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
