Multiple Same Choices (MSC) is a voting method that merges parts of Approval,
Condorcet and NOTA for single or multiple executive and judicial offices.
A voter could vote 0, 1, 2, etc. for more than 1 candidate.
Why permit a voter to cast the same numerical vote (0, 1, 2 etc.) more than
Answer. In a single winner case, a voter may (a) equally approve 2 or more
candidates for the office at a given choice level (first, second, etc.)
and/or (b) not want his/her second (or later) choice vote(s) to help elect a
person who he/she regards as an extremist (and may only "strategically" vote
for such an extremist to offset some other voter's vote for some even worse
The voter thus has maximum choices of his/her choosing.
or arranging in number order--
Extreme cases would be voting 0 (disapprove all- same as NOTA) or 1 (approve
all) for all candidates.
The "1" votes would be totalled for all candidates.
If there is no majority winner or two or more majority winners, then
Condorcet pair comparisons could be done as follows.
TW = Test Winner, TL = Test Loser, TOL = Test Other Losers, NT = Not
Transferred, V = Voter
If a voter voted at a level (1, 2, etc.) for TW and/or TL, then the votes for
TW and/or TL at such level would count for TW and/or TL and the ballot stops.
Otherwise the ballot continues to the next lower level (as when a voter voted
his/her first choices for one or more TOL only). Due to the "if" condition
computers would generally be needed to do the math.
TW TL NT TOL1 TOL2
V1 1 1
V2 1 1 1
The ballots of V3 and V4 would go to the second choice level. If there is a
"2" vote on the ballots or V3 and/or V4 for TW and/or TL the ballot would
stop. Otherwise the ballot would go to the third choice level and so forth.
The above also applies if 2 or more executive or judicial officers are being
elected (i.e. 2 or more test winners in any Condorcet tests).
On Mon, 26 Feb 1996 [hidden email] wrote:
> Answer. In a single winner case, a voter may (a) equally approve 2 or more
> candidates for the office at a given choice level (first, second, etc.)
> and/or (b) not want his/her second (or later) choice vote(s) to help elect a
> person who he/she regards as an extremist (and may only "strategically" vote
> for such an extremist to offset some other voter's vote for some even worse
A voter who considers someone an extremist won't rank that candidate, or
rank them very low, only above someone who is *even more* extreme. Their
vote will never help that candidate get elected, except when the
alternative is even more extreme, in which case it is exactly what that
Reason (a) is a good reason. Reason (b) is not.
I'm trying to understand the mathematical implications of this as they
apply to the algorithm I use to calculate the Condorcet winner. Right
now I count the votes against a candidate in determining their worst
defeat. Identical rankings throws a bit of a monkey wrench into the
Joe Left 1
Sally Middle 1
Martha Right 2
What should the pairwise set of results be?
I'm inclined to say B, because the implication is that this voter would
not have voted in an election between Left and Middle. Therefore, neither
Left nor Middle should have even half of a vote counted against them in
this particular election.
Since unranked candidates are on the "same level", my program already has
to deal with this issue. When tabulating a specific pair of candidates,
any ballot where neither candidate is ranked isn't counted. So, in the
name of consistancy, ballots where both candidates are ranked identically
should also not be counted (in this pairwise calculation).
My brain hurts.
In reply to this post by Craig Carey-2
Demorep, the method you describe has been "Point Assignment", or
"Point Ratings". Where the points can be from 0 to 10, it could
be called the Olympic 0-10 point system, or "Olympic 0-10".
As I've said, Olympic 0-10 would be a good choice if voters would
only accept something that's already familiar to them. But 0-10,
or any other such point assignment method is nowhere near as good
as Condorcet's method, because, as you yourself said, Point Assignment
requires defensive strategic voting, and Condorcet's method never requires
it under plausible conditions.
Again, you propose (presumably an implementation of) Pairwise-Count
only if Point Assignment returns a tie or doesn't show a majority. And,
again, I ask why you want to use a not-as-good method as the primary
method, and a better method only as the tie-breaker.
You still continue to not address the comments about the need for
defensive strategy, and the electoral reformers' wish for a method
that lets us cast a full-strength, reliably-counted vote for Compromise
over Worst, while still casting one for Best over everyone.
And, about your implementation for picking a candidate who beats
everyone, why do you still propose that procedure, with its long
& wordy definition, when the candidate who beats everyone can
be picked by a very simply & briefly stated rule? Whether the
procedure you describe, with test winner, test losers, & test
other losers, would actually accomplish that, I admit I don't
In reply to this post by Rob Lanphier-2
I agree that B) is the correct interpretation of that ballot:
The voter hasn't expressed any preference between Joe & Sally, and
has voted each of them over Martha.
By not voting for either Sally or Joe over the other, that voter
not only isn't counted as voting for 1 to beat the other, but also
, in Condorcet's method's circular tie solution, that ranking
wouldn't be counted as a vote for anyone over Joe or Sally, for the
purposes of carrying out Condorcet's method's rule for circular ties.
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