Assume 5 voters and 3 candidates A, B, C.
Suppose the ranked votes are:
Then the "Condorcet winner" (beats all rivals in man-to-man two-choice elections) is A.
Suppose we interpret the first and third voting patterns to mean that the sincere evaluations of the 3 candidates are "excellent," "very good," "terrible," and assign corresponding weights 9, 8, 0, and that we interpret the second and fourth patterns to mean "excellent," "terrible," "worst," then the weights might be 9, 1, 0.
The corresponding range votes would be:
|2||A=9, B=8, C=0|
|1||A=1, B=9, C=0|
|1||A=0, B=8, C=9|
|1||A=1, B=0, C=9|
Then the Range Voting winner is B (summed-score 33=2×8+9+8+0).
For Approval Voting it would be reasonable to consider 8 as approval and 1 as not approved, so the Approval Vote would look like:
|2||A=1, B=1, C=0|
|1||A=0, B=1, C=1|
|1||A=0, B=0, C=1|
The Approval Voting winner is B.
But if we took the same ranked votes and, for lines 1 and 3 considered the rankings to be "excellent," "terrible," "worse," and line 2 to mean "excellent," "very good," and "terrible," we would find, when the weights are changed appropriately, that A would win both the Range and Approval elections.
This illustrates how both Range and Approval can have more discriminating power then any ranked system. Or we could say that Approval and Range both allow the voter to be more precise.
For ranking systems where equal ranking is permitted (at present, most do not permit that), then we can think of Approval as the simplest case of such systems and therefore such ranking systems are more powerful than Approval (but remain weaker than Range). But then we have to look at complexity. For both voters, and, more important, for the tallying process, Range and Approval both are much simpler.
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