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Assuming there are no overthrows and no extras (like no-balls, wide balls, etc), what is the highest a batsman score in an over?

UPDATE: If the ball first hits the bat, then the helmet and goes for a boundary, he will score 5+4 =9 runs/ball. So, in an over, 9*6 = 54 runs. Is this logic correct?

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I don't know if he gets 9 (5+4) runs in one ball. But if I am a captain, and if the ball touches the helmet back to back (or twice), I will send the helmet into the pavilion! – hims056 Jan 22 '13 at 11:06
@hims056 hehe yeah...but it is possible..may be! – joey rohan Jan 22 '13 at 11:08
The 5 runs for hitting the helmet are penalty runs. They don't go to the batsman's personal score. – Peter Eisentraut Jan 22 '13 at 17:05
I don't 9 runs is possible.theoretically,consider this scenario, the batsmen has already ran 6 max possible(if ball not last) then the fielder throws the ball on to the helmet , the ball gets deflected and reaches the boundary then according to you the team would get 6+5+4=15 runs. I guess the ball will be called dead once it touches the helmet – gout Jan 30 '13 at 7:14
Penalty runs are also classed as extras, and as @PeterEisentraut points out, they do not count towards the batsman's score. Therefore 6*6 = 36 runs would be the maximum a batsman could score from a 6-ball over, without overthrows or running more than 6. – Spinner Aug 16 '13 at 13:03
up vote 10 down vote accepted

The number of runs you can score of a single delivery is in principle unlimited, because the batsmen can keep running forever. This was especially possible before the introduction of current Lost Ball rule, for example when the ball got stuck in a tree.

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I am not sure, I could follow you properly. How can a batsman get unlimited run of a single delivery? – Mistu4u Jan 23 '13 at 17:12
@Mistu4u - there is no limit to the number of runs the batsmen can run between the wickets, as long as someone doesn't run them out - 6 is not a limit. This was especially true before the Lost Ball rule - some grounds have trees within the boundary, where a ball could theoretically get lost in. For most circumstances however 6 runs is the limit per ball. – iandotkelly Jan 23 '13 at 20:42

If all balls are fair there will be 6 balls in the over. If one batter faces all of these and gets the maximum score possible each ball, which is 6 runs for hitting the ball on the full over the boundary, then they will score 36 runs in the over.

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If the ball does not cross the boundary, however unlikely, is it possible for the batsmen to run more than 6 runs off one delivery? – Orangecrush Jan 22 '13 at 9:18
True, it is possible, though the exclusion of overthrows makes it even less likely. However I guess you're right, if all 11 players in the fielding side spontaneously break their legs running for the ball the batsmen could, in theory, keep running for ever.... – Bogdanovist Jan 22 '13 at 9:20
I think the relevant rules are here… I can't see anything that limits the number of runs scored by running between wickets, so yes I think you're correct, you could in theory score more than 6 runs in a ball in this way. – Bogdanovist Jan 22 '13 at 9:32
@Orangecrush see update – joey rohan Jan 22 '13 at 10:54
And yeah,not those kind of tricks,running forever and stuff.. – joey rohan Jan 22 '13 at 10:55

i think so that there can be 48 made in a single over...four run by running nd while running the fourth run if filder overthrow nd it goes out for boundry then by ths way there is 8 runs per ball and 8*6=48

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Thanks for your answer, but the question specifies that "there are no overthrows." – Spinner Nov 26 '13 at 20:54

Its 1751 in 50 overs 5 sixes every over + 5 runs running to get strike in next over = 35 Total runs = 35*49 + 36 ( 6 sixes in last over) = 1751

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The question is: "How many runs can be scored in a single over?" – hims056 Aug 7 '13 at 6:45

5*6 runs + 3 runs in last ball and last over 6*6 runs = 1653

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The math does not appear to total 1653. Where does that number come from? – Michael Myers Oct 9 '13 at 17:17
The question is: "How many runs can be scored in a single over?" – hims056 Oct 10 '13 at 4:30
@MichaelMyers I believe sachin means ((5 [first 5 balls of each over] x 6) + 3 [3 runs off the last ball of each over except the 50th]) x 49 [overs]) + (6 x 6 [all 6 balls of the 50th]) = 1653 total runs in the team's innings. However, as hims056 pointed out, this is not the answer to the original question. – Spinner Oct 10 '13 at 15:20

protected by ᴍᴀsᴛᴇʀᴍɪɴᴅ_ᴇᴅ Jan 10 '14 at 14:26

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