Math Alive Voting 1

Detailed Explanation for Sequential Runoff Example

Labs: Voting and Social Choice. Part 1.

Samuel Adams

	group of 18	group of 12	group of 10	group of 9	group of 4	group of 2
Killians	5	1	2	4	2	4
Molson	1	5	5	5	5	5
Samuel Adams	2	3	4	1	3	3
Guinness	4	4	1	2	4	2
Meister Brau	3	2	3	3	1	1

Meister Brau has the lowest number of first-place votes (6), and is therefore the first beer to be eliminated. After eliminating Meister Brau and renumbering, the reduced preference schedule becomes:

Continuing this process, we eliminate Samuel Adams, because it has the fewest first-place votes (9). The next reduced preference schedule is then

	18	12	10	9	4	2
Killians	3	1	2	2	1	2
Molson	1	3	3	3	3	3
Guinness	2	2	1	1	2	1

and we now must eliminate Killians. Finally we are down to a race with two contenders only:

	18	12	10	9	4	2
Molson	1	2	2	2	2	2
Guinness	2	1	1	1	1	1

and Guinness, with 37 first-place votes against Molson's 18, wins.

Back to Sequential Runoff

Last Modified: August 2008