In the last trip between the two groups last season, Mumbai won by 42 runs.
In the 65th match of the Indian Premier League (IPL) 2022, Mumbai Indians (MI) will take on Sunrisers Hyderabad (SRH) on Tuesday, May 17, at 7:30 PM at the Wankhede Stadium in Mumbai.
The five-time IPL champions have become the first team to be ruled out of consecutive IPL seasons. After losing eight games in a row, the opposition’s best group did not track their musicality throughout the season and surrendered to their difficulties. A short circle is back as Mumbai Indians have won three apparatuses in their last four matches and will jump at the chance to complete the season with an optimistic outlook. Kishor, Tilak Verma are finding time for MI and will depend on the southpaw to dominate them in the future. The Mumbai establishment played 12 games, lost nine and won three times. Mumbai has secured the tenth position with six places.
Again, Sunrisers Hyderabad have almost lost their chance of meeting all the requirements for end of season games. The 2016 owners had a significant circle and secured a spot in the main four after dominating five consecutive matches prior to the season. SRH could not figure out how to stay with the force going forward as they are still on a five-match long string of failures. Kane Williamson’s absence of commitment with the bat has been the key explanation for the group’s disappointing batting performance. The Hyderabad-based establishment has played 12 games so far, of which it has won five and lost seven. SRH have included the eighth position with 10 places.
In a direct rivalry between the two groups, MI has a slight advantage. The two groups crashed in 17 events, with Mumbai dominating nine matches, while Hyderabad were successful in eight events.
In the last tour between the two groups last season, Mumbai had won by 42 runs.