[EM] Generalization of Forest's problem
Richard Moore
rmoore4 at home.com
Thu Dec 20 17:04:32 PST 2001
Forest Simmons wrote:
> 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.]
>
<snip>
>>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.
I'm not sure the two methods are equivalent. The idea of a mixed strategy
is to maximize uncertainty among the other groups, so that they cannot
deliberately strategize in a way that causes maximum harm to your faction.
The optimum mix might be bimodal (for instance, all voters in the B group
vote B or all vote BC, with a 60% chance for the first strategy and a 40%
chance for the second strategy). But if you randomize the individual
votes, the result will be 60% B votes and 40% BC votes, plus or minus
some uncertainty.
-- Richard
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