GT first, CSK second, LSG third, one spot up for grabs now

LSG’s narrow 1 run win vs KKR on Saturday officially knocked the two time champions out of the playoffs race and guaranteed LSG the third spot on the points table.
Three teams have now sealed a berth in the final four. The one remaining berth will go to either RCB, MI or RR.
TOI’s Shankar Raghuraman does the number crunching to determine all possible scenarios as of Sunday, May 21 morning, in 10 points:
The equation is now simple:
1) Irrespective of the outcome of Sunday’s games, GT will finish first, CSK second and LSG third.
2) The race for the fourth slot is between RCB, MI and RR.
3) If either of RCB or MI win, RR will be out of the race.
4) If both of them win, they will be tied on 16 points and NRR will decide the outcome. In that case, RCB will have a huge advantage. Even if RCB score, say, 220 and win by just 1 run, MI will need to win by 78 runs (if they also bat first and score 220) to equal RCB’s NRR.
5) If only one of RCB or Mumbai wins, that team will be the fourth play-off qualifier.
6) If both lose, the fourth place will be decided between RCB and RR by NRR (MI will be out of contention on NRR). As of now RCB has the better NRR. But even if they were to bat first, score 180 and lose with just three balls to spare, their NRR would drop to 0.147, just below RR’s 0.148. Similarly, if they chase 180 and fall short by just six runs, their NRR would drop below RR’s.
7) KKR will finish seventh.
8) If SRH lose to MI, PBKS finishes eighth, DC ninth and SRH tenth.
9) If SRH beats MI, the eighth and ninth spots will be decided by NRR between PBKS and SRH, with the former having the advantage as things stand, and DC will finish last.
10) We will not know the fourth qualifier till the last league game between RCB and GT.

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