[EM] Number Votes - 10 April 2016

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Number Votes - 10 April 2016

1. Use Number Votes (1, 2, 3, etc.) for each choice.
A Voter MUST put a number vote for ALL choices.

2. Vote YES or NO for each choice.
3. Ballot Form -
Numbered Boxes 1 2 3 etc. (i.e. total choices)
YES/NO Boxes  Name
4. Do Head to Head combinations math (would require computer voting in any large election).
Test Winner(s) (TW) -- Test Loser (TL) -- Test Others (TO)
Each TO 2nd or later choice vote goes to a TW or TL.

5. If a TW wins in all combinations, then it wins.

6. If a TW loses in all combinations, then it loses (and repeat step 5).
7. Tiebreaker - If no TW, then add 1st plus 2nd plus etc. choices to get a Droop Quota (majority for 1 seat winner).  If 2 or more get a Droop Quota, then any other non-Droop Quota choices lose (and repeat step 5).
8. OR -- have the lowest YES choice lose (and repeat step 5).
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Legislative bodies - each final winner would have a voting power equal to the final votes received.

For multiple executive/judicial offices (e.g. elect 3 judges) the N highest Number Votes count - for TO transfer purposes.
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Theory - Condorcet in France in the 1780s (repeat 1780s) noted that a third choice could beat two existing choices head to head --
A beats B
C comes along.
C beats A and C beats B.
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Divided majority example -
26 AB
25 BA
49 Z

The Z voters (i.e. all voters) would be required to make number votes for ALL choices.

i.e. example might result in --
26 ABZ
25 BAZ
26 ZBA
23 ZAB
Z is beat by both A and B and thus loses.
51 B beats 49 A.
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Possible circular tie -
34 ABC
33 BCA
32 CAB
99

Adding 1st plus 2nd votes -
34 + 32 = 66 A
33 + 34 = 67 B
32 + 33 = 65 C  Loses.
66 A beats 33 B.

OR use Approval Votes - Lowest would lose.
One of the two remaining would win.

Note - 
Number Votes are *relative* only.
Approval Votes are *absolute* only.
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