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Former India cricketer Irfan Pathan picked the side's playing XI for the first T20I of the five-match series against Australia that will take place on October 29 in Canberra. He decided to make a big change in the team and dropped star spinner Kuldeep Yadav for Harshit Rana.
Pathan's reason behind including Harshit
The left-arm spinner had been the Men in Blue's top performer in the Asia Cup, picking up 17 wickets in seven matches. However, Pathan feels that Harshit can go on to contribute with the bat as well and this can see the team management select him in the XI. He mentioned the pacer's 24-run cameo against Australia in the second ODI to point out that he can indeed bat. Moreover, he opted to go for three pace and two spin options and decided to make Varun Chakravarthy the frontline spinner over Kuldeep.
“Harshit Rana’s cameo in the second ODI would have given the Indian team management a lot of confidence. So if he plays at No.8, India have their order till No.7 sorted, with two all-rounders in Dube and Axar, who can give you at least 6 overs. Then you can also play 2 pacers in Arshdeep and Jasprit, but who will the frontline spinner be?" Pathan said on his YouTube channel.
“I am thinking towards Chakaravarthy because he can bowl a couple of overs in the powerplay as well, if needed. India also need that bit of a mystery factor in their playing XI," he added.
Coming to the batting order, Pathan decided to go for the same top 6 as the Asia Cup. He also gave a chance to left-arm pacer Arshdeep Singh in his squad. Notably, he had only played two matches in the Asia Cup, but with Australian conditions generally favouring the pacers, India may need his services. The Men in Blue can also use Shivam Dube in the bowling department if required.
Irfan Pathan's playing XI for 1st T20I against Australia: Abhishek Sharma, Shubman Gill, Suryakumar Yadav, Tilak Varma, Sanju Samson, Shivam Dube, Axar Patel, Harshit Rana, Arshdeep Singh, Jasprit Bumrah, Varun Chakravarthy
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