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  #1  
Old 04-18-2005, 05:55 PM
Jeff W Jeff W is offline
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Default Two Six-Packs of Beer on a Game Theory Wager

I have a wager with my friend on a point of game theory. Can someone verify my claim below? If you have a link to a credible source, I would appreciate it.

JeffW Okay, my wager is
JeffW In a two person zero sum game
JeffW The player who is second to act
JeffW Always wins or ties under the following conditions:
JeffW 1. He has perfect information about the game.
JeffW 2. He plays perfectly under all contingencies.
JeffW okay, 2 six packs of beer valued at no more than $15 total.
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  #2  
Old 04-18-2005, 06:19 PM
xCEO xCEO is offline
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Default Re: Two Six-Packs of Beer on a Game Theory Wager

if he allways wins or ties, is this a zero-sum game?
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  #3  
Old 04-18-2005, 06:25 PM
MCS MCS is offline
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Default Re: Two Six-Packs of Beer on a Game Theory Wager

[ QUOTE ]
if he allways wins or ties, is this a zero-sum game?

[/ QUOTE ]

Sure. Let's say the game is: you pick an integer between zero and five, then I pick an integer between zero and five. Then, I receive 10 points and you receive -10 points. That's a zero-sum game I always win.

Which actually leads me to the question of whether you can have the same trivial case serve as a counterexample to your claim (where the result is always 10 for player one and -10 for player two).
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  #4  
Old 04-18-2005, 06:31 PM
Jeff W Jeff W is offline
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Default Re: Two Six-Packs of Beer on a Game Theory Wager

I think I am missing an assumption somewhere. The basic argument was that in Chess, according to game theory, black must either win or tie if he plays perfectly. He cannot lose.
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  #5  
Old 04-18-2005, 06:41 PM
MCS MCS is offline
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Default Re: Two Six-Packs of Beer on a Game Theory Wager

Hmmmm. I would think that white should always win or tie, not black.

In tic-tac-toe on a 4x4 board where you still only need three in a row, the first player always wins.

And yeah, chess is a game of complete information and thus theoretically "solvable."
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  #6  
Old 04-18-2005, 06:52 PM
WhiteWolf WhiteWolf is offline
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Default Re: Two Six-Packs of Beer on a Game Theory Wager

I do not think this follows from game theory. I'll make up a trivial game: Put four beans in a jar. Two players alternate moves. For each move, you can take out either one or two beans. The player who removes the last bean wins. In this game, the smart first player will remove one bean, leaving three beans in the jar. No matter what move the second player makes (removing one bean to leave two, or two beans to leave one), the first player will be able to win, thus disproving the hypothesis that moving second is always an advantage....
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  #7  
Old 04-18-2005, 06:54 PM
bobbyi bobbyi is offline
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Default Re: Two Six-Packs of Beer on a Game Theory Wager

You are definitely missing some part of the description of what you are trying to say, because as it stands now it is trivial to come up with games under your conditions that do not have first mover advantage. Suppose we play iterative roshambo. You say "rock", "paper" or "scissors". Then I do. Obviously, this game is a win for the second player.
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  #8  
Old 04-18-2005, 07:18 PM
Iceman Iceman is offline
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Default Re: Two Six-Packs of Beer on a Game Theory Wager

[ QUOTE ]
I have a wager with my friend on a point of game theory. Can someone verify my claim below? If you have a link to a credible source, I would appreciate it.

JeffW Okay, my wager is
JeffW In a two person zero sum game
JeffW The player who is second to act
JeffW Always wins or ties under the following conditions:
JeffW 1. He has perfect information about the game.
JeffW 2. He plays perfectly under all contingencies.
JeffW okay, 2 six packs of beer valued at no more than $15 total.

[/ QUOTE ]

In Connect Four, if both players play perfectly the first player always wins. The game Hex is also a first player win with perfect play.

John Nash, the subject of "A Beautiful Mind", proved that Hex is a win for the first player:

http://maarup.net/thomas/hex/
(scroll down to "first player wins")
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  #9  
Old 04-18-2005, 07:21 PM
marv marv is offline
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Default Re: Two Six-Packs of Beer on a Game Theory Wager


Just because the game has perfect information and is zero sum still allows lots of flexibility since the game is necessarily asymmetric.

Silly example:

We flip a coin.
Both players are told the result.
Player 1 chooses heads or tails.
Player 2 has no actions.
If player 1 guesses correctly he wins USD 1 from player 2.
If player 1 guesses incorrectly he gives USD 1 to player 2.

This is zero sum but under optimal play the second player always loses.

Similarly we can construct second-player advantage games.

Marv
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  #10  
Old 04-18-2005, 08:54 PM
pzhon pzhon is offline
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Default Re: Two Six-Packs of Beer on a Game Theory Wager

[ QUOTE ]
I think I am missing an assumption somewhere.

[/ QUOTE ]
Yes. You lose the bet.

[ QUOTE ]
The basic argument was that in Chess, according to game theory, black must either win or tie if he plays perfectly. He cannot lose.

[/ QUOTE ]
That is not known. It is not a simple consequence of basic principles of game theory. It is conceivable that the opening position could be an instance of zugzwang.

Some games like hex or chomp have a structure that allows you to do at least as well as the first player as you could as the second player. In those games, the first player has a strategy that at least ties. However, not all games have this property, and it has not been proven that White can at least draw in chess.
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