Transferring Votes in Condorcet
DEMOREP1 at aol.com
DEMOREP1 at aol.com
Mon Apr 15 22:35:06 PDT 1996
The use of the Condorcet method is the functional equivalent of the transfer
of votes.
Example
3 Candidates A, B and C
Voting summary
5 voters vote A, B, C
4 voters vote B, C, A
2 voters vote C, B, A
A v. B 5 to 6 B beats A
A v. C 5 to 6 C beats A
B v. C 9 to 2 B beats C
B wins.
A voter's vote is in effect transferred to the candidate (a) who is one of
the candidates being paired and (b) who is ranked earliest by the voter.
Even the use of computers doing the results of all the pairings on each
ballot can be thought as transferring votes. Namely, a vote is tranferred
from each pairing on a ballot to a matrix of all the combinations of
candidates.
Example 5 Candidates J, K, L, M, N
A voter votes M, K, J, N, L
The computer transfers votes as follows
1 to M in the M v. K, M v. J, M v. N and M v. L pairings
1 to K in the K v. J, K v. N and K v. L pairings
1 to J in the J v. N and J v. L pairings
1 to N in the N v. L pairing
Such transfers are of course additions to the matrix cells.
For an election with a large number of voters and/or candidates, a computer
would have to do the vote transfers/additions, of course.
A summary of how many voters voted for each possible combination (assuming no
write-in votes) would have to be made as a check to show the summary math-
that is how many voters voted J, K, L, M, N; how many voters voted K, L, M,
N, J; etc.
With X number of candidates there are X factorial combinations (assuming no
truncation of votes).
Example 5 candidates
The first choice may be one of 5 candidates, the second choice may be one of
the 4 remaining candidates, the third choice may be one of the 3 remaining
candidates, the fourth choice may be one of the 2 remaining candidates, the
fifth choice is the remaining candidate.
Thus, there are 5 x 4 x 3 x 2 x 1 or 120 combinations
Truncated votes produce shortened factorial combinations.
Namely in the example
1st choice only- 5
1st and 2nd choices only- 5 x 4 or 20
1st, 2nd, 3rd choices only- 5 x 4 x 3 or 60
1st, 2nd, 3rd, 4th choices only- 5 x 4 x 3 x 2 or 120
Grand total- 325 ways a voter could vote (assuming that the same choice
ranking is only used once). Thus the need for computers !!!
More information about the Election-Methods
mailing list