RCB, DC seek top-two finish.
SRH have to win their last IPL 2020 match against MI.
Only two days are left for the end of the league stage of the Indian Premier League (IPL) 2020, but only one team – Mumbai Indians (MI) – have booked their place in the playoffs and a top-two finish.
Three teams – Chennai Super Kings, Kings XI Punjab and Rajasthan Royals – have been knocked out of the tournament while the other four teams have a chance to seal the remaining three positions. The match between Delhi Capitals (DC) and Royal Challengers Bangalore (RCB) on Monday will be a shoot-out for the second spot.
Here’s how both DC and RCB can qualify for the playoffs
The winner of DC vs RCB match will get to 16 points and be guaranteed a top-two finish, which ensures two games to reach the final. The losing team can still progress to the playoffs but needs to avoid a heavy defeat so that their Net Run Rate (NRR) doesn’t fall below that of fourth-placed Kolkata Knight Riders (KKR).
If DC loses by 18 or more runs, KKR will go through and RCB occupying the second spot. If SRH beats MI on Tuesday, DC will bow out.
If RCB loses by 22 runs or more, KKR will qualify and DC would be at the second spot. If Sunrisers beat MI, Kohli’s RCB will be knocked out.
Also, while chasing if RCB gets to the target in less than 18.4 overs, then DC’s NRR will fall below KKR’s, and the SRH win over MI will knock Shreyas Iyer & Co. out of the tournament. Conversely, if DC chases down the total in less than 18 overs, RCB’s NRR will fall below KKR’s, and SRH victory will end the playoff road for RCB.
That said, if MI beats Warner’s men, DC, RCB and KKR will advance to the playoffs.