Condorcet Basics

Thu Oct 15 20:25:53 PDT 1998

Some new folks on the EM list may not know about the basics of the plain
Condorcet method.

A voter votes 1, 2, etc. for his/her first, second, etc. choices.

For N choices, there are N factorial (N!) possible combinations if ALL voters
number vote for ALL choices.

Example- 4 choices- A,B,C,D
N! = 4 possible first choices x 3 possible second choices for each first
choice x 2  possible third choices for each second choice x 1 possible
remaining fourth choice for each third choice = 24 combinations of A,B,C and D

There are additional combinations if the voters do not vote for all of the
possible choices (i.e. truncate) - such as only voting for 1 of 4 choices, 2
of 4 choices or 3 of 4 choices.

Namely--  N!/(N-1)!,   N!/(N-2)!,   N!/(N-3)!,   etc. combinations.

Example- 4 choices
4 + 12 + 24 = 40 

Thus, a voting matrix can be arranged ---
 Choices                 Horizontal
voted for                Choices 1 to N





Thus for 4 choices, there would be 24 + 24 + 12 + 4 = 64 vertical

In each pair of choices, the votes that rank each choice over the other choice
are compared from all of the combinations.
Example-  100 voters, 54 rank C over A, 45 voters rank A over C.  C thus has a
net 9 votes over A.

Thus, the winner in each pairing may or may not get a majority of all of the
Example-  100 voters, 48 B over D, 45 D over B, net 3 B over D.

Note- if a voter does not vote for all choices, then the choices not voted for
may be deemed to be in last place for pairing purposes.

Example- 4 choices, a voter votes D > A

The vote would be deemed D > A > blank > [B=C] so that D and A both beat both
B and C on such ballot.

One choice may win in all of his/her pairings and is called a Condorcet Winner
A choice may lose in all of his/her pairings and is called a Condorcet Loser

If there is no choice that wins in all of his/her pairings, then such a case
is called a "circular tie" and needs a tiebreaker.

In a more complex variation, a voter could vote 2 or choices in the same
Example- 6 choices
A voter votes   D=E=F > blank > blank > C > blank > [A=B]

In some variations each of choices having an equal vote would have a half vote
(0.5) in their pairings.
Example- in the preceding, D, E and F would each get a half vote when paired
with each other. Using the half votes results in having the winners in all
pairings getting a majority of all of the voters (assuming no 50-50 type

Of some interest in one tiebreaker is the number of votes that each choice
gets at the N/2 (N even) or (N+1)/2 (N odd) horizontal choices level (both
left to right and right to left).

Other folks may wish to give *simple* examples of their tiebreakers.

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