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New Zealand all out for 185 in 40.2 overs at Eden Park after Rohit's diving run-out of Mitchell
Rohit Sharma's dive and throw ran out Daryl Mitchell for 28, and India's bowlers kept New Zealand's lower order under pressure to bowl the hosts out for 185 in 40.2 overs of the first ODI in Auckland. India need 186 to win.
Key points
- New Zealand were all out for 185 in 40.2 overs; India need 186 to win.
- Rohit Sharma ran out Daryl Mitchell for 28 in the 22nd over.
- The tail added 70 runs in ten overs after 115 for 7.
New Zealand’s lower order put up a fight in the first ODI at Eden Park. After slumping to 115 for 7 in the 29th over, the hosts reached 185 for 9 after 40 overs, and the last wicket fell two balls later: New Zealand all out for 185 in 40.2 overs. India need 186 to win.
The middle of the innings had belonged to India’s fielders. In the 22nd over Daryl Mitchell, on 28, pushed the ball into the covers and set off for a run. Rohit Sharma dived, gathered and threw to wicketkeeper KL Rahul, who completed the run-out. Mitchell had been New Zealand’s best hope of building an innings after the new-ball burst from Arshdeep Singh and Gurnoor Brar.
That moment came between two others that will be replayed tonight: Brar’s one-handed return catch to remove Tom Latham and Virat Kohli’s diving catch to send back Nathan Smith. New Zealand were 137 for 8 by the 35th over, and a total of 200 looked a long way off.
The tail hits back
The last pair of wickets has proved harder to shift. New Zealand added 70 runs in about ten overs from 115 for 7, using the short straight boundaries that make Eden Park one of the trickiest grounds in the world to defend on.
Shubman Gill chose to bowl after a rain-delayed toss, and his attack has done most of what he asked. India will now want to finish the innings quickly and give Rohit, Gill and Kohli a chase they can pace on a bouncy surface under lights.