1. As the fourth voter, with utilities A = 100, B = 70, and C = 0,
how should you vote: A or AB?
As I was saying, vote only for A, if you're letting them flip coins.
2. Explain how you arrived at the answer to #1.
For each of the 8 equiprobable ways that those 3 people could all vote,
I might have said "Add the expressions and solve for Ub", when
I meant to say "Add the expressions for the 8 situations, set their
sum equal to zero, and solve for Ub."
Mike Ossipoff
_
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I have ambiguous feelings about my two favorite methods.
Sometimes it seems like Condorcet is over-elaborate, making fine
distinctions between the deck chairs on the sinking Titanic. (How's that
for a mixed metaphor?)
On the other hand, there are times when Approval strategy leaves me in a
On Tues. eve., suitably enfeebled, I turned to Richard's Approval Strategy
puzzle and argued basically as Mike suggests. Voter #4 should vote both A
and B if B has utility Ub and vote just A if B has utility Ub, where -
if my arithmetic was correct - Ub = (5/6)*100 = 83 1/3.
Joe
Forest Simmons wrote:
I have ambiguous feelings about my two favorite methods.
Sometimes it seems like Condorcet is over-elaborate, making fine
distinctions between the deck chairs on the sinking Titanic. (How's that
for a mixed metaphor?)
On the other hand, there are times when
Forest Simmons wrote:
I have a couple of suggested compromises starting with Condorcet and
moving towards Approval, but stopping short of ordinary Approval.
Looking at just the first for now...
I. Condorcet. I prefer the version of Condorcet that allows voters to
give a partial preference
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
Now that we've calculated the raw probabilities, what do I mean by
"strategy matrix" and all those delta Ps?
With the probabilities known for each outcome, we can calculate how
a specific way of voting affects the probabilities. In particular, we
are
interested in the difference between voting
Don wrote:
Your two person co-chair election should be conducted as follows:
1) Divide the total votes by the number of seats. This is known as
the Hare Quota. You should get two halves. One half will end up on one
candidate and the other half will end up on another candidate.
Do
Joe wrote (in relation to Forest's Condorcet//Approval compromise):
I like it - was considering the same thing myself (Smith//Approval, I
guess it'd be called). If there's only strategy in Condorcet methods if
there's a tie, then it'd make sense to resolve the tie using methods
which have
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