It was the 2018 IPL final all over again. Sunrisers Hyderabad put up a total in excess of 170. They started their defence well with Bhuvneshwar Kumar bowling a maiden first over to Shane Watson. And they ended it wondering why this keeps happening to them?

Watson was lucky to survive the first over from Bhuvneshwar in that final. But luck perhaps didn't have as big a role to play in Watson surviving the first over in this match. It waited until later in the innings.

Watson was on 33 off 26 balls when wicketkeeper Jonny Bairstow dropped him off Sandeep Sharma. When Bairstow did take his catch later on, Watson had made 96 runs off 53 balls. After he was dropped, Watson scored 63 runs off 27 balls at a strike rate of 233.

Had that catch been taken, Sunrisers would have got two wickets in two overs. With 96 runs required off the last ten overs, they would have had two new batsmen to target and things could have panned out differently.

ESPNcricinfo's Luck Index estimates that the drop cost them 31 runs. The runs cost is calculated by allotting the deliveries faced by Watson after the reprieve to the batsmen who follow him, and simulating the rest of the innings to estimate how many runs they would have scored off those deliveries. Luck Index estimates that given there were five overs left from Rashid Khan and Bhuvneshwar, the Super Kings batsmen to follow would have managed only 32 runs off the 27 extra balls that Watson faced.

Considering that the match was won only off the penultimate ball, Sunrisers could have ended up on the right side of the result had Bairstow taken the chance.