17

This is also a comment, along the lines of usul's comment but a little different. I think it is best not to phrase your question in terms of "no matter what input type is used" but "no matter how apathetic the voters are." To be more clear, the way I think of Arrow's theorem is not that if you design a ballot with a total order then you are in trouble; ...

answered Jul 15 '14 at 20:03

Timothy Chow

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13

Here are the fairest sequences with $v_0\ge v_1$ for small $n = $ $1,$ $2,$ $\dots,$ $14$, according to an exhaustive search, where $v_b$ denotes the expected score for the player who can choose when the binary digit is $b$.
The reverse Thue-Morse sequence do look better than the Thue-Morse sequence but its score is far from the fairest. The Thue-Morse ...

11

Just a remark : with your weights (0,...,n) you have an simple formula to calculate the expectation.
$$v_1(1b_1b_2\cdots b_n)=1+v_1(b_1\cdots b_n) $$
$$v_1(0b_1b_2\cdots b_n)=\frac{1}{n+2}+\frac{n+3}{n+2}v_1(b_1 \cdots b_n) $$
Proof :
Let us call $Y$ the value obtained by the first player with a sequence $b_1 \cdots b_n$. Now consider the same sequence ...

8

Here is a counter-intuitive result that gets us as far as possible from the Thue-Morse sequence. Infinitely many sequences which are alternating only once are among the fairest sequences of all. Their score differences are 0.
Let's use free-monoid notations and write $1^40^2$ for $111100$. Then
$v_0(1^40^2) = v_1(1^40^2) = 5$
$v_0(1^{12}0^4) = v_1(1^{12}0^...

6

This is a complete rewrite of my original answer, combined with my comments on the original question and various other answers.
Suppose we have a finite set $C$ of candidates, and a finite set $E$ of electors (with $|C|>1$ and $|E|>2$, to avoid some degenerate cases). I will assume that we are just modelling what happens in some kind of secret ballot ...

answered Jul 15 '14 at 12:38

Neil Strickland

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6

Approval voting may well be the reasonable solution you are looking for. In approval voting, the voter either approves or disapproves every candidate on the ballot. The winner is the candidate that gets the most approvals. This method is not subject to Arrow's theorem because all approved candidates (and disapproved candidates) on a particular ballot are ...

5

For a sufficiently loose definition of "voting" - one that includes spending money - there is a solution, coming from the field of mechanism design.
The voting system works like this: Each voter must assign to each of the candidates a monetary value. The individual values are not meaningful, but the differences are - they represent the difference in the ...

5

This is an interesting question, but in fact it mixes together several different issues, each of which is the subject of its own research program within the social choice literature.
One question is: Can we design a voting rule which prevents or somehow minimizes strategic voting? The Gibbard-Satterthwaite Theorem is the original "impossibility theorem" ...

2

Here is a no-goish theorem regarding the more general question of how to aggregate the preferences of a collection of agents. Assume the agents are von Neumann-Morgenstern rational, so that their preferences are represented by (equivalence classes of) utility functions $u_1, u_2, \dots u_n$. These utility functions take as input a possible state of the world,...

answered Jul 18 '14 at 19:53

Qiaochu Yuan

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2

Extended comment. The primary challenge in the question is how one models the voters, which relates directly to what types of inputs they can provide and how one evaluates the solution. Your question seems to presume some nice answer to these questions, but in fact I think these are very thorny.
To begin, suppose that voters have a well-defined preference ...

1

There exists adaptions of Arrow's theorem to von Neumann-Morgenstern preferences. see for example Theorem 4.3 here.
the weighted utilitarianism you propose violates the independence axiom, and one can multiply each of the utility functions by some positive number. This changes the SWF, but not the preferences over lotterie represented.
There is an ...

answered Sep 30 '13 at 11:25

Michael Greinecker

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