Looney Labs Icehouse Mailing list Archive

Re: [Icehouse] Your favorite (and least favorite) 3 (or 7) games

  • FromDennis Duquette <dennisdduquette@xxxxxxxxx>
  • DateSun, 22 Mar 2009 09:26:44 -0700 (PDT)
Hello. I'm also sticking to only Icehouse games I've played.

Zendo is my favorite icehouse game, by far.

Martian Chess is fantastic! I prefer 4 player to 2 player. I've some foam board and printouts of ee0r's Martian Chessboard Sections (http://www.ee0r.com/tri-chess/index.html), but have not yet constructed my half dozen of these to try 3 player.

I never got my head around Zarcana or Gnostica, but I love Zark City! I'm one of those who likes to blend luck with strategy: It's much better practice for real life.

I like Homeworlds. With other adults I'd much rather play any of the above 3 games, but I'd not turn down a game of Homeworlds.

Most of my gaming lately is with toddler to elementary school age players. We have a preschooler and toddler twins, and visits with elementary school aged nieces, friends' kids, etc. With these and Icehouse newbies I like Treehouse, Martian Coasters, Moonshot and Drip (in about that order). Our kids' favorite is Drip, or The Icehouse Null Game (http://icehousegames.com/contest/icedes-1/ting/ting.html) without a TUG :) They're only allowed to play with pyramids under close supervision, so every time is a treat.

That's 8 games, but there are 3 types of mathematicians, those who can count and those who can't.

I'm also fond of the pyramids for their other uses. I've used them for scoring, and pseudo-scoring counters, (e.g. hit points, fatigue and mana in role-playing games) in other games. They are also visually very appealing link markers for Illuminati New World Order (now sadly out of print).

I shall never be drunk enough that spicklehead (http://www.dangermouse.net/games/icehouse/spicklehead.html) will sound like a good idea.

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