Looney Labs Icehouse Mailing list Archive

Re: [Icehouse] Number of players / Game popularity

  • FromDennis Duquette <dennisdduquette@xxxxxxxxx>
  • DateFri, 20 Feb 2009 10:18:03 -0800 (PST)
--- On Thu, 2/19/09, David L. Willson <DLWillson@xxxxxxxxxx> wrote:

> I wonder how a modification of Zendo would work out:
> Any player with a guessing stone may guess ~every~ time a
> koan is completed.
> Guessing priority begins with the player who built the
> koan.  That player may take one guess, if he has a guessing
> stone, or pass. Then each other player, in turn order, may
> spend a guessing stone.  The player who built the koan gets
> a second opportunity to guess, then all the other players,
> in turn order, until all players pass.  At that point, the
> "turn" moves to the next player in turn order who
> begins creating their koan.
> It seems to me that that would keep players more involved. 
> They'd have more frequent opportunities to guess, and
> watching the koans as they happen would be much more
> important.

I'm going to try this :D

However, for competitive Zendo experiences, I suggest a different guessing mechanic (think of guessing as initiating a guessing phase with an instantaneous auction):

After any player incorrectly guesses, the other player with the most guessing stones may guess or pass. Ties for the most stones are resolved by an auction where players reveal simultaneously as many guessing stones as they are willing to pay for the next guess. The guessing phase ends and koan building resumes where it had been left off for any ties on the guessing auction or when every player with guessing stones has passed.

Dennis D Duquette


All numbers are equal.

Choose arbitrary a and b, and let t = a + b.
(a + b)(a - b) = t(a - b)
a^2 - b^2 = ta - tb
a^2 - ta = b^2 - tb
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
(a - t/2)^2 = (b - t/2)^2
a - t/2 = b - t/2
a = b