[EM] Generalization of Forest's problem

[Forest Simmons]

On Wed, 19 Dec 2001, Richard Moore wrote:

> 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.

For the stochastic cases (where randomization is needed for optimal
strategy) why not just let a, b, and c be the respective probabilities
for the CA, AB, and BC options (versus the corresponding bullet votes).

Each voter gets a random number between zero and one from a random number
generator.  If the member is in the C faction, then that member approves A
if and only if her random number is less than a.

[Note that this is like drawing colored balls from a bag with replacement,
whereas the non random use of a, b, c proposed above is like drawing
colored balls from a bag (having respectively C, A, B total balls) without
replacement.]

Now all calculations below can be based on expected utilities.

The cases where the optimal values of a, b, and c end up being zeros and
ones are the cases where mixed (stochastic) strategies are not needed for
optimization.

However, I must confess that this kind of thinking led me to look for
stable solutions among the pure strategies first, and so I ended up
over-looking Richard's (more stable) mixed strategy in my example. 


> 
> 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. 

Or just replace the slip, shuffle and draw again.

However, see my comment above for more drastic simplification.

Forest

> 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
> 
>