[EM] D2MAC can be much more efficient than Range Voting

Jobst Heitzig heitzig-j at web.de
Wed Mar 7 02:06:47 PST 2007

Hello folks,

it is frequently claimed that methods which involve randomness may be
fairer than other methods but will give "worse" results.

Here's evidence for just the contrary:

A typical voting situation is

55%:  A>C>B
45%:  B>C>A

with C being considered a good compromise by all voters
(in the sense that all voters would definitely prefer C strongly
to tossing a coin between A and B).

At first, it seems that this is exactly the situation where
methods which claim to maximize "social utility" or "social benefit"
should lead to the "right" answer C.

If all voters were sincere, indeed both Approval Voting and
Range Voting will elect C, since the ballots would then
look like this:

Approval Voting:
  55%: A,C
  45%: B,C
Winner: C

Range Voting:
  55%: A 100, C 50+whatever, B 0
  45%: B 100, C 50+whatever, A 0
Winner: C

The problem is, rational voters just *won't* vote sincere in this
situation. And since Approval and Range Voting are *majoritarian*
methods, the real outcome will rather look like this:

Approval Voting:
  55%: A
  45%: B,C
Winner: A

Range Voting:
  55%: A 100, C 0, B 0
  45%: B 100, C 99, A 0
Winner: A

So, both these methods *fail* to do just what they were apparently
constructed to do!

Now let us look how D2MAC performs in this typical situation,
a democratic, non-majoritarian method:

Recall that in D2MAC you specify a favourite and as many "also approved"
options as you want. Then two ballots are drawn and the winner is the
most approved option amoung those that are approved on both ballots
(if such an option exists), or else the favourite option of the first

If voters are sincere, the result will be this:
  55%: favourite A, also approved C
  45%: favourite B, also approved C
Winner: C

Can the A-faction improve their result upon this by voting differently?

If they switch to
  55%: favourite A, none also approved
then A will win whenever an A-ballot is drawn first, i.e., with 55%
probability. However, B will win in the remaining cases, i.e., with 45%
probability. For the A-supporters, this "almost coin tossing" is not
preferable to C, so the strategy won't help them.

Thus, the "obvious" A-strategy cannot destroy the compromise under D2MAC!

The only way the A-voters could have to strategically improve the result
is to switch to something like this:
  55-x%: favourite A, none also approved
  x%:    favourite A, also approved C

With a very small x, they can perhaps make sure that C ends up with a 
approval score than A so that the winning probabilities will turn out as

  A: 55%
  B: 45/100 * 55/100 ~ 25%
  C: 45/100 * 45/100 ~ 20%

This lottery may or may not be preferable to C for the A-voters. When C is
really a good compromise which the A-voters still prefer to, say, the 
75%A + 25%B, then they will not use the above strategy.

With larger x, the analysis is somewhat more complicated and will have to
involve the possible strategic reactions by the B-voters.

The result is that voting sincerely will happen and C will be the sure 

(i)  the A-voters prefer C to the lottery 77.5%A + 22.5%B, and
(ii) the B-voters prefer C to the lottery 72.5%B + 27.5%A.

In terms of individual utility scaled from 0 to 100, this means that

(i)  C must have a utility of at least 77.5 for the A-voters, and
(ii) C must have a utility of at least 72.5 for the B-voters

for C to be elected with certainty.

In general, it can be shown that under D2MAC a good compromise will win with
certainty as long as it is "good enough" in a sense similar to the above.

Conclusion: it is not true that "efficiency" and "equality" are hard to 

To the contrary: D2MAC elects the good compromise C *because* it is 
in the sense that it allows each voter to control and *trade* her share 
of the
winning probability!

Yours, Jobst


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