Williamson wins toss, opts to bowl against India in first ODI

PTI| Auckland | United Kingdom

Updated: 25-11-2022 06:46 IST | Created: 25-11-2022 06:46 IST

New Zealand skipper Kane Williamson won the toss and elected to bowl against India in the first One-Day International, here on Friday.

India have handed debuts to Arshdeep Singh and Umran Malik and are fielding both Rishabh Pant and Sanju Samson.

The visitors are playing with three specialist fast bowlers while the hosts are going in with four of them. Williamson had missed the final match of the T20I series due to a medical appointment but is back to lead the side.

India had won the three-match T20 series 1-0.

Teams: India: Shikhar Dhawan (C), Shubman Gill, Rishabh Pant, 4 Shreyas Iyer, Suryakumar Yadav, Sanju Samson, Washington Sundar, Shardul Thakur, Umran Malik, Arshdeep Singh and Yuzvendra Chahal.

New Zealand: Kane Williamson (C), Finn Allen, Devon Conway, Tom Latham, Daryl Mitchell, Glenn Phillips, Mitchell Santner, Adam Milne, Matt Henry, Tim Southee and Lockie Ferguson.

(This story has not been edited by Devdiscourse staff and is auto-generated from a syndicated feed.)

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Sanju SamsonSinghLockie FergusonDevon ConwayRishabh Pant and Sanju SamsonWilliamsonFinn AllenTom LathamRishabh PantMitchellNew ZealandAdam MilneKane WilliamsonMatt HenryDhawanWashington SundarIndiaShardul ThakurSoutheeShreyas Iyer

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