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Two batsmen are batting on 94 not out. The team needs 7 runs to win with only 3 balls left. Both batsmen remain unbeaten on 100. How is it possible?

Hi Shin,

Here is the corrected version of your input:

Ball 1: 1st batsman - got six and reached 100.

Ball 2: 1st batsman - trying for six but lost the wicket by a catch. Both batsmen crossed each other (no runs, still need 1 run to win, but 2nd batsman on strike with 94).

Ball 3: 2nd batsman got six and reached 100 - won the match too.

Hope this would be the answer.

Regards,
Shin

Two possibilities:

First scenario:
The 1st batsman hits the ball and runs for 3 runs, but one run is a short run. The fielder throws the ball, and it is overthrown for 4 runs - actual runs scored = 6 because of 1 short run. Now, the 2nd batsman is on the crease, and he hits a six on the next ball. Both batsmen reach 100, not beaten.

Second scenario:
It starts raining at 46.5 overs, and the game is stopped for some time. According to the Duckworth-Lewis rule, they have to score 7 runs off 3 balls to overcome the lost time. The first batsman hits a 6 and reaches a hundred. At the start of the new over, the 2nd batsman is at the crease and hits a six, making it a century for him. Both batsmen remain unbeaten on 100. :)

Mahesh,

Hi.

I believe there are many possibilities and ways to solve it.

A. One six and other balls just played without getting any run, thus making 94+6+0+0 = 100.

B. 2nd Solution - score of 3 balls as under: 4 - 2 - 0.

C. 3rd Solution - score of remaining 3 balls as: 4 - 1 - 1.

Supriya's answer is apt too.

These are the possible answers that I could think of. I stand to be corrected! Well, what's the answer?

Hi Shalu,

It seems you did not read the question properly. BOTH batsmen reached 100 UNBEATEN. So, your answers are not correct, but well tried. :)

Yeah, of course Shalu... trying is more important than anything else. You tried well... :)

Mahesh,

Hi.

I believe there are many possibilities and ways to solve it.

As three balls are remaining:

Ball 1 = 4 runs

Ball 2 = Wide ball + 2 runs

Batsman 1 = 100 runs

Ball 2 = Byes batsman cross 1 run to the team.

Ball 3 = Batsman 2 hits a six.

Hope it's correct.

Supriya's answer is apt too, right?

Dear Mahesh,

Really, a great one, in fact, which I had received through SMS long back, but I just left it unanswered. Today, it was not the same for me. From the moment I saw the question, I was trying to find out the solution. Well, here is my reply:

1st Ball --- The 1st Batsman hits the ball for SIX and completes his 100.
2nd Ball --- The 1st Batsman hits the ball and runs for a short quick single, but after crossing over, he collides with the fielder, gets injured, and has to retire hurt. Here, the fielding side will not appeal as it was a case of injury to the batsman (I hope you all remember the recent incident related to this that happened in the England vs. New Zealand match). Therefore, a new batsman comes to the middle.
3rd Ball --- The 2nd Batsman hits the ball for SIX to complete his 100 and also the match.

This is one such possibility that I thought of. In case of any other scenarios that can happen, I will post those as well. I hope this will not be a shortened match as mentioned by Supriya. Let me know about my reply.

Regards,
Amith R.

Well, according to me --

The 1st batsman hits a 4 on a "No Ball," which is equal to 5 runs, and on the next ball takes a single run. Now, for the last ball, the 2nd batsman comes on the crease to bat, and he hits a six. The match is won by them; both batsmen are on 100 UNBEATEN.

Mahesh, waiting for your reply...

Dear Chandrakanth,

Great try, but one thing I want to point out here is that when you have described Ball 2 as a wide ball, those runs will not be counted for the batsman. Furthermore, if you consider it as a No Ball, the total runs scored will exceed 7, which is the required number of runs for the team to win. I hope you agree with this clarification.

Regards,
Amith R.

Dear Swathi,

Well, I would say the same again - great try. However, when the first batsman hits a four off the first ball, it will only be 4 runs for him, and any No Ball runs will only be counted towards the batting side. If he takes a single off the next ball, he will remain not out on 99 and will not reach 100.

Regards,
Amith R.

Hi Chandrakant,

According to your answer:

"Ball 1 = 4 runs Ball 2 = Wide ball + 2 runs Batsman 1 = 100 runs Ball 2 = bys batsman cross 1 run to team."

Match would end here only, my dear, since only 7 runs are required...

Hi, I am Neer.

First batsman scored six on the first ball, and the second ball is a no ball. The batsman took a single run. Then the second batsman scored either a six or a maximum of four plus two runs.

Could you please confirm if this is correct?

Hi Amit, you have assumed the shortened innings of the batsman, and I have assumed a shortened match due to rain! :)

The first batsman scored six on the first ball, and the second ball is a no-ball. The batsman took a single run (leg bye), which is not counted on the player's account.

It seems Supriya and Amar are right. Special congrats to Amar because you came up with a different solution. More answers may be out there... Try out. Good luck.

Dear Supriya,

Yes, you are right. In fact, I am thinking about your first possibility. Anyways, great try and keep posting if you get any other options...

Thanks, Mahesh. I will certainly try to come up with different solutions, and if we do, we will post it.

Regards,
Amith R.

Dear Beniwalneer,

In your case, if the first ball goes for a SIX and the second is a NO BALL, then the match would end, and the second batsman will not get a chance to complete his 100. Nevertheless, great try, which is important.

Regards,
Amith R.

What if the situation is "2 balls left instead of 3"? Can anybody suggest any other options than those suggested by me?

Thanks Supriya for thinking out of the box. Good question!

But for my previous question, many answers came. Thanks to all who replied. Some of the answers are given:

1st ball: wide and a 4.
Total now is 5 runs.
2nd ball: the batsman takes a single. Total is 6 runs.

3rd ball: the 2nd batsman on strike scores a six, and hence he gets his century. So both get their centuries! Hope I guessed it right.

Note: Here the wide goes to the team, not the batsman. So the criteria of both batsmen hitting 100 is not met.

Another answer was:

Batsman A, whose score is 94, scores and reaches his hundred. (If the ball hits the bat and then the helmet placed behind the keeper, the batsman is awarded 5 runs.)

Batsman B is now at the crease. The bowler is going to bowl the last ball of the innings. The team needs 1 run to win, and the batsman hits a huge six and gets up his hundred. What a winning moment!

Note: Many answers include:
- No-balls. Even one no-ball will mean that the match is over as soon as one batsman gets his century, not giving the other one a chance.
- Ball hitting a helmet on the field. The batting team would be awarded five penalty runs, which are awarded as extras, not to the batsman's score.

Now let us come to Supriya's question:

- 7 to win, 2 balls to go, Batsman A (94*) on strike:
Ball is struck, the batsmen run three, but one is disallowed as an accidental short run (because one of the batsmen accidentally fails to make his ground before turning and running back). The fielder chasing the ball then tries to return the ball to the wicket-keeper (the ball is still live at this point), but the keeper doesn't get hold of it, and the ball continues to the boundary for four overthrows. Thus, the batsman scores six for his century, and Batsman B is now at the crease.
- 1 to win, 1 ball to go, Batsman B (94*) on strike:
Ball is hit for six. Batsman B gets his century, and his team wins the match.

In another way: Batsman A takes 7 runs by running out of which he unintentionally makes a short run. It gives only 6 countable runs. A moves to the non-striking end, A finishes his ton, and his team requires 1 more run to win.

Now, B does what he needs to do to finish his 100 runs and what his team requires.

Note - the short run must be accidental, since in an intentional short run, all runs are disallowed.

Ball 1 - One batsman edges the ball, and it hits the helmet behind the keeper, so five runs as per the rules and not byes because it's edged from the bat of the batsman, so it's counted as his runs and not byes.

Ball 2 - Single run taken by Batsman 1, and he reaches his hundred.

Ball 3 - The next Batsman hits it for six, and he also reaches a hundred.

Note: As ball 1 is edged and hits the helmet, it's counted as the batsman's runs and not byes as per the rules. It's as simple as an edge not taken by the keeper and goes to the boundary; it is counted as four by the batsman and not four byes.

Regards,
Kamal Modi

Dear Friends,

Great thinking by Supriya, and I guess the second answer from Mahesh and Kamal is also apt to be the correct answer. However, I believe in the first answer, where the first ball is a wide, you cannot hit it for four. Wide will be given when the ball is out of the reach of the batsman. Anyways, great question! Let's keep it going to see if we can find any other solutions for the same.

Regards,
Amith R.

Hi Kamal, good try, but penalty points by any means go to the team only and would not be counted in the batsman's account. Anyways, great try, keep posting.

I have figured out one more solution...

1st ball: Batsman A hits the ball for 6 and gets his muscle pulled in doing so, but since he's hit a six, he completes his century. He is retired hurt. New Batsman takes over.

2nd ball: The new batsman hits the ball and runs for 1 run but is caught out. However, the two batsmen have already crossed each other, so now Batsman B is on strike.

3rd Ball: Batsman B hits the ball for 6, completes the century, and wins the game. :)

Dear Supriya,

This is really a great answer. But with Kamal's answer, I guess the runs will certainly go to the batsman also. Let me get it clarified. Nevertheless, another great answer...

Regards,
Amith R.

Dear Supriya,

As the ball's edged by the batsman, it's always counted as the batsman's runs rather than extras. If he would have missed the ball and it would have hit the helmet, then it would have been five byes. Good alternate answer by Supriya again. It shows that a problem can be solved by various means; the only difference is to put in an effort to do so.

Kamal.

First batsman hit a six off the last ball of the running over. When the over changed, the batsmen swapped; the first batsman moved to the non-striking end, and the second batsman hit another six. As a result, both batsmen remained not out and completed a century with one ball in hand.

Sorry to disappoint you, Amith, but the runs would not go into the batsman's account; rather, they will be in the team's account only. I am sure about it. Please check again from some sources.

Can you guess which festival this is...? The answer is on the way.

Regards,
Mahesh

Dear Supriya,

No problem, I don't see anything to get disappointed with this. Let me check and come back to you on this.

Regards,
Amith R.

Sure Amith, if you find something different than this, please do increase my knowledge as well.

Regards,
Supriya