Here’s how Mumbai Indians hold edge against Pune Supergiant

Indians have entered the final of IPL 2017 after their commanding six-wicket win over Kolkata Knight Riders in the second qualifier at the M Chinnaswamy stadium in Bangalore on Friday.

Rohit Sharma’s boys have entered the IPL final for the fourth time and they will be aiming to become the first team to win the title three times. However, they will first have to break their losing streak against Steve Smith’s Rising Pune Supergiant, against whom they have lost all three matches in this edition.

Heading into the final, Mumbai Indians hold a queer statistical advantage. This is the third time in IPL history that one team will play each other four times in one edition. Oddly, Mumbai Indians have featured in all three of them. Their first two were against Chennai Super Kings in 2013 and 2015. On both occasions, they won the title and both were held in Kolkata. This time, they will meet Rising Pune Supergiant in the final in Hyderabad.

In 2013 IPL, Mumbai Indians defeated Chennai Super Kings by nine runs in their first encounter at Chepauk. In the return leg at the Wankhede stadium, Mumbai Indians won by 60 runs as Chennai Super Kings were bowled out for 79. In the first qualifier, Dhoni’s men had their revenge as they secured a 48-run win but in the final, Mumbai Indians secured the title for the first time by winning by 23 runs. Thus, the head-to-head in the entire season was 3-1 in Mumbai Indians’ favour.

In 2015, Chennai Super Kings struck the first blow by beating Rohit Sharma’s team by six wickets at the Wankhede. However, MI came back on level terms with a six-wicket win in Chennai. The 2013 champions secured another win in the first qualifier in Mumbai when they won by 25 runs and continued the winning streak by winning the final in Eden Gardens by 48 runs to secure the title for the second time.