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Topic: a number picking game (Read 518 times) |
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Slayer
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a number picking game
« on: Feb 23rd, 2009, 8:42am » |
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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.
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Slayer
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Re: a number picking game
« Reply #2 on: Feb 23rd, 2009, 9:22am » |
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on Feb 23rd, 2009, 8:59am, towr wrote:Each round randomly pick 1, 2 or 3 |
| 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.
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Grimbal
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Re: a number picking game
« Reply #3 on: Feb 23rd, 2009, 9:58am » |
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Just play whatever you like. Nowhere does it say how you can loose money. If the collaborate, one plays 1,2,1,2,... the other 2,1,2,1,..., they both win $1 per round on average.
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towr
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Re: a number picking game
« Reply #4 on: Feb 23rd, 2009, 10:12am » |
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on Feb 23rd, 2009, 9:22am, 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 Don't play Threaten player 2
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« Last Edit: Feb 23rd, 2009, 10:13am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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howard roark
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Re: a number picking game
« Reply #5 on: Feb 23rd, 2009, 11:48am » |
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I think equilibrium exists only when actions available to players is finite....which doesnt seem to be true in this case
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howard roark
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Re: a number picking game
« Reply #6 on: Feb 23rd, 2009, 11:49am » |
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What happens when both chose the same number?
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rmsgrey
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Re: a number picking game
« Reply #7 on: Feb 23rd, 2009, 2:17pm » |
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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...
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howard roark
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Re: a number picking game
« Reply #8 on: Feb 23rd, 2009, 5:21pm » |
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on Feb 23rd, 2009, 2:17pm, rmsgrey wrote: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... |
| Is there a formal way to prove that?
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rmsgrey
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Re: a number picking game
« Reply #9 on: Feb 24th, 2009, 3:55am » |
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on Feb 23rd, 2009, 5:21pm, howard roark wrote: Is there a formal way to prove that? |
| Yes. I just don't know what it is (Okay, strictly, I have been told there exists a formal proof by someone who claimed to have seen it...)
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greenfin1
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Re: a number picking game
« Reply #10 on: Feb 24th, 2009, 4:22am » |
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its really confusing for me, tell the detail to play this game:
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Grimbal
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Re: a number picking game
« Reply #11 on: Feb 24th, 2009, 7:24am » |
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on Feb 24th, 2009, 3:55am, 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.
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rmsgrey
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Re: a number picking game
« Reply #12 on: Feb 25th, 2009, 2:06pm » |
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on Feb 24th, 2009, 7:24am, 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. |
| Proof by urban myth?
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