(b) If a runner is forced to advance by reason of the batter becoming a runner and two runners are touching a base to which the following runner is forced, the following runner is entitled to the base and the preceding runner shall be out when tagged or when a fielder possesses the ball and touches the base to which such preceding runner is forced.
Yes. Since the batter is forced to run to first base the other runners are forced to advance. However, once the batter is put out at first base the other runners may attempt to go back to their original base.