First batsman hits the ball and run for a run. The ball hits on the helmet kept behind the wicket keeper before fielded. Thus the batsman gets 6 runs and he completes century.
The next ball is faced by other batsman who has crossed over. He hits the next ball for a six. He also completes century and the team wins.
2007-04-30 06:01:43
·
answer #1
·
answered by vakayil k 7
·
2⤊
1⤋
I can only think of one way that this is possible. I note that you do not specify how many wickets have fallen. I'm assuming less than nine have fallen.
Batsman #1 on strike hits the ball towards the boundary and the batsman run 3 runs and continue for a fourth run. The ball is cutoff by a fielder who throws in wildly, and the ball is overthrown. The runners complete the fourth run and continue running completing 2 more runs (6 runs) and continuing for an additional run. A fielder fields the overthrow and throws in knocking the stumps of the non-strikers end before the batsmen complete the seventh run. Batsman #1 has scored six (completing his century) but is runout going for the seventh run.
A new batsman comes in for the last ball, but he is at the non-strikers end. Batsmen #2 is now at the strikers end to face the last ball needing only one run to win, and he hits a six, completing his century and winning the game.
Vakayil, you're idea is good, but I believe if the ball hit the helmet behind the wicket keeper the batting team would receive 5 PENALTY runs, which would not be attributed to the batters total.
Malfoy, if there is one wide or no ball then the first batsman to hit a six wins the game. There is no way that they can both hit a six in that scenario.
Damsel, if the first batsmen hits a six, he will still be onstrike for the last ball. How will the other batsmen get the strike without the last ball being bowled or another run (including and extra) being scored. I don't think your scenario works either.
2007-04-30 08:26:54
·
answer #2
·
answered by dsl67 4
·
0⤊
0⤋
Batsman #1 on strike hits the ball towards the boundary and the batsman run 3 runs and continue for a fourth run. The ball is cutoff by a fielder who throws in wildly, and the ball is overthrown. The runners complete the fourth run and continue running completing 2 more runs (6 runs) and continuing for an additional run. A fielder fields the overthrow and throws in knocking the stumps of the non-strikers end before the batsmen complete the seventh run. Batsman #1 has scored six (completing his century) but is runout going for the seventh run.
A new batsman comes in for the last ball, but he is at the non-strikers end. Batsmen #2 is now at the strikers end to face the last ball needing only one run to win, and he hits a six, completing his century and winning the game.
2007-04-30 09:10:44
·
answer #3
·
answered by ? 3
·
0⤊
1⤋
u havent mentioned the score is individually 94 or their partnership is of 94 if it is partnership then simple if individual then the first played the ball take 1 run . but suddenly due to over throw they ball misses stumps and crosses boundary. hence now there are two balls and one is on 99 and other is on 94 and runs needed are now 2 runs , so he takes a single run and completes century , the other one hits a six and completes his century . HOPE THIS IS THE ONLY ANSWER . AND FIRST ANSWERED BY ME SO THE BONUS IS TO ME
2016-05-17 08:31:02
·
answer #4
·
answered by ? 3
·
0⤊
0⤋
both hit a sixer on each ball n they crossed on a dead ball or they made a run which was denied due to running on pitch.
2007-04-30 07:36:49
·
answer #5
·
answered by Anonymous
·
1⤊
1⤋
Wides and no balls and each of them hits a six.
2007-04-30 04:49:53
·
answer #6
·
answered by Malfoy vs Potter 5
·
0⤊
1⤋
Not possible man. Is this a question for us or u already know the answer and u're asking us?
2007-04-30 04:29:30
·
answer #7
·
answered by ? 2
·
0⤊
1⤋
That's not possible.
2007-04-30 05:16:01
·
answer #8
·
answered by www.webhosting.uk.com 1
·
0⤊
1⤋
miracle:D
2007-04-30 03:48:17
·
answer #9
·
answered by Ariba 2
·
0⤊
1⤋
it cannot happen
2007-04-30 03:41:40
·
answer #10
·
answered by Titan 4
·
1⤊
1⤋