Kingstown, St Vincent, Mar 21: This is one of the rarest of rare instances in the game of cricket. A batsman was bowled twice in two successive balls, but still he was not out.
Wondering how this happened? Yes it is true. This happened in the One Day International between Australia and West Indies at the Arnos Vale Sports Complex on Tuesday. And the batsmen who was so lucky was West Indies' Andre Russel.
The match itself was a thrilling affair as it ended in a tie in the last over. West Indies, chasing 221 to win, were all out for 220 in 49.4 overs when captain Darren Sammy was run out.
This interesting incident of Russell took place in the 40th over.
Russell was first bowled by Australian skipper Shane Watson. As he celebrated his 150th ODI wicket, umpire had doubt whether Watson had delivered a legal ball or not.
He sent the request to third umpire when the replay showed Watson to have overstepped. It was a no-ball. A chance for Russell on the first ball of the 40th over. Watson broke through Russell's defence in the very next ball, but that was not counted as out as it was a free hit where a batsman cannot be dismissed in this manner other than a run out. Russell's chances came when he was batting on 31.
"It was very disappointing especially at that crucial time as well," Watson said after the match, talking about the incident.
"To bowl a guy out twice in two balls and actually have zero result out of it and giving him a run, very disappointing. When I checked the footmarks, I knew I stepped over. That was not a good feeling ... it's not good enough," Watson added.
Russell finally was out for 37, caught behind by wicketkeeper Matthew Wade off Clint McKay.