[EM] 3 Choices - Condorcet - Approval tiebreaker

DNOW1 at aol.com DNOW1 at aol.com
Sat Apr 26 20:04:26 PDT 2014


Early polling shows -- 

50 A > 49 B

Choice C (1 of the 99) does NOT like the A or B choices and decides to run.

C obviously needs 50 votes to defeat A -
i.e. 50 C > 49 A.

i.e. to get a range of from 1 A plus 49 B to 49 A plus 1 B (assuming choice 
A will again vote for A) -- perhaps even 25 A and 25 B.

To defeat C (50 A > 49 C), A would need at least 1 vote from B ---
49 A plus 1 from B = 50.

Thus - The *sincere* voting stuff is worthless at the margins.
All sides can play the last second switch games with bribes and threats.

The election system must act on the votes actually cast.
*******
Again - Condorcet Head to Head math - all elections - legislative and 
executive/judicial.
Use Number Votes for choices.
Use Approval Voting -- YES/NO (default) as a tiebreaker.

Test Winner(s) vs. Test Loser -- Other Losers (later Number Votes go to a 
TW or a TL)

If a TW wins in all combinations, then the TW is a Condorcet Winner (CW).
If a TL loses in all combinations, then the TL is a Condorcet Loser (CL).
Remove the CL with the lowest Approval.
Redo the head to head combinations.
Repeat, if necessary.

If not all wins or loses for remaining choices, then remove the remaining 
choice with the lowest Approval.
Redo the head to head combinations.
Repeat, if necessary.
----------
Legislative - each final winner would have a voting power in the 
legislative body equal to the final votes that he/she gets.

Executive/Judicial - for multiple offices (e.g. 3 judges) the top M Number 
Votes would count in doing the combination math.

Computerized voting would be necessary in larger elections to do the 
combinations math.
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