Re: [spoon-discuss] Re: [Spoon-business] Let's play the Voting Game!

[Rob Speer <rspeer@xxxxxxx>]

On Fri, Jan 30, 2004 at 04:43:31PM -0600, Baron von Skippy wrote:
> It'th quite thimple, really. If we all vote for Foo, Baz, Qux, and Spam, and againtht Bar, we all get -10 + 0 + 0 + 50 + 30 = 70 pointh. But if one of uth betrayth the rest and voteth for Bar abd againtht Foo, that player geth 30 + 40 + 30 = 100 points, a much nyther package, if I do thay tho, and I would know. But here'th where it geth tricky. If we /all/ do that, then we all get 0 + 0 + 0 + 0 + 30 = 30 pointh. Tho it'th quite thimple, babeth: We mutht all vote for Foo, Baz, Qux, and Spam, and all vote againtht Bar. Or a /thet/ number could switch their voteth on Foo and Bar, and we thplit the pointh, but I'm afraid that thuch a thing ith jutht too rithky in a game like thith. Thorry, all. I love you like my family, but you're thuch thelf-interethted little monkeyth.
> 
> Anyway, Wonko, thince you're doing tho well tho far, I'd thay you're front-runner to get the Thcarf at the end of the nweek.-

I don't see why you count Foo as -10 for voting for it, instead of 10
for voting against it. That just makes our numbers off by 10 - you say
70 when I say 80.

I can't see a way to get 100 points. Since voting against Bar is the
dominant strategy, Bar won't pass, so the most points you can get from
the first four proposals is 50 (whether it's 10+40 or 50). With Spam,
that makes the best total 80. So there isn't even a prisoner's dilemma,
though it would be cool if there were one.

In fact, you can't really make a true prisoner's dilemma for the entire
Nomic, because winning the Nomic is zero-sum, even though the scores
aren't. If you lose 50 points but so does everyone else, you haven't
really lost anything. (The same goes for gaining points, unless you
happen to have 9xx points, but prisoners' dilemmas are usually phrased
in terms of losing rather than winning.)

-- 
Rob Speer

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