Reverse Bucklin vs. Instant Runoff
DEMOREP1 at aol.com
DEMOREP1 at aol.com
Mon Jun 22 20:35:55 PDT 1998
A larger tie example
15 A>B>C>D>E>F
16 B>C>D>E>F>A
17 C>D>E>F>A>B
18 D>E>F>A>B>C
19 E>F>A>B>C>D
20 F>A>B>C>D>E
105
Assume each is acceptable to a majority of the voters using a YES/NO vote.
89 A/B 16
88 B/C 17
87 C/D 18
86 D/E 19
85 E/F 20
90 F/A 15
72 A/C 33
70 B/D 35
68 C/E 37
66 D/F 39
70 E/A 35
74 F/B 31
54 A/D 51
51 B/E 54
48 C/F 57
A>B>C>D>E>F>A
If the 3 last choices are added using Reverse Bucklin, then --
A 51
B 54 loses
C 57 loses
D 54 loses
E 51
F 48
315
70 E/A 35
90 F/A 15
85 E/F 20
E>F>A, E wins
Should the 51 first choice voters for B, C and D have to lose their first
choices?
If no, then an obvious tie breaker is to have the 15 A first choice voters
lose their first choices (i.e. use the Instant Runoff tiebreaker)
88 B/C 17
87 C/D 18
86 D/E 19
85 E/F 20
70 B/D 35
68 C/E 37
66 D/F 39
74 F/B 31
51 B/E 54
48 C/F 57
B>C>D>E>F>B
First choice votes--
31 B
17 C loses
18 D
19 E
20 F
105
86 D/E 19
85 E/F 20
70 B/D 35
66 D/F 39
74 F/B 31
51 B/E 54
D>E>F>B>D
First choice votes--
31 B
35 D
19 E loses
20 F
105
70 B/D 35
66 D/F 39
74 F/B 31
B>D>F>B
First choice votes--
31 B loses
35 D
39 F
105
66 D/F 39, D wins.
Note the relation of D and F at the top. In a real election of course, the
votes would probably be much more mixed (i.e. only 6 of the possible 6
factorial (720) combinations are in the example).
Thus, should instant run-off be used repeatedly as the tiebreaker in circular
ties ?
The theory is that the smallest number of voters should have to change their
choices repeatedly to break a tie.
More information about the Election-Methods
mailing list