Obstruction
Bases are loaded with no outs. The batter bunts a fair ball ball up the first base side. As the batter-runner is leaving the batters box, the batter-runner is obstructed by the catcher who has no play on the ball. The home plate umpire signals obstruction. Then, the pitcher gets the ball and throws it back to the catcher who steps on home plate before the runner from 3rd gets home. The catcher then immediately throws the ball to 1st base who steps on 1st base before the batter-runner gets to 1st base. Finally, 1st base, being alert, throws the ball across the diamond to 3rd base who tags the runner (starting at 2nd base and rounding 3rd towards home) out while diving back to 3rd base. How many outs do you have? Back to Quiz
One.
When the pitcher throws to the catcher and the catcher steps on home plate, there is one forced out of the runner coming from 3rd base.
When the catcher throws to 1st base, there is a play on the obstructed batter-runner and the umpire should immediately declare "DEAD BALL", kill the play, and award the obstructed batter-runner the base they would have made had there been no obstruction (rule 8-5-B.2 EFFECT). The umpire would put the batter-runner on 1st, forcing the runner starting at 1st to 2nd, and forcing the runner starting at 2nd to 3rd. Bases would still be loaded. However, there can not be an out at 1st because the obstructed runner is protected between home and 1st.
Also, there can not be an out at 3rd because the ball is "dead" once the play is made on the obstructed batter-runner at 1st base. There can not be any additional plays made once the ball is "dead".
It is interesting to note that if the catcher had thrown to 3rd base instead of 1st base, and 3rd gets the force out, and then 3rd throws to 2nd, and 2nd gets the forced out, that, in this case, there would be a total of 3 outs on the triple play! |
