Cricket Puzzle of the Day!

[Zeeshan]

This is equivalent to Fermat's Last Theorem of Cricket. (Found on the net)

Two batsmen on 94* each. 2 balls remaining. 7 runs to win. Both gets his/her century and wins the match in the process. How?

[zinatf]

with the help of wd/no balls

[Naimul_Hd]

1 ball, batsman X takes 3 runs and fielder overthrows to 4. However, batsmen run short 1 in the process, So, total runs 6 (goes to batsman X). Now target 1 off 1 ball and batsman Y is on the strike (batsman X 100, batsman Y 94)

Last ball, batsman Y hits 6. Match over. Batsman X 100, Batsman Y 100


Eazzy Pizzy

[Naimul_Hd]

Another scenario :

1 ball, batsman X edges and keeper drops the ball and ball hits the helmet on the ground (behind the keeper). However, before the ball hits helmet, batsman takes 1 run. So, batsman X gets 1 + 5 penalty =6 runs. (Batsman X 100, Batsman Y 94) Now, batsman Y is on the strike. Target 1 run off 1 ball.

Last ball, Batsman Y hits 6. Match over, Batsman X 100, Batsman Y 100.