[EM] Generalization of Forest's problem

[Richard Moore]

Here's Forest's problem in a more general form.

I'm still assuming 3 factions with the utilities Forest proposed.

The sizes of the factions are constants A, B, and C:

A: 
A(100) 
B(50) 
C(0)
B: 
B(100) 
C(50) 
A(0)
C: 
C(100) 
A(50) 
B(0)

The fraction of A voters who vote AB will be designated as b; the
fraction of B voters who vote BC is c; and the fraction of C voters
who vote CA is a.

Total votes for each candidate:

A gets A + Ca votes;
B gets B + Ab votes;
C gets C + Bc votes.

Now suppose the B voters are trying to evaluate their *unilateral*
Approval strategy.

The two independent variables they need to consider are a and b. They
need to determine the optimum value of c.

If (A + Ca) == (B + Ab), then their favorite candidate B is tied with
their least favorite candidate A. If there is a 50/50 tie-breaker, then
their expected utility is the same whether they vote to ensure their
compromise wins, or whether they vote to ensure their compromise loses
and gamble on the tie-breaker. So it's a wash, though there may be
psychological factors beyond the stated utilities involved in each
voter's choice.

For all cases where (A + Ca) > ( B + Ab), there is no way B (their favorite)
can win, so they want their compromise (C) to beat candidate A. They
should choose c such that:

	(C + Bc) > (A + Ca), or

	c > (A/B) + (C/B)(a-1).

In other words, the larger a is, the larger c needs to be. Cost of picking
too small a value of c is 50 points (least favorite beats compromise).

For all cases where (A + Ca) < ( B + Ab), there is no way A (their least
favorite) can win, so they want to minimize the reliance on their 
compromise.
They should choose c such that:

	(C + Bc) < (B + Ab), or

	c < 1 - (C/B) + (A/B)b.

In other words, the smaller b is, provided the conditional inequality is
still met, then the smaller c needs to be. Cost of picking too large a
value of c is 50 points (compromise beats favorite).

Case 1: The B group has a majority. In this case, the B group should set
c < 1 - (C/B). The A and C groups together are a minority so they cannot
defeat B without help from the B group. Therefore the B group should use
c = 0.

Case 2: The A group has a majority. Since the A group will probably set
b = 0 (due to the rationale in Case 1), the B group should set c = 1.
This will not defeat A, unless there is a poor turnout from the A group,
but it has symbolic value as a statement against A. Also, if the A group
is inexplicably altruistic and casts some AB votes, C will not win
anyway, so B would then have a chance to win.

Case 3: The C group has a majority. They will probably set a = 0 (Case 1),
and the A group will probably set b = 1 (Case 2). The best bet for the B
group is then to set c = 0.

Case 4: No group has a majority. The B group has no idea what values of
a and b the other two groups will use. Unfortunately, there is no overlap
between the two regions defined by the above inequalities:

Region 1: a > (B/C) + (A/C)(b-1)
           c > (A/B) + (C/B)(a-1)

and

Region 2: a < (B/C) + (A/C)(b-1)
           c < 1 - (C/B) + (A/B)b

though the two regions touch along a line L:

L: a = (B/C) + (A/C)(b-1)
    c = (A/B) + (C/B)(a-1) = 1 - (C/B) + (A/B)b

(NOTE: a, b, and c are confined to the range [0,1] inclusive.)

This is the difficult case. A mixed strategy is needed. Suppose the B
faction's strategy options are c = 0, c = 1/30, c = 2/30, c = 3/30,
..., c = 29/30, c = 1. They calculate the optimum probability for each
value of c (I confess I don't know how to do this, but I think game
theory does provide a method). They give each of their voters a unique
number from 1 to 30. They then put slips with numbers from 0 to 30
into a barrel. Each of the numbers may occur multiple times, in
proportions equivalent to the calculated probabilities. They draw one
slip at random, and tell all voters with a number less than or equal
to the number on the slip to vote BC, and all other voters to vote B
only.

Of course, that might not be practical in a public election, but in
a public election, the distribution of utilities would never be so
simplistic either.

I'll read up on game theory to see how the mixed strategy should be
calculated. I know how it works in simple cases but having three
variables complicates things.

  -- Richard