# [EM] Solution to approval strategy puzzle, Part 1

Richard Moore rmoore4 at home.com
Thu Mar 1 18:59:58 PST 2001

```I am posting this in two parts. The first part discusses the puzzle's
solution and the second part uses the problem as the basis for an
example of what I called "strategy matrix" in an earlier post. I'm
not sure if I previously made it clear what I meant by that phrase.

First, thanks to Mike and Joe for their correct answers.

Mike's initial response was interesting. But I was not looking for
a solution involving subterfuge or a change of methods, since I
phrased the question as "how should you vote" rather than "how
can you manipulate the process". His second response told
how to arrive at the solution. Joe gave the answer for the
break-even utility for B.

My solution follows. I've elaborated much more than Mike and
Joe, so that I can use the results to develop my strategy matrix
example.

> An approval election is held with three candidates (A, B, C)
> and four voters. The first three voters have the following
> probabilities of voting various ways:
>
> Voter 1    A (50%) or AB (50%)
> Voter 2    B (50%) or BC (50%)
> Voter 3    C (50%) or CA (50%)

> Assume ties will be broken by a fair and random lottery.

The probability of each possible outcome, with only
these three voters, is:

3-way tie    1/4    (A-B-C or AB-BC-CA)
AB tie         1/8   (AB-B-CA)
BC tie        1/8    (AB-BC-C)
CA tie        1/8    (A-BC-CA)
A wins        1/8    (A-B-CA)
B wins        1/8    (AB-B-C)
C wins        1/8    (A-BC-C)

Using the lottery to break ties, each candidate
gets a (1/12 + 1/16 + 1/16 + 1/8) = 1/3 chance
of winning (naturally, for this example!).

Your A vote would break any tie in which A was
a contender, convert the B and C wins to ties, or
convert the BC tie into a 3-way tie:

A wins    1/4    (A-B-C-A or AB-BC-CA-A)
A wins    1/8   (AB-B-CA-A)
3-way tie 1/8    (AB-BC-C-A)
A wins    1/8    (A-BC-CA-A)
A wins    1/8    (A-B-CA-A)
AB tie     1/8    (AB-B-C-A)
AC tie     1/8    (A-BC-C-A)

After possible tie-breakers:

A    1/4 + 1/8 + 1/24 + 1/8 + 1/8 + 1/16 + 1/16 = 19/24
B    1/24 + 1/16 = 5/48
C    1/24 + 1/16 = 5/48

An AB vote would convert the results as follows:

AB tie        1/4    (A-B-C-AB or AB-BC-CA-AB)
AB tie        1/8   (AB-B-CA-AB)
B wins        1/8    (AB-BC-C-AB)
A wins        1/8    (A-BC-CA-AB)
A wins        1/8    (A-B-CA-AB)
B wins        1/8    (AB-B-C-AB)
3-way tie    1/8    (A-BC-C-AB)

After tie-breakers:

A    1/8 + 1/16 + 1/8 + 1/8 + 1/24 = 23/48
B    1/8 + 1/16 + 1/8 + 1/8 + 1/24 = 23/48
C    1/24

1st election's expected utility = 33.33 + 23.33 = 53.67 (no 4th voter)
2nd election's expected utility = 79.17 + 7.29 = 86.46 (4th vote = A)
3rd election's expected utility = 47.92 + 33.54 = 81.46 (4th vote = AB)

So the best strategy is to vote A only.

This is different from the usual rule of using the mean utility as a
threshold,
because the electorate is so small that one voter has a considerable
chance
of becoming a tie-breaker, and therefore has a lot of power. The mean
utility rule is a valid approximation for large populations.

> At what B utility (keeping A and C constant) do your A and
> AB votes' strategic values become equal?

(19/24)*100 + (5/48)*X = (23/48)*(100+X)

(1500/48) - (18/48)*X = 0

X = 1500/18 = 83.33

This could explain why some IRV fanatics complain that approval voting
will lead to bullet voting. This opinion is only validd for elections
involving very small populations, where each voter has a lot of power.
Anyone who claims it will happen with a large electorate has failed to
understand what happens in approval voting with a large electorate: the
break-even utilities move towards the mean, which motivates most
voters to cast their ballots for more than one candidate.

-- Richard

```