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"Discussion behind this answer
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.
Sorry but its not possible without cheats.
Well, first comes the first inning. After that usually comes the second inning, followed by the third inning. Provided it isn't raining, then come the fourth, fifth and sixth innings, respectvely. Innings seven, eight and nine come last, in that order.
He sleeps at night
the answer is 0wont run at all without windingIt can't.