1. GAME THEORY Day 5

2. Minimax and Maximin Step 1. Write down the minimum entry in each row. Which one is the largest? Maximin Step 2. Write down the maximum entry in each column. Which one is the smallest? Minimax Rose\ColinABCD A4325 B-1020 C7523 D08-4-5 Row Minimum 2 -10 2 5 Column Maximum78 2 5 Maximin Minimax Saddle Points

3. Mixed Strategy  Yesterday, we discussed dominance and saddle points.  Games where ONE strategy can be played with certainty to receive the best outcome are pure strategy games  Contrastingly, when there is no saddle point (or multiple saddle points) where neither player would want to play a single strategy with certainty, it is called a mixed strategy game Example. Row Minimum: -3 and 0 Maximin 0 Column Maximum: 2 and 3 Minimax 2 Rose\ColinAB A2-3 B03

4. Expected Value  To analyze the effect of one or both players using mixed strategies, we can use the concept of expected value.  Let’s say in our example Colin will flip a coin to decide between strategy A and B.  Colin A and Colin B have probability of ½ each  Example.  Expected Value. Rose A: ½ (2) + ½ (-3) = - ½ Rose B: ½ (0) + ½ (3) = 3/2  Thus, if Colin is playing the mixed strategy ½ A, ½ B, then Rose should play Rose B. Rose\ColinAB A2-3 B03

5. If you know that your opponent is playing a given mixed strategy, and will continue to play it regardless of what you do, you should play your strategy which has the largest expected value. Expected Value Principle

6. Colin’s Point of View  How can Colin choose a mixed strategy such that Rose cannot take advantage of it!  Let Colin play a mixed strategy of probability x for strategy A and (1- x) for B where 0<x<1.  Thus, Colin can should play mixed strategy ¾ A, ¼ B  Expected Value. Rose A: x(2) + (1-x)(-3) = -3 + 5x Rose B: x (0) + (1-x)(3) = 3 - 3x  What can we do with this information? -3 + 5x = 3 – 3x So x = ¾  What would Rose’s average payoff be? Rose\ColinAB A2-3 B03 Rose A: ¾ (2) + ¼ (-3) = ¾ Rose B: ¾ (0) + ¼ (3) = ¾ ¾

7. What about 3x3?  Colin has strategies A, B, and C with probabilities (x, y, 1 – x – y)  Solve using a system of two equations in two unknowns. Rose A: Rose B: Rose C: Rose\ColinABC A122 B212 C220 (2/5, 2/5, ½)