[EM] IRV tiebreaker in P.R. elections

DEMOREP1 at aol.com DEMOREP1 at aol.com
Fri Mar 24 17:51:38 PST 2000


Proportional representation election.  4 to be elected.

22 ABCDE
21 BCDEA
20 CDEAB
19 DEABC
18 EABCD
100

No CW or CL.

Which group of voters should move to their second choice ? 

If E loses (i.e. an IRV tiebreaker), then

40 A
21 B
20 C
19 D

100

Each legislator would have a voting power in the legislative body equal to 
the number of votes that he/she receives.  Doing this avoids all the 
fractional math connected with STV.

Same votes. Elect 2

22 ABCDE
21 BCDEA
20 CDEAB
19 DEABC
18 EABCD
100

The general head to head case is--

Test Winner(s)   Test Loser    Test Other Loser(s)

TW                      TL                   TOL            First choice Votes
TOL to TW           TOL to TL       -TOL           + Later TOL choices
TW                      TL                  zero            Totals

A TW is a CW, CL or in a tie.

A TW which gets a Droop Quota in all of his/her combinations obviously is a 
CW.

Sample combinations---
TW        TL      TOL 

A,B        C        D,E
22,21     20     19,18
19                  -19
18                       -18
59, 21    20      0,0

A,B        E         C,D
22,21    18       20,19
              20     -20
             19           -19
22,21    57       0,0

If E loses (due to the various ties), then--

40 A       (more than Droop Quota)
21 BCDA
20 CDAB
19 DABC
100

A, B>C
A, C>D
A, D>B

If D loses, then

59 A       (more than Droop Quota)
21 BCA
20 CAB
100

A,B > C
Final
79 A, 21 B



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