On Thu, Mar 27, 2008 at 9:20 PM, Ian Kelly <[EMAIL PROTECTED]> wrote:
>  I think you're wrong.  ceiling(50 - ((50N-1)/N)) = ceiling(1/N) = 1.
Oh.. you're right.  Well, in that case, you'd get positive sum, S >
50N.  With zero-sum, it would only be possible for all parties to
leave if each had exactly 50 VP; if anyone had 51 VP, there wouldn't
be enough VP among the rest to go around; not so with positive-sum.
That's not the end of the world, but I would prefer a real zero-sum
game.

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