Beaten thrice, Mumbai Indians still hold the edge against Supergiant

HYDERABAD: Mumbai Indians have entered their fourth IPL final, and will aim to become the first team to win the title for a third time when they face Rising Pune Supergiant at the Rajiv Gandhi Stadium here on Sunday.

However, Rohit Sharma’s boys will have to end their losing streak against Steve Smith’s Rising Pune Supergiant, having lost all three matches in this edition.

Heading into the final, Mumbai Indians hold an odd statistical advantage. For the third time, two teams will face off for the fourth time in an edition. Mumbai Indians have featured in all three. The first two was against Chennai Super Kings in 2013 and 2015 and on both occasions, Rohit Sharma’s team won the title.

In 2013, the head-to-head for the season ended 3-1 in Mumbai Indians’ favour. In 2015 too, Mumbai Indians finished with a 3-1 head-to-head record over Chennai Super Kings, rallying after losing the first meeting.

That shows no IPL team has won four matches in a single edition against another team. With Rising Pune Supergiant having won the first three, two league matches and the Qualifier 1, Mumbai Indians will be hoping the head-to-head tradition continues on Sunday. Breaking the Australian jinx However, Rohit Sharma will have to break another jinx to lift the trophy a third time. In 10 editions of the IPL, no team led by an Australian has lost in the final. Rajasthan Royals, captained by Shane Warne, won the inaugural edition in 2008 against Chennai Super Kings while Adam Gilchrist helped Deccan Chargers clinch victory in IPL 2009. The domination by Australian captains continued in IPL 2016 when David Warner, leading Sunrisers Hyderabad, got the better of Virat Kohli’s RCB.

This time, it will be Steve Smith’s side that Rohit Sharma’s Mumbai Indians will have to beat. Will they do it? The answer will come on Sunday.