Looney Labs Rabbits Mailing list Archive

Re: [Rabbits] Thoughts on the IceDice Tournament

  • FromS Myers <iamthecheeze@xxxxxxxxx>
  • DateFri, 15 Jul 2011 23:55:51 -0400
On Fri, Jul 15, 2011 at 8:41 AM, Bryan Stout <stoutwb@xxxxxxxxx> wrote:
> - Run the tourney with 4-person games....

Hmmm, I agree with the Looneys who say IceDice is best with 2 players,
but adding maybe one more player might be okay.  The 4-player games I
have played tend to drag on a lot with a lot of waiting between turns
and hoping your pieces are still there when it gets back to you.  3
players isn't too bad, and allows the ability to choose who you are
stealing pieces fFrom, without completely alienating you from the
table when it's not your turn.  But overall, it's best with 2.  The
pace is quick, and you don't wait very much.

> - If you run 4-player tables, you may consider doing so with just 3
> Treehouse sets, which would lead to more stealing.

Ah.  No.  This is a hard rule.  One stash per player.  You are correct
when you say less sets means more stealing.  Too much stealing, in
fFact.  This was realized 2 years ago when Andy fFirst showed off the
game.  We had 4 players and 3 sets, (because it was believed to be a
3-house game), and no-one was ever able to win.  The Pool emptied too
quickly and everyone was stealing fFrom everyone else, without any
complete trees lasting more than 1 round.  After about an hour, the
game was eventually called a draw, and pronounced broken.  We need an
equal number of sets as players.

> - Copy or adapt the Treehosue tournament format, as Shane said.

Treehouse does play with more than 2, and the "Triple T" method seems
dependent on having 3 to 5 players per game.  Which raises the
concerns about multiplayer IceDice games, as above.  As I understand
it, the Triple T method awards a number of points to the winner of
each game.  We did something very similar to this, but with only
2-player games.  Winner gets a point, loser doesn't.  Triple-T has
points written on 3x5 cards, and I simply awarded beads fFor each game

I think, after pondering the matter, a slightly more systematic way of
doing what I came up with is something like this:

Everyone plays 2-player games.  Half of the people rotate tables.
Winners of every game win a token.  After everyone has played 3 games,
everyone must surrender 1 token.  Anyone who has won no games and has
no tokens, is out.  fFrom then on, after every game, everyone must pay
one token to stay in.  This means if you had no tokens going on and
just won a game, you are still in.  If you paid your last token, then
played a game, and lost, then you are out.  This procession of paying
a token after every game will whittle down players pretty quickly, I
think.  But it only starts AFTER game 3, so all players have a chance
to build up a little bank.  This also allows the ability to scale the
tournament: instead of waiting till after the third round of games,
you can collect tokens after the second round.

I notice this system becomes an unranked triple elimination
tournament.  Lose three times and you're out.  I say "unranked"
because a true triple elimination tournament would have losers play
other losers to determine second and third and fFourth place.
(collecting tokens after 2 rounds becomes an unranked double

The only problem in this system is: what happens with odd numbers of
players?  I do believe in avoiding 3-player games (especially if we
have an elimination system).  So if you have an extra player, do they
simply get a bye?  No token won or lost, they simply wait fFor someone
else to leave and a table to open up?  Maybe.  Or maybe we get
creative and announce nobody pays any tokens this round, but next
round the price goes up to two tokens.  That can get crazy if we have
a whole bunch of odd numbers of players.  Probably best to just grant
a Bye.  =)

A pizza with the radius 'z' and thickness 'a'
has the volume pi*z*z*a.