MATH 131--Class Notes--Spring 08

PART II: Voting Theory

 

Monday, 02/25/08: We started discussing the following example:

 

A group of students stuck in Tahoe (with only one watchable TV) in a snow storm needs to decide what to watch on TV to pass the time

 

until the roads clear. The options are:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A = American Idol

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B = Big Brother

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C = CSI

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

D = Deal or No Deal

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E = Extreme Makeover

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The group decides to vote and every member of the group writes down their respective preferences. The preference lists are gathered and organized.

14 voters preferred A over B over C over D over E, 10 voters preferred C over B over D over E over A, etc.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

# OF VOTERS

 

 

14

10

8

4

2

2

2

2

2

2

1

RANKING

1st choice

A

C

D

B

E

E

E

E

E

E

C

2nd choice

B

B

C

D

B

C

B

C

D

D

D

3rd choice

C

D

B

C

C

B

D

D

B

C

B

4th choice

D

E

A

E

D

D

C

B

C

B

A

5th choice

E

A

E

A

A

A

A

A

A

A

E

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Question: Which program should the group watch? Why?

 

 

 

 

 

 

 

 

 

Two answers came out of the discussion in your groups:

 

1)      14 people prefer A, 12 people prefer E, 11 people prefer C, 8 people prefer D and 4 people prefer B. This makes A the most

preferred choice and  A should be the winner. However, there are three good reasons why we should question the wisdom of the

choice of A. For one, A received only 28% (14 our 49) of the 1st choice votes. Second, 53% (26 out of 49) of the voters rank A

as their LAST choice. Third, by using only the 1st choice votes to determine a winner we are ignoring the wealth of information

expressed in the whole table.

2)      One way to use this wealth of information is to assign points to the different levels of votes. For example, a first place vote is

worth 5 points, a second place vote is worth 4 points, a third place vote is worth 3 points, a fourth place vote is worth 2 points,

and a fifth place vote is worth 1 point. Add up all the points for each candidate and the candidate with the most points wins. (Other

schemes that were suggested were 4,3,2,1,0 points, or reversing the appoints 1,2,3,4,5 and then the winner is the candidate with

the fewest points.) All of those produce the same winner albeit with different point totals.

 

Let’s carry this out for the above table:

 

Candidate A ranks first with 14 voters. A first place vote is worth 5 points each. Thus the 14 voters who rank A first amount to a

Total of 14 x 5 = 70 points. We proceed in this fashion with all candidates and all voters and we get:

 

Candidate A:  14x5+10x1+8x2+4x1+2x1+2x1+2x1+2x1+2x1+2x1+1x2 = 114

     Candidate B:  14x4+10x4+8x3+4x5+2x4+2x3+2x4+2x2+2x3+2x2+1x3 = 179

 

     Candidate C:  14x3+10x5+8x4+4x3+2x3+2x4+2x2+2x4+2x2+2x3+1x5 = 177

 

     Candidate D:  14x2+10x3+8x5+4x4+2x2+2x2+2x3+2x3+2x4+2x4+1x4 = 151

 

     Candidate E:  14x1+10x2+8x1+4x2+2x5+2x5+2x5+2x5+2x5+2x5+1x1 = 111

 

In this method candidate A isn’t even close to winning the lection. The fact that it is ranked last by 26 voters is now reflected in the point total. Candidate B is the winner with 179 points. What’s the problem with this method? Well, candidate B has the fewest first choice votes with 8% (4 out 49). It seems a bit strange to elect a candidate who has such a low percentage of first place votes.

 

     We’ll continue with this example on Wednesday.

 

     Note: The following words are used interchangeably in the context of voting theory: candidate, choice, alternative.

 

Friday, 02/29/08: We tried to find other voting methods that seem reasonable. Suggestions from the class included:

 

In order to avoid really bad choices we should count 3rd place votes. Those alternatives aren’t great but aren’t terrible either.

The problem with this method is that people would figure out pretty quickly that 3rd place is really 1st place and would

vote accordingly.

 

Count only votes for 1st, 2nd, and 3rd choice and ignore votes for 4th and 5th choice. This is a variation of the Borda Count.

Assign 1 pt to each 1st, 2nd, and 3rd choice vote and 0 points to every 4th and 5th choice vote.

 

There were other variations of Borda Count that came up, e.g., assign 1,2,4,8,16 pts, a geometric series rather than the

usual arithmetic series 1,2,3,4,5 pts.

 

One suggestion that we will start with on Monday stems from the questions that was raised by D’ana: Can we have a run-off

with three candidates? Thinks about how that might work. IS there anything special about three candidates?

 

Monday, 03/03/08, and Wednesday, 03/0508: We discussed the plurality-with-sequential-elimination method of voting. It is a 
generalization of the plurality with run-off method. Instead of taking just the top to first choice vote getters into a run-off we 
take all candidates except the one with the fewest first place votes into a second round. We then re-distribute the votes and
again eliminate the candidate with the fewest first place votes. And we repeat the process until we have a winner. 
 
Here’s how this works in our example:
 

 

 

# OF VOTERS

 

 

14

10

8

4

2

2

2

2

2

2

1

RANKING

1st choice

A

C

D

B

E

E

E

E

E

E

C

2nd choice

B

B

C

D

B

C

B

C

D

D

D

3rd choice

C

D

B

C

C

B

D

D

B

C

B

4th choice

D

E