IPL 2021, MI vs SRH: Rohit Sharma elects to bat

Last updated on Apr 17, 2021, 07:03 pm
IPL 2021, MI vs SRH: Rohit Sharma elects to bat

Mumbai Indians are squaring off with Sunrisers Hyderabad in the ninth game of Indian Premier League (IPL) 2021. Rohit Sharma has won the toss and elected to bat in the riveting encounter at Chepauk. The defending champions will be high on confidence after stealing a victory against Kolkata Knight Riders, while SRH are still in pursuit of their maiden win this season. Here is more.

Details

Chasing has been difficult at Chepauk

The MA Chidambaram Stadium in Chennai is hosting the match. Interestingly, the teams haven't been able to chase on this venue in the last three games. SRH (vs RCB) and KKR (vs MI) lost from winning spots in the previous two games respectively. The wicket in the incumbent game once again will be on the slower side, thereby making the spinners effective.

H2H

A look at the head-to-head record

The competition between Mumbai Indians and Sunrisers Hyderabad has been neck-to-neck in the IPL. Both the sides have won eight times each, having faced each other in a total of 16 games. Notably, MI and SRH won a game against each other last season. MI won the first game by 34 runs, while SRH won the next one by 10 wickets.

Stats

SRH vs MI: Stats that matter

Leg-spinner Rashid Khan has had a wood over Hardik Pandya in the IPL, having dismissed him twice. Meanwhile, Pandya has scored only 18 (33) against Rashid. SRH skipper Warner could become the first batsman with 50 half-centuries in the IPL. Meanwhile, Kieron Pollard needs two maximums to become the fifth player with 200 sixes in the IPL.

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