You cannot get 8 batters in 1 half inning, the most batters (roster spots) you can get to the plate in 1 half inning is 7 (although everyone here says 6, that is incorrect) -- in a half inning you can have 7 batter slots in the line up come up by:
- 3 runners on, reaching in any manner (3 batters)
- 2 batters get out (5 batters total)
- batter puts ball in play everyone advances a base, including the guy from 3rd to home (6 batters)
- batter is announced (making his appearance official), then without throwing a pitch, the defense team appeals a call (say the guy didnt touch home) -- ump calls him out
there you have it -- 7 batting slots in the lineup make official batters (because if a 7th batter is a pinch hitter, even though he never sees a pitch he is officially in the game, and takes the roster spot of the player he was subbed in for), no runs scored
You can only have a maximum of 6 batters appear who receive official plate appearances in a half inning without scoring a run. As stated above, you get 3 runners on in any manner, and then the next 3 batters get out without anyone scoring (i.e strikeouts), this would be 6 batters total receiving a plate appearance. In the scenario above, the pinch hitter or the 7th 'batter' would not receive any official stat other then "+1 in games played"
This must be from the guy who asks how may ways a batter can score from third without a hit. 45 is the answer. Example. First batter singles. Second batter hits into a double play. Next three batters all have infield singles to load the bases. Sixth batter's ground ball strikes the base runner going from 1st-2nd. The batter is out, the ball is dead runners may not advance and the batter gets credited with a base hit. So ... five hits in the inning. The team plays eight more that go the same way. 9x5=45.
For this explanation, I will assume one batter is equal to one plate appearance and since the number of innings is not specified, I will assume a 9 inning game. These assumptions would make the answer 112.The maximum number of batters in a half inning where no runs were scored would be 6 (3 outs, 3 runners left on base), the maximum number of batters in a half inning, other than the bottom of the final inning, where 1 run was scored would be 7 (3 outs, 3 runners left on base, 1 run scored), and the maximum of batters in a half inning, other than the bottom of the final inning, where 2 runs were scored would be 8 (3 outs, 3 runners left on base, 2 runs scored).The visiting team would score 2 runs in 1 inning and no runs in 8 innings or 1 run in 2 innings and no runs in 7 innings. This would give them a total of 56 batters regardless of whether they scored 1 run in 2 innings ((7 * 6) + (2 * 7) = 56) or 2 runs in 1 inning ((8 * 6) + (1 * 8) = 56).Through the home team's first eight at bats, they would score 2 runs in 1 inning and no runs in 7 innings or 1 run in 2 innings and no runs in 6 innings. This would give them a total of 50 batters entering the bottom of the ninth regardless of whether they scored 1 run in 2 innings ((6 * 6) + (2 * 7) = 50) or 2 runs in 1 inning ((7 * 6) + (1 * 8) = 50). In the bottom of the ninth, the home team would score the run to give them the 3-2 win. The maximum number of batters would be 6 (2 outs, 3 runners left on base, 1 run scored). This would give the home team a total of 56 batters.Both teams would send 56 batters to the plate for a game total of 112.
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