[EM] Cardinal Condorcet based on lotteries: seems like it doesn't work
km_elmet at t-online.de
Thu Nov 4 16:16:18 PDT 2021
So I've been tinkering with the idea of a cardinal Condorcet method
based on lottery information as a way to merge Condorcet and cardinal
voting. I found a paper on randomized Condorcet that seemed to be
promising, because it made use of a preference relation defined on
lotteries, so it should be a simple matter to replace their inferred
lottery preferences with explicit lottery preferences from cardinal
voting, while retaining the idea that society prefers one lottery to
another if a majority of the voters do so.
And, well, it is. It is relatively straightforward to replace the
implied lottery with an explicit one, and to define the relation for
lotteries with any number of candidates.
But doing so invalidates the paper's proof that there's always a lottery
that's a Condorcet winner, and my experimentation seems to show that
even for relatively "easy" elections (the "bad centrist" example),
there's no Condorcet winner among the lotteries.
Then we're in trouble, for what does it mean for two lotteries to be in
the Smith set? I don't think we could simply average the lotteries
together because the preference relation isn't linear. The step function
from "just short of a majority" to "just above a majority" is what's
causing all the trouble -- like in ordinary Condorcet.
What I can do, then, is to document what I've done in case it'll be
useful to others. But unless I missed something or my experimental
evidence is flawed, I think this is it for the idea to use randomized
Condorcet but replacing lotteries.
There seems to be a tension between the majoritarianism of Condorcet and
the strength of preference of cardinal methods: either the method is
easily coerced into the Condorcet side, or the method behaves rather
more like Range. I'll write another post about that.
As experiment, I ran a grid search test on the bad centrist example:
consider a turn-based game where both players choose a lottery. If the
first player's lottery is prefered to the second, the first player gets
a payoff equal to the number of voters who did so; and similarly but
negated for the second player. If there is a CW, then the optimum play
is for both players to choose the CW, and then the payoff (value of the
game) is zero. But if either player always wins, then there is no CW.
And that's what the grid search shows. I've added the code.
As for constructing the preference relation between lotteries, suppose
first that there are three candidates with nonzero probability of election.
If the lotteries to be compared are (p, q, r) vs (s, t, u) for
candidates A, B, and C respectively, and a voter has indicated A>B>C
with indifference probability p_B, so that he is indifferent between B
being elected with certainty and a p_B chance of A being elected, 1-p_B
of C being elected, then from a utilitarian perspective, that means:
The voter values A at av_A+b, B at av_B+b, C at av_C+b, so that
p_B(av_A+b) + (1-p_B)(av_C+b) = av_B+b. Under the incommensurability
model, we don't actually know the scaling values a and b, but we can
just divide out a (assuming it's nonzero). Expanding and gathering the b
terms on one side also cancels them out:
p_Bv_A + p_Bb + (1-p_B)av_C + (1-p_B)b = av_B + b
p_Bv_A + (1-p_B)av_C = av_B + b - p_Bb - (1-p_B)b = av_B
By one man, one vote, we may set v_A and v_C to fixed values; in this
case it makes sense to set v_A = 1, v_C = 0 since the voter prefers A to
v_B = p_B + (1-p_B)*0, i.e. v_B = p_B
So the voter prefers lottery (p, q, r) to (s, t, u) if p+p_Bq > s +
p_Bt. And the preference relation between (p, q, r) and (s, t, u) is
simply whether a majority prefers (p, q, r) to (s, t, u); a simple
generalization of pairwise comparisons.
Fair enough. For fewer candidates, we can simply consider a lottery with
three candidates where the win probability of some of these are set to
zero. (Omitted here: if each lottery has only a single candidate with
nonzero probability, then the comparison is the ordinary Condorcet one.)
And for more than three candidates, lotteries of three can be combined.
E.g. if the voter ranks A>B>C>D and is indifferent between lotteries (0,
1, 0, 0) and (p_B, 0, 1-p_B, 0); and also indifferent between lotteries
(0, 0, 1, 0) and (0, p_C, 0, 1-p_C), then we can construct similar
virtual ratings by the linear set of equations:
p_B * v_A + (1-p_B) * v_C = v_B
p_C * v_B + (1-p_C) * v_D = v_C
and just set the extremes (v_A and v_D) to 1 and 0 as before, i.e.
p_B + (1-p_B) * v_C = v_B
p_C * v_B = v_C
For instance, for p_B = 0.1, p_C = 0.5, this gives
v_B = 2/11, v_C = 1/11
which finally answers what the one honest vote in a rated system is! (Ha
ha, only serious.) The one honest continuous Range vote consistent with
that ranking and those indifference lotteries is (1, 2/11, 1/11, 0).
And that voter prefers lottery (p,q,r,s) to (t,u,v,w) if p + q2/11 +
r1/11 > t + u2/11 + v1/11.
(Unfortunately, Range doesn't incentivize this kind of honest voting.)
-------------- next part --------------
A non-text attachment was scrubbed...
Size: 2069 bytes
Desc: not available
More information about the Election-Methods