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Sequential Runoff

We continue with the same table:

Preference Schedule

  group of 18 group of 12 group of 10 group of 9 group of 4 group of 2
Killians 512424
Molson 155555
Samuel Adams234133
Guinness 441242
Meister Brau323311

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:

Killians 311212
Molson 133333
Samuel Adams222121

then we would have to eliminate Samuel Adams in the first step, ending up with the reduced preference schedule

Killians 211111
Molson 122222

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.

Sequential Runoff

Which beer would win in the end, with this sequential runoff method? Do not forget to press "Enter".

Click here if you need an explanation for the example above.

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