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 Science, Math, and Philosophy Discussions regarding science, math, and/or philosophy.

08-03-2012, 12:41 AM   #31
journeyman

Join Date: Jun 2010
Location: Sydney
Posts: 303
Re: What is the optimal strategy for this game?

Quote:
 Originally Posted by Rhaegar P.S. That's one ****ty game.
It is a great game. We just never played it with 2 players. The minimum of players was 4 and going first seemed to be a big disadvantage.

 08-03-2012, 06:39 AM #32 Pooh-Bah     Join Date: Aug 2006 Location: you got it Posts: 4,047 Re: What is the optimal strategy for this game? I was going to ask for the multi-player strategy but assumed it would be horrendously complicated.
08-03-2012, 02:40 PM   #33
Carpal \'Tunnel

Join Date: Sep 2002
Location: Henderson, NV
Posts: 21,429
Re: What is the optimal strategy for this game?

Quote:
 Originally Posted by lastcardcharlie I was going to ask for the multi-player strategy but assumed it would be horrendously complicated.
I would guess that it starts with every player picking their values at random.

It probably also has Player 1 guessing either 4 or 5 at random (50/50) since these are the two most common values given that everyone picks randomly.

Player 2 will need some sort of mixed/balanced strategy to hide information from Player 3, and that's probably where things get ugly.

 08-03-2012, 05:23 PM #34 veteran     Join Date: Feb 2007 Location: Razzville Posts: 2,108 Re: What is the optimal strategy for this game? I have played this game 10 handed. Just trying to remember which numbers have already been taken is enough of a job.
 08-03-2012, 06:06 PM #35 Pooh-Bah     Join Date: Sep 2004 Posts: 4,346 Re: What is the optimal strategy for this game? P3 shouldn't do worse than half the pot (in a repeated game)- he and P2 can implicitly collude to chop it no matter what P1 does (if P1 guesses 5 or less, P2 and P3 agree to both choose 3 and guess 6-9 randomly, if P1 guesses 6+, P2 and P3 agree to both choose 0 and guess 0-3 randomly). This is clearly better than his noncooperative equity choosing 50% 3 and guessing 6/50% 0 and guessing 3 after P1 and P2 choose randomly and guess 4 and 5. Since P1+P2 random has equity: P3 (choose 3 guess 6/choose 0 guess 3): 4 of 16 winning combos P2 (choose 4): 2 or 3 of 16 winning combos P1 (choose 5): 3 or 2 of 16 winning combos P2 has 5/18 equity not cooperating, so there's a collusion/defection region where P2 and P3 are haggling over the extra 4/18 of the pot. I'm not sure what happens in there.

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