# [Election-Methods] RE : Re: Is the Condorcet winner always the best?

Jonathan Lundell jlundell at pobox.com
Thu Dec 13 21:56:56 PST 2007

```On Dec 13, 2007, at 6:52 PM, Dave Ketchum wrote:

>> Cycles don't enter into it (and if A is guaranteed a solid win,
>> then  of course strategy is irrelevant).
>> My argument is about expected utility. Let's go back to Diego's
>> scenario:
>> 46: A >> B > C
>> 5: B >> A > C
>> 5:  B >> C > A
>> 44: C >> B > A
>> Suppose the A voters' utility for {A, B, C} is {100, 10, 0}, and
>> B's  likewise, mutatis mutandis; their estimated probability of A's
>> (B's)  election must be low indeed before it's rational for them to
>> approve B  (or to rank B in a Condorcet election).
> Lets assume:
>     B voters are going to vote as described (avoids debating how
> they might change).
>     There are 90 major party voters, desperately wishing to win, but
> willing to let B win on near ties, rather than risking opponents
> winning without earning full backing.
>     No cycles (which require full 3-way competition, while this is
> mostly 2-way.

This may be the source of our disagreement. I'm looking at expected
utility.

Assume the same utility vector as above for B's voters, and the above
breakdown, and calculate the aggregate utility of the three outcomes.

A: 4600+50+0+0 = 4650
B: 460+500+500+440 = 1900
C: 0+0+50+4400 = 4450

Alternatively, consider the A & C voters if they know in advance,
let's say from polling, the approximate shape of the profile.

If they bury B, they figure a tossup between A & C; expected utility:
50.
If they don't, B is a shoo-in; expected utility: 10.

Even one chance in five for an A (or C) voter beats a B sure thing.

So why rank (or approve) B?

>
>
> The pattern above works for this:
>    A or C may get more than 50 votes - and win.
>    Best of A or C may be less than 50 votes, letting B take a turn
> as winner.
>    A or C can get 50 (of the 90) - making a tie in A>B or C>B - have
> fun unless this gets resolved.
>
> Note that the pattern is neutral as to A vs C, and provides for hand-
> off to B on near A-C ties WITHOUT needing to estimate chances of a
> tie.
>
> A and C actually using it makes sense to me for:
>    Whichever is strong on election day wins.
>    If neither is strong then, better B than the enemy.
>
> I vote against "Random Ballot" for this task.

```

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