Tuesday, October 23, 2007

Trick question

Two batsmen are on 94 not out each.

Three balls left in the match.

Seven runs to win.

Both finish on 100 not out.

How?

6 comments:

Boredomboy said...

Lol... I can't belive I am bored enough to answer this.

Batsman 1 hits a six, reaches century. The next ball is a leg bye (or a bye) and the batsmen cross and then Batsman 2 hits a six.

OR

The second last ball is a wide where both batsmen cross and the next one is a six and the last a dot ball or the next ball is a dot and the last is a six.

OR

Ok I m bored there are too many scenarios

The Not So Talkative Man said...

Incorrect

1) the batsmen take the bye/lb after hitting a six, GAME OVER. It's the seventh run, and that's all there's left to play for.

2) In the 2nd scenario, both batsmen don't reach their hundreds.

Australopithecus said...

batsman 1 hits a 4..then they run 3..but its one short..so 6 runs to basman 1. 1 run to win. batsman 2 hits six. reaches 100. ?

The Not So Talkative Man said...

Sahi jawaaaab!

The Not So Talkative Man said...

There's another way possible.

Batsman A hits 3, and then gets 4 overthrows. Umpire calls one short. A reaches 100 not out.

Strike rotated. B takes strike, scores level. He hits six. 100 not out.

Game over.

falguncontractor87 said...

1st ball: wicket keeper tries to stop ball with helmet or thowin gloves so\(5 run penlty)
2nd ball: he takes 1 run
3rd ball : other bats man hits six