IPL 2019 final: McClenaghan in for Jayant, MI bat against unchanged CSK
Either MS Dhoni or Rohit Sharma will lift the IPL title for the fourth time. (AFP Image)

In the big final of the IPL 2019, it’s the Mumbai Indians who have won the toss and opted to bat first against an unchanged Chennai Super Kings. MI skipper Rohit mentioned that the side has made one change, bringing in fast bowler Mitchell McClenaghan ahead of offspinner Jayant Yadav. (ALSO READ: IPL Final Live Updates)

“We are going to bat first. It’s a big game, and we prefer to bat. Good pitch, there might be some help first for the spinners and fast bowlers, but we are prepared. We have batted on tacky wickets throughout the season, so we don’t mind batting first,” Rohit said at the toss.

“We have been on the road for a while. We came here right after the qualifier in Chennai. A lot of guys have an off-day, many came out for training. I would look at that as a positive. A few days off have done us some good. We have one change: Mitch McCleneghan comes in for Jayant Yadav – the conditions demand a fast bowler ahead of a spinner, so an unforced change.”

It wasn’t a bad toss to lose for MS Dhoni, the CSK skipper, who revealed he’s wanted to bowl. “We were looking to bowl first. If the result is in your favour, then it’s fine, otherwise they’ll say the guys are fatigued,” he said.


Chennai Super Kings (Playing XI): Faf du Plessis, Shane Watson, Suresh Raina, Ambati Rayudu, MS Dhoni (capt/wk), Dwayne Bravo, Ravindra Jadeja, Harbhajan Singh, Deepak Chahar, Shardul Thakur, Imran Tahir

Mumbai Indians (Playing XI): Rohit Sharma (capt), Quinton de Kock (wk), Suryakumar Yadav, Ishan Kishan, Krunal Pandya, Hardik Pandya, Kieron Pollard, Rahul Chahar, Mitchell McClenaghan, Jasprit Bumrah, Lasith Malinga