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        riddles >> medium >> Hand Game
        
(Message started by: JP05 on Sep 18^th, 2007, 8:35pm)

Title: Hand Game
Post by JP05 on Sep 18^th, 2007, 8:35pm

Two players A,B engage in a game. A move consists in each showing an open, O, or closed, C, hand. If two O's show, A wins $3; if two C's show, A wins $1; if an O and a C show, B wins $2.

(1) Is there a winning strategy for A? for B?
(2) If there is one, is it unique?

Title: Re: Hand Game
Post by srn347 on Sep 18^th, 2007, 10:36pm

A wants them to choose the same, and b wants them to choose opposites. A prefers that the one they both choose is O, which b knows, which a knows b knows. It's paradoxial.

Title: Re: Hand Game
Post by Sameer on Sep 18^th, 2007, 10:42pm

Does the game go like this? A and B both have hands behind their backs and on count of three they either bring 1 or both of their hands forward either in open or closed position?

Title: Re: Hand Game
Post by JP05 on Sep 18^th, 2007, 10:56pm

You must know that it does not matter if they use two hands to convey the O or C, but that they convey one or the other.

Perhaps I don't understand your question but you might be thinking of something having to do with symmetry but that has nothing to do with the problem at all.

Title: Re: Hand Game
Post by towr on Sep 18^th, 2007, 11:31pm

It seems like the prisoners dilemma. To win they have to cooperate, but if they cooperate A might benefit from defection. However it seems B is in control: if he gives O, but doesn't get enough C is return from A, he can punish A by only giving Cs (which means A's average is much lower than when A cooperates).

Not quite sure if they would reach a nash equilibrium though. (not enough time to think on it now either)

Title: Re: Hand Game
Post by srn347 on Sep 19^th, 2007, 6:36am

Oh, I forgot the winning strategy for b. Choosing O limits what a gets, so A would have considered it and chose O, so b would know that and choose A(paradox again).

Title: Re: Hand Game
Post by Aryabhatta on Sep 19^th, 2007, 10:52am

I haven't gone through all the motions, but it looks like A must select O with probability 1/3.

Isn't this a typical Game Theory question (or Prisoner's Dilemma as towr mentioned)?

Title: Re: Hand Game
Post by Michael_Dagg on Sep 19^th, 2007, 11:12am

One can consider that the payout goes to the opponent which lets
you work with minimums in each case. A has no winning strategy.
B does and that probability lies in (1/3,2/5) and can't be unique but
there seems to be an optimal number in that interval as well.

Title: Re: Hand Game
Post by Joe Fendel on Sep 21^st, 2007, 3:10pm

I don't think there is any true equilibrium here. Clearly a strategy must be "mixed", picking open a certain random percentage of the time. We can first approach the problem by assuming one player decides and announces his mixed strategy before the other chooses.

If A decides first, he will choose open 49.9% of the time (randomly). That way, B does best to choose open 100% of the time, winning $2 about 50.1% of the time (averaging slightly over $1 per game). The remaining 49.9% of the time, A will win $3, which averages to about $1.50 per game for A.

If B decides first, he will choose open 25.1% of the time. That way A should choose open 100% of the time, winning $3 25.1% of the time (averaging slightly over $0.75 per game), while B wins $2 74.9% of the time (averaging $1.50 per game).

If A picks open 50% of the time, and B picks open 25% of the time, then 1/8 of the time A wins $3, 3/8 of the time A wins $1, and 1/2 the time B wins $2. Thus A wins $0.75 per game, and B wins $1 per game, an outcome worse for both players than either outcome above. Maybe this is an equilibrium of sorts, but it is highly unstable.

If A begins with a strategy of picking open 49.9% of the time, and B begins with a strategy of picking open 25.1% of the time, this becomes a game of economic "chicken." Players goad each other, saying "Come on, choose open! I'm not gonna budge, and you can win more on average if you do!"

However, I think that A has the leverage here. If B refuses to raise his open percentage, A can begin lowering his (which is slightly to his detriment in the short run), until B has no real option except to increase his open %, which will allow A to follow suit, at least back up to 49.9%. B, on the other hand, cannot really afford to lower his open %, because by crossing the 25% mark, he actually incents A to reduce his open %. Since A has this "punishment mechanism" available to him that B does not, I think A has leverage.

So my intuition, which I can't quite prove, is that the strategy of A choosing open 49.9% of the time and B choosing open 100% of the time is the long-run equilibrium. A simply has to maintain the discipline to not pass that 50% mark, and he'll continue to rake in $3 every other game.

Title: Re: Hand Game
Post by towr on Sep 22^nd, 2007, 3:25am

on 09/21/07 at 15:10:17, Joe Fendel wrote:

Well, if they can announce their strategy, as B I'd just announce
I'll repeatedly pick OCCOC, as long as you (A) pick OOOOO; if you don't then I'll pick C untill you have picked O again often enough to compensate. This way we both get the most from the game (each 1.2 per round on average).

If A doesn't cooperate he'll get only 1. But then, B wouldn't get anything if A doesn't cooperate, so you can argue against the rationality for B; which is why rationality is overrated when fairness is concerned. If your opponent won't play fairly, don't play; getting nothing is better than getting an unfair cut.

Title: Re: Hand Game
Post by Michael_Dagg on Sep 22^nd, 2007, 10:41am

Take P,Q as the probabilities that A and B show O. A
needs to choose P so that, regardless of what B shows,
the (expected) payoff to A will be > 0. i.e,
min{3P - 2(1 - P), (1 - P) - 2P} > 0. But this says that
P > 2/5 and P < 1/3 which can't happen so A has no
winning strategy.

B needs to pick Q so that min{2(1 - Q) - 3Q, 2Q - (1 - Q)} > 0.
This says 1/3 < Q < 2/5. So the the strategy for B is
not unique but best when Q = 3/8.