[ad_1]
Suryakumar Yadav led India to a two-wicket over Australia in the first T20I game at Visakhapatnam on Thursday. It was a dramatic game, which saw India losing three wickets — Axar Patel, Ravi Bishnoi and Arshdeep Singh — in the final over as they chased a total of 209.
With the fate of the game on the line, the hosts needed one run off the final ball. Rinku Singh, batting on 22, was on strike, facing Sean Abbott.
As it transpired, Rinku slammed the ball from Abbott over long on for a maximum as Indian fans erupted in celebration. But the umpire then signalled a no ball because Abbott had overstepped.
While India’s win was secure, the six runs were not added to Rinku’s total.
Here’s why Rinku Singh’s last-ball six against Australia will not count
According to ICC Men’s T20I Playing Conditions, clause 16.5.1: “As soon as a result is reached as defined in clauses 16.1, 16.2 or 16.3.1, the match is at an end. Nothing that happens thereafter, except as in clause 41.17.2 (Penalty runs), shall be regarded as part of it.”
This rule means that as soon as the no ball was bowled, India had reached its target and the six runs wouldn’t stand.
Unfortunately, for Rinku if India had needed more than one run to win, the six runs would have stood.
The rulebook adds: “If a boundary is scored before the batters have completed sufficient runs to win the match, the whole of the boundary allowance shall be credited to the side’s total and, in the case of a hit by the bat, to the striker’s score.”
[ad_2]