Re: Twenty Bank Robbers -- Game theory:)
I think the best way to approach this problem is to first try to solve it assuming there are only two robbers rather than 20. Then once you have that figured out, try it for three, then four, and so on. Keep in mind that 50% support is enough for a proposed distribution to pass, you don't need a strict majority. Hal
At 4:05 PM -0700 7/25/96, Hal wrote:
I think the best way to approach this problem is to first try to solve it assuming there are only two robbers rather than 20. Then once you have that figured out, try it for three, then four, and so on. Keep in mind that 50% support is enough for a proposed distribution to pass, you don't need a strict majority.
Exactly. I arrived at the solution the same way. Note that there is another assumption needed--that the selection of a proposer is by lot at each new stage. If the ordering of proposers is known in advance, a different solution results. A further assumption is that a certainty gain of 1/n of the total sum is preferred to a 1/n probability of gaining the entire sum and a (1-1/n) probability of gaining nothing.. David
David Sternlight wrote:
At 4:05 PM -0700 7/25/96, Hal wrote:
I think the best way to approach this problem is to first try to solve it assuming there are only two robbers rather than 20. Then once you have that figured out, try it for three, then four, and so on. Keep in mind that 50% support is enough for a proposed distribution to pass, you don't need a strict majority.
Exactly. I arrived at the solution the same way. Note that there is another assumption needed--that the selection of a proposer is by lot at each new stage. If the ordering of proposers is known in advance, a different solution results.
Yes, the cypherpunk robbers are ordered by alphabet. igor
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