Sunday, September 11, 2011

Game Theory: Part II: Dominating Strategies

And now on to the next item in the discussion of game theory: strategies that are absolutely better than other strategies.

The following material is flagged Green Level. It is intended to reflect material that the author believes to be a matter of consensus among experts in the field. This belief may be incorrect, however; and as the author is not an expert and does not have an expert fact-checking the article, errors may creep in.
Suppose that I offer you two possible bargains: either I will give you $1, or I will flip a coin and give you $1 if it lands heads, nothing otherwise. It would make sense for you to pick the first offer, right?

Right. The first offer a) is never worse than the second offer, and b) is sometimes better than the second offer. Similar situations can happen in game theory. Let us look at the following:


A

B
A $1 $2
B$-2$0
You would obviously be better off picking choice A. No matter what I pick, A is always better for you than B. So what will I pick?

That's right. Since I know you will pick A, I will minimize my losses and pick A as well. If you were to change your choice in response to my choosing A, I would only be pleasantly surprised, and the same would still be true if I were to pick B. Thus, you are always better off picking A.

And this leads to the main subject of this post: dominating strategies. A strategy (the game-theory term for a particular move or action) is considered dominating over another one if it is a) never worse for the player choosing it, and b) better at least once for the player choosing it.

So, here's where the core principle of game theory gets into it. Because your opponent is acting intelligently, (or at least non-randomly), your opponent is allowed to choose strategies. Barring unusual circumstances (such as the iterated prisoners' dilemma) that will be mentioned later, a rational opponent (that is, one acting in its own best interest) will never choose a dominated strategy. In a zero-sum game, it is generally best to assume that your opponent will never choose a strategy if one that always hurts you at least as much is available.

So, given the following game, what do you do?



A

B
C
A$0$0-$1
B$1$1$1
C$1$0$2
Let's look at this. Your opponent will never play A, because A is dominated by B.




B
C
A
$0-$1
B
$1$1
C     $0$2
 You will never play A, because A is dominated by all other strategies.





B
C




B
$1$1
C     $0$2
Your opponent will never play C, because C is dominated by B.





B





B
$1
C     $0     
You will never play C, because C is dominated by B.




B





B
$1

    
     

And so we have the final outcome: both players will play B, resulting in you winning $1. Thus, for you, the value of this game is $1, and the value for your opponent is -$1.
<<Notation and Basic Concepts | Game Theory | Utility Functions>>

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