Mathematics in Action

Plurality with Elimination in the MAS Election

Preference Schedule for the MAS Election
Number of voters 14 10 8 4 1
1st choice A C D B C
2nd choice B B C D D
3rd choice C D B C B
4th choice D A A A A

Recall in Plurality with elimination, in each round we are only concerned with the number of 1st choice votes.

Set Up:
Before we begin, we must first determine what would constitute a majority. There are a total of 14 + 10 + 8 + 4 + 1 = 37 votes. 37/2 = 18.5 This means that to have a majority, a candidate must have 19 or more votes.

Round 1:
A:14, B:4, C:11, D:8
No one has a majority. B has the least amount of 1st choice votes, so B is eliminated. We get a new table as follows:

First, remove all of the B's:

Removing the B's
Number of voters 14 10 8 4 1
A C D C
C D D
C D C
D A A A A

Then, make this into a new table.

Making a B-less preference table
Number of voters 14 10 8 4 1
1st choice A C D D C
2nd choice C D C C D
3rd choice D A A A A

To make this table much nicer, we may combine identical rows
The 2nd and 5th rows are identical, and the 3rd and 4th rows are identical.

We get:

End of Round 1
Number of voters 14 11 12
1st choice A C D
2nd choice C D C
3rd choice D A A

Adding colors:

End of Round 1
Number of voters 14 11 12
1st choice A C D
2nd choice C D C
3rd choice D A A

Round 2:
A:14, C:11, D:12
No one has a majority, so we eliminate C with the least amount of 1st choice votes. We need to remove C from the table and see what happens.

First, we remove C.

Removing C
Number of voters 14 11 12
A D
D
D A A

Now, we make this into an actual preference schedule.

Making a C-less preference table
Number of voters 14 11 12
1st choice A D D
2nd choice D A A

Or Equivalently:

End of Round 2
Number of voters 14 23
1st choice A D
2nd choice D A

With Color:

End of Round 2
Number of voters 14 23
1st choice A D
2nd choice D A

Now A has 14 first choice votes and D has 23 first choice votes. D has a majority, so D is the winner!!!!