Mathematical utility expectation maximization in Approval

[Bart Ingles]

MIKE OSSIPOFF wrote:
>
> Bart wrote:
> >
> >Example:
> >- You have candidates A, B, C, D, and E,
> >- A has an 90% chance of winning,
> >- B and C each have a 4% chance of winning,
> >- D and E each have a 1% chance of winning,
> >then A, B, and C are considered the frontrunners
> , since they are equally
> >likely to be involved in a tie for first place (either 2-way or 3-way).
> 
> But what I meant by the 2 frontrunners is: the 2 candidates who,
> in the final election count, have the highest vote totals.
> 
> Candidate A is more likely than B or C to be in a tie for 1st place,
> because he has a greater win probability. Numbers proportional to
> these candidates' individual frontrunner probabilities are:
> A: .95, B: .2, C: .2   There's a .04 probability that B & C will be
> the 2 frontrunners, but there's a .19 probability of A & B being the 2
> frontrunners in the final election count. Those are estimates based
> on Tideman's square root suggestion.

My point was that, given a tie exists for 1st place, then it is most
likely between A&B or A&C, with a lesser chance of it being between A&D
or A&E.  Assuming the chance of a tie not involving A is negligable,
along with the chance of a multi-way tie, then Pt(AB) = Pt(AC) = approx.
0.4, and Pt(AD) = Pt(AE) = approx. 0.1

This assumes that the probability of an underdog being in a tie, given
one exists, is the same as the probability of an underdog winning an
upset, given one exists.  I'm not 100% certain of this assumption, but
it seems plausible.

I was incorrect in saying that A, B, and C are equally likely to be in a
tie, since the probability of A being in a tie is approx. 1.0, vs. 0.4
each for B and C.  But the probability of B or C being in a tie, given a
tie exists, is not necessarily proportional to the likelihood of B or C
winning.

Bart