3.4 Game Theory
Part C: Solving 2 × 2 Games

Part A: Zero-Sum Games and Mixed Strategies
Part B: Fundamental Principle of Game Theory and Reduction by Dominance

(This topic is also in Section 3.4 in Finite Mathematics and Finite Mathematics and Applied Calculus)

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Recall the following from the previous tutorials: For every mixed strategy one player uses, the other player has a best pure counter-strategy. A player using the minimax chooses a mixed strategy that minimizes the effect of the other player's best pure counterstrategy. to solve the game (using this criterion) means to find each player's optimal mixed strategy using the minimax criterion. So far, we can only sole a game if it happens to reduce to a 1×1 game by dominance. If the game is larger than 2×2, it can be solved using linear progamming as described in Section 4.5 of Finite Mathematics and Finite Mathematics and Applied Calculus. In this tutorial we discuss a procedure to solve a 2×2 game or a game that reduces to a 2×2 one. To best illustrate how this procedure works, we look at a specific example:

Notice that this game cannot reduce to a 1×1 game by dominance.

Finding the Row Player's Optimal Strategy

The row player's optimal strategy is a mixed strategy that minimizes the worst damage that can be caused by a pure strategy used by the column player. Right now we don't know what the optimal row strategy is, so let us call it

Our job is to find the value or values of x that minimize the worst damage the column player can inflict using a pure strategy. Now there are only two possible pure strategies the column player can use:
Case 1: C =
  1  
0
To measure the result of this pure strategy, we calculate the value of the game: Of course, this answer depends on x.
Case 2: C =
  0  
1
Here:

Here are the graphs corresponding to the two cases above. As we see, for each value of x, there are (usually) two values of e corresponding to the two pure strategies the column player can use, one lower (worse for the column player) than the other:


When x = 0.8 the two values of e are shown.
The higher one corresponds to the Case 2 column strategy.
the lower one corresponds to the Case 1 column strategy.

Now the fundamental principle of game theory tells us that, for every x contemplated by the row player (for instance x = 0.8) the column player can be expected to choose the column strategy that does the most harm to the row player (in this case, the Case 1 strategy [1   0]T). Now this is true for every choice of x. Here we see the darker part of the above graph representing these choices for every x.


For each choice of x, the heavy part of the graph gives the corresponding worse outcome for the row player.

Q In view of this graph, which x should the row player use?
A The row player iss seeking to minimize her potential losses (or maximize her potential gain) assuming the worst (that is, assuming the value of the game is given by the heavy blue part of the graph). Thus, all we need to do is determine the value of x for which the blue part of the graph is highest, and that occurs at the point of intersection of the two lines (x = 0.4 and e = 0.2).

We conclude that the row player should take x = 0.4; in other words, the row player should use the mixed strategy:

for an expected value of the game of e = 0.2.   Press here to see how we got this?

We have solved the game for the row player, and we now need to solve the game for the column player.

Q Oh no! we have to go through the whole thing again to solve for the column player!
A Probably, and that is the way the book discusses it too. However, here is a little secret that allows you to get to the equations of the two lines quickly: First notice that if we take the payoff matrix and replace Row 1 by Row 1 - Row 2, we get the following:

Taking xR1 + R2 now gives the two equations for e:

For the corresponding equations for the column player, use columns instead of rows:
  1. Take the payoff matrix and replace Column 1 by Column 1 - Column 2,
  2. Take xC1 + C2 to get
So the two equations for e are: Ploting them gives the graph shown on the left, with the largest values of e shown in heavy blue on the right (remember the column player hates high numbers.)

From the information and graphs above, the optimal column strategy is:

The value of the game is obtained by substituting the value x = 0.6 into either equation:

the same as for the row player (the value of the game is always the same for the row and column player).

Now try some of the exercises in Section 3.4 of Finite Mathematics and Finite Mathematics and Applied Calculus that pertain to solving 2×2 games.

Last Updated: August, 2006
Copyright © 1999, 2003, 2006 Stefan Waner

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