[EM] Suggestions for choice of Best Frontrunner versions

[MIKE OSSIPOFF]

  I'm talking here about the Best Frontrunner (BF) versions that assume that 
at least 1 of the 2 expected frontrunners will be in the tie or near-tie if 
there is one.

But, first, referring to the BF version that doesn't make that assumption, I 
used it to determine how to vote if some actual public election were by 
Approval, and found that its estimates were easy, if there are 2 clear 
expected frontrunners. I like it and would likely use it under those 
conditions.

  But, back to the BF versions that assume that at least 1 of the 2 expected 
frontrunners will be in the tie or near-tie if there is one:

   Here are the ones that I compare:

1.  The one that considers the probabity that F1 will outpoll F2, and the 
probability that F2 will outpoll F1. I'll call that BF(probability-weighted)

2. Rob LeGrand's strategy that puts the cutoff point next to the expected 
top votegetter, on the side toward the expected #2 votegetter. I'll call 
that BF(1st)

3. The ordinary BF, that says to vote for the best of the 2 expected 
frontrunners, and for everyone whom one likes better. I'll call that 
BF(plain)

4. The BF that says to vote for everyone who is better than the mean of F1 & 
F2. I'll call that
   BF(mean)

So it's:

1. BF(probability-weighted)
2. BF(1st)
3. BF(plain)
4. BF(mean)

Of course BF(probability-weighted) is the ultimate of these, and is the 
clear best--if one estimates the cardinal utilities needed, and the 
probability that F1 will outpoll F2, and the probability that F2 will 
outpoll F1.

  But the need for those estimates could be called a disadvantage, if the 
estimates aren't obvious.

  So what remains is to compare BF(plain), BF(1st) and BF(mean).

  As for the matter of how far the cutoff point could be off by, compared to 
where
  BF(probability-weighted) would put it, BF(plain) comes in last, because it 
could put the cutoff point off by F1-F2. BF(1st) & BF(mean) could only put 
it off by half that much.

  So then it's a matter of comparing BF(1st) to BF(mean).

   If you consider F1 & F2 to be equally likely to outpoll eachother, or if 
you don't have an estimate about which is more likely, then use BF(mean). If 
you believe that one is far more likely to outpoll the other, then use 
BF(1st).

Of course, BF(mean) has greater applicability, because it's the one to use 
if you have no idea which of F1 or F2 is more likely to outpoll the other.

  But, in another way, BF(1st) could be said to have more applicability, 
because it doeesn't require as much in the way of utility estimates: , it  
doesn't require judging if candidates are better than the mean of F1 & F2. 
BF(1st), as BF(mean) does. It only asks about ordinal utility relations.

  So: if it's clear who's better than the mean of F1 & F2, and  (if F1 & F2 
are about equally likely to outpoll eachother, or if you have no idea about 
that), then use BF(mean)

  But if it isn't easy to judge whether a candidate is halfway between the 
utilitiles of F1 & F2, or if one of {F1,F2) is by far the more likely to 
outpoll the other, then use BF(1st).

Mike Ossipoff

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