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        riddles >> easy >> a number picking game
        
(Message started by: Slayer on Feb 23^rd, 2009, 8:42am)

Title: a number picking game
Post by Slayer on Feb 23^rd, 2009, 8:42am

the game to be repeatedly played:
Player 1 and Player 2 each pick any natural number (zero is not natural).
player Lowest number wins $1 EXCEPT
when Lowest number = HighestNumber - 1, then player with Highest number wins $2.

Problem: Develop a long-term strategy for Player 1 such that Player 1 does not lose even if Player 2 knows the strategy Player 1 is using.

Title: Re: a number picking game
Post by towr on Feb 23^rd, 2009, 8:59am

[hide]Each round randomly pick 1, 2 or 3[/hide]

Title: Re: a number picking game
Post by Slayer on Feb 23^rd, 2009, 9:22am

on 02/23/09 at 08:59:46, towr wrote:

[hide]Each round randomly pick 1, 2 or 3[/hide]

Then player B can always choose 3, his expected gain is 1/3*(-1)+1/3*2=1/3 which means player A loses 1/3 on average.

Title: Re: a number picking game
Post by Grimbal on Feb 23^rd, 2009, 9:58am

Just play whatever you like.
Nowhere does it say how you can loose money. :P

If the collaborate, one plays 1,2,1,2,... the other 2,1,2,1,..., they both win $1 per round on average.

Title: Re: a number picking game
Post by towr on Feb 23^rd, 2009, 10:12am

on 02/23/09 at 09:22:13, Slayer wrote:

Then player B can always choose 3, his expected gain is 1/3*(-1)+1/3*2=1/3 which means player A loses 1/3 on average.

Hmm. :-/

Well, some other strategies I can think of are
[hide]Don't play[/hide]
[hide]Threaten player 2[/hide]

Title: Re: a number picking game
Post by howard roark on Feb 23^rd, 2009, 11:48am

I think equilibrium exists only when actions available to players is finite....which doesnt seem to be true in this case

Title: Re: a number picking game
Post by howard roark on Feb 23^rd, 2009, 11:49am

What happens when both chose the same number?

Title: Re: a number picking game
Post by rmsgrey on Feb 23^rd, 2009, 2:17pm

There's a venerable variant on this, which again has no upper limit on what can be played, in which playing higher numbers scores more points (though, again, only if your opponent plays at least 2 higher or 1 below)

In that variant, as I recall, the unbeatable strategy involves playing nothing over 6...

Despite the fact that playing one more than your opponent's number is best possible, if your opponent never plays higher than, say, 5, and only rarely plays that, then it's trivially not worth playing 7 or higher, and probably not worth playing 6...

Title: Re: a number picking game
Post by howard roark on Feb 23^rd, 2009, 5:21pm

on 02/23/09 at 14:17:38, rmsgrey wrote:

Is there a formal way to prove that?

Title: Re: a number picking game
Post by rmsgrey on Feb 24^th, 2009, 3:55am

on 02/23/09 at 17:21:14, howard roark wrote:

Is there a formal way to prove that?

Yes.

I just don't know what it is :P

(Okay, strictly, I have been told there exists a formal proof by someone who claimed to have seen it...)

Title: Re: a number picking game
Post by greenfin1 on Feb 24^th, 2009, 4:22am

its really confusing for me, tell the detail to play this game:

Title: Re: a number picking game
Post by Grimbal on Feb 24^th, 2009, 7:24am

on 02/24/09 at 03:55:02, rmsgrey wrote:

(Okay, strictly, I have been told there exists a formal proof by someone who claimed to have seen it...)

OK. That's something. If I tell someone about the problem I can tell that I read on a forum about someone who has been told there exists a formal proof by someone who claimed to have seen it. :P

Title: Re: a number picking game
Post by rmsgrey on Feb 25^th, 2009, 2:06pm

on 02/24/09 at 07:24:20, Grimbal wrote:

OK. That's something. If I tell someone about the problem I can tell that I read on a forum about someone who has been told there exists a formal proof by someone who claimed to have seen it. :P

Proof by urban myth?