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In a course I'm taking, the professor mentioned that a zero sum games are only interesting for 2 players.
Can someone explain me that?

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It's an online course with thousands of students, the support will be limited. He explained that a n-1 players general game can be expressed as a n players zero sum game (which I understand). But I don't see as obvious that any n players zs game can be reduced to a 2 players zs game. –  julio.g Mar 17 '12 at 21:27
    
The course by Jackson and Shoham? –  Michael Greinecker Mar 17 '12 at 21:52
    
Hey, I know the coursera game theory course! I know lots of people doing that one - I think it will be pretty slick. –  mixedmath Mar 17 '12 at 21:59
    
@MichaelGreinecker, mixedmath: Yes, that's exactly the one: coursera.org/gametheory/class –  julio.g Mar 18 '12 at 0:53

3 Answers 3

By definition of the zero-sum game, the losses of player p1 are entirely added the wining of player p2 and vice verse. If you introduce a third player, p3, and say if p1 losses 5 points, we can't move the 5 points to both players p2 and p3 simultaneously, otherwise the game will not be a zero-sum game. Also, we can't introduce a new distribution rule because the current methods will not then be valid.

For more insight take a look at the Google Books book:Quantitative tech. for management decisions

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Are you saying that there can't be a n-player zero sum game? –  julio.g Mar 18 '12 at 5:15
    
There are zero-sum games for two and more players, however, according the source quoted in the answer body, "Solution for n-person games is still being developed". –  Emmad Kareem Mar 18 '12 at 7:58

My guess on why n-person zerosum games are "not interesting" is that we don't have nice general results such as the minimax theorem(of 2-player zerosum games) for n-player zerosum games. I think one reason why the 2-player cases are so widely studied is because there are many interesting results on the whole class of such games.

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Take any $n$-player game. Add a $n+1$th dummy player who has a single action available and a payoff function $u_{n+1}$ given by $u_{n+1}(a_1,\ldots,a_n,a_{n+1})=-\sum_{i=1}^n u_i(a_1,\ldots,a_n)$. This $n+1$-player game is a zero-sum game that is strategically equivalent to the original $n$-player game.

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Michael, that I understand, what I can't get is how to go from that to the 2-players zs game's conclusion our teacher mentioned. –  julio.g Mar 18 '12 at 0:41
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One cannot reduce any n-player 0-sum game to a two player 0-sum game. The point is simply that 0-sum is an assumption with little interesting consequences for general games. –  Michael Greinecker Mar 18 '12 at 1:02

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