India have named Shardul Thakur in their XI for the WTC Final against Australia at The Kia Oval, which starts today (June 7), leaving out R Ashwin from their playing XI.

India have decided to go in with Thakur over Ashwin, with Rohit Sharma citing at the toss that it was a difficult decision to leave a ‘match-winner’ like Ashwin out, but they had to do so looking at the conditions on offer. On commentary, former coach Ravi Shastri said it was a 50-50 decision, and depended on the weather in the morning.

IND vs AUS, live score: Follow the live scorecard here. India have won the toss and are fielding first.

There were many questions and debates around India’s team selection ahead of the WTC Final. More specifically, the discussions concerned whether India should pick a second spinner in Ravichandran Ashwin or go in with four seam bowlers with Shardul Thakur being the fourth one.

India, however, decided to stick with their policy of four seamers and one spinner outside the subcontinent which they adopted after the last WTC Final against New Zealand.

Overall, R Ashwin has taken 114 wickets at 28.36 against Australia in Tests, with seven five-fors. In England, he’s taken 18 wickets in 7 Tests at 28.11.

“It is tough to leave Ashwin out. He has been a match-winner for us, it is not a great occasion to leave him out, but you have got to do what is best for the team in the given conditions,” said Rohit Sharma at the toss.

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As anticipated, this selection call has already garnered lots of divided reactions on social media, and will continue to do so based on how Thakur goes in the game.

 

Related tags:

India R AshwinShardul ThakurSeries EditorialWorld Test Championship 2021/23