Introduction to Game Theory: a Discovery Approach

In this section, we will use the idea of expected value to find the equilibrium mixed strategies for repeated two-person zero-sum games.

One of the significant drawbacks of the graphical solution from the previous sections is that it can only solve 2 X 2 matrix games. If each player has 3 options, we would need to graph in three dimensions. Technically this is possible, but rather complicated. If each player has more than 3 options, since we can’t graph in four or more dimensions, we are at a complete loss. So we need to think about an alternate way to solve for the mixed strategies. Although we will begin with 2 X 2 games, this method will easily generalize to larger games.

We could use our previous graphical method to determine the expected value of the game (you might quickly try this just to verify your prediction). However, as we have noted, a major drawback of the graphical solution is that if our players have 3 or more options, then we would need to graph an equation in 3 or more variables; which, I hope you agree, we don’t want to do. Although we will continue to focus on \(2 \times 2\) games, we will develop a new method which can more easily be used to solve larger games.

From Activity 3.4.8 we now know that Player 2 wants \(E_1(H)=E_1(T)\text{,}\) we can use a little algebra to compute the best defensive strategy for Player 2. Remember, we want to assume that Player 1 will always be able to chose the strategy that will be best for her, and thus Player 2 wants to protect himself. Let’s find the probabilities with which Player 2 should play H and T.

So in general, to solve for Player 2’s strategy, we want to write an equation for each of Player 1’s pure strategy expected values in terms of Player 2’s probabilities. For example, \(E_1(H)\) and \(E_1(T)\) in terms of variables \(P_2(H)\) and \(P_2(T)\text{.}\) We then set the expected values equal to each other. We now have an equation just in terms of Player 2’s probabilities.

In order to solve for the probabilities, we also need to use the fact that Player 2’s probabilities sum to 1. For example, \(P_2(H)+P_2(T)=1\text{.}\) For a \(2 \times 2\) game, you now have 2 equations with 2 unknowns (\(P_2(H)\) and \(P_2(T)\)). Use an algebraic method such as substitution or elimination to solve the system of equations.

Now can you use a similar process to find Player 1’s strategy? Whose expected values should you use? Whose probabliities?

We now have a new method for finding the best mixed strategies for Players 1 and 2, assuming that each player is playing defensively against an ideal player. But how can we find the value of the game? For Player 2, we assumed \(E_1(H)=E_1(T)\text{.}\) In other words, we found the situation in which Player 1’s expected value is the same no matter which pure strategy she plays. Thus, Player 1 is indifferent to which pure strategy she plays. If she were not indifferent, then she would play the strategy with a higher expected value. But, as we saw, this would be bad for Player 2. So Player 1 can assume that Player 2 will play the equilibrium mixed strategy. Thus, as long as Player 1 plays a mixed strategy, it doesn’t matter whether at any given time, she plays H or T (this is the idea that she is indifferent to them). Hence, the expected value of the game for Player 1 is the same as \(E_1(H)\text{,}\) which is the same as \(E_1(T)\text{.}\) Similarly, we find that the expected value of the game for Player 2 is \(E_2(H)\) (or \(E_2(T)\)). This is a pretty complicated idea. You may need to think about it for a while. In the meantime, use the probabilities you found for each player and the equations for \(E_1(H)\) and \(E_2(H)\) to find the value of the game for each player.

Keep practicing with the expected value method on some other games.

If you found this last activity to be algebraically challenging, don’t worry, we will be able to use technology to help us solve equations with several variables.