Hare-Clark preferential voting system

General Faculty Preferential Voting

For detailed information about the preferential voting system used in General Faculty elections, see the General Faculty Election Rules. This preferential system is a version of the Hare-Cla rk system, which is a refinement of the Hare system used at UT Austin for many years. The Hare system has the inherent disadvantage that a random proce ss is involved in transferring any excess votes from candidates who have been elected; different choices can affect the results of the election. The Hare-Clark system removes this defect by using fractional weighting for transferred excess votes. In addition, fractional weighting takes into account more of the information provided by the voters' indicated preferences. Counting votes electronically makes fractional weighting easy. 

In the Hare-Clark system, you need not rank all the candidates on the ballot, but skipping a number for those you do rank is not allowed. As a matter of strategy, there is one basic rule: if you prefer one candidate to another, then you should rank the one above the other, or at least rank the one who is perferred.

Hare-Clark Example 

This example illustrates the preferential voting system used in General Faculty elections. The process, a variation of the Hare-Clark system, is described in Paragraph 3 of Policy Memorandum 1.301 (By-Laws of the Faculty Council), revised. The example uses the following assumptions:

That 100 voters cast final ballots to fill 5 positions from 10 nominees.

That those casting ballots are divided clearly into two groups, 60 conservatives and 40 liberals, whose members vote along strict party lines. The final ballot has 6 conservatives (A, B, C, D, E, F) and 4 liberals (W, X, Y, Z).

To make the details simple, assume:

  • 30 vote ACDEFB (That is, A first, C second, and so on.)
  • 30 vote BCDEFA
  • 10 vote WXYZ
  • 10 vote XYZW
  • 10 vote YZWX
  • 10 vote ZWXY

(The conservatives are evenly split with strong differing opinions on A and B, and are in agreement on the others. The liberals vote in such a way that the final result is bound to depend on random selection.)

Because of the 60-40 split, we should expect that 3 conservatives and 2 liberals will be elected.

Quota

If n denotes the number of ballots cast and p denotes the number of positions to be filled, then the electoral quota is

q = (n/(p + 1)) + 1 = (100/(5 + 1)) + 1 = 16 2/3 + 1 rounded down, which gives q = 17.

 

After the First Count
Candidate
Vote
A
30 (Candidate A elected)
B
30 (Candidate B elected)
W
10
X
10
Y
10
Z
10
C
0
D
0
E
0
F
0
After first transfer (down from A and B) 
Candidate
Vote
C
26 [(13/30)30 = 13 from
each of A and B.] 
(Candidate C elected)
W
10
X
10
Y
10
Z
10
D
0
E
0
F
0
After second transfer (down from C) 
Candidate
Vote
D
9 [(9/26)(13/30)60 = 9]
W
10
X
10
Y
10
Z
10
E
0
F
0
Choose one of E or F randomly to eliminate. 
Assume it is E. (Candidate E eliminated.) 
After third transfer (up from E)
Candidate
Vote
D
9
W
10
X
10
Y
10
Z
10
F
0
(Candidate F eliminated.) 
After fourth transfer (up from F)
Candidate
Vote
D
9
W
10
X
10
Y
10
Z
10
(Candidate D eliminated.) 

After fifth transfer (up from D)

Candidate
Vote
W
10
X
10
Y
10
Z
10
Choose one randomly to eliminate. 
Assume it is Z. 
(Candidate Z eliminated.) 
After sixth transfer (up from Z)
Candidate
Vote
W
20 [10 + 10 (from Z)] 
(Candidate W elected)
X
10
Y
10
After seventh transfer (down from W)
Candidate
Vote
X
13 [10 + (3/20)20 (from W)]
Y
10
(Candidate Y eliminated.) 
Final result: Candidates A, B, C, W, and X elected.