We continue with the same table:
The sequential run-off scheme eliminates candidates one by one. First, we eliminate the candidate with the fewest first-place votes. In this case, that would be Meister Brau. Then we make up a preference schedule in which only the remaining candidates are listed, in this case Killians, Molson, Samuel Adams and Guinness; we renumber the ranking in every column so that the 4 remaining candidates have now ranks labeled by 1, 2, 3 and 4.
If we had started with only 3 beers, Killians, Molson and Samuel Adams, with the preference schedule:
then we would have to eliminate Samuel Adams in the first step, ending up with the reduced preference schedule
and then, in the next round, Killians, with 12+10+9+4+2=37 first places would have won.
In the situation of the preference schedule at the top, we have 5 contenders originally, and we need more rounds. After the first round, we renumber the rankings so that we get a preference schedule with only 4 contenders left, and in each column the rankings are labeled with the numbers 1, 2, 3 and 4. We then repeat the procedure: we eliminate again the candidate with the fewest 1st place rankings, and renumber for the remaining 3 contenders, and so on.
Note: if at some step, two (or more) candidates have the same number of first place votes, look to their second-place (or if necessary, their third-place, etc.) votes to decide how to rank them.
Which beer would win in the end, with this sequential runoff method? Do not forget to press "Enter".