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.