# NON - zero sum games.

## Presentation on theme: "NON - zero sum games."— Presentation transcript:

NON - zero sum games

1.7 2-Person, Non-Zero Sum Games (Bimatrix games)
In general more difficult and more interesting to analyze. We have already met one case: constant - sum games. We have shown that they can be converted into equivalent zero-sum games. So these are really of no further interest here, EXCEPT, REMEMBER TO CHECK and SEE IF HAVE THIS CASE, BEFORE GOING FURTHER!

Reminder Example The “Constant Sum” is equal to 5.
Suppose we subtract 5/2 from each entry (this will not change the optimal strategies). The the new equivalent game is as follows:

This is a zero sum game! Thus we can use the standard convention and represent it thus:

In fact, we can now add back 5/2 to each entry and solve the equivalent zero-sum game:
In short, If we have a constant sum game, we can treat it as a zero-sum game (for the purpose of obtaining the optimal strategies).

Comment Zero-sum: One player’s loss is the other player’s gain. The game is purely competitive. Non-zero-sum: Not purely competitive. Both may gain or both lose if they play certain strategies.

The Most Famous Game Prisoner’s Dilemma: Prosecutor talking to two prisoners We have enough evidence to convict you both on a lesser charge. If you both plead innocent, you'll be convicted and each receive a two year sentence. However, if you help us we'll reward you. Admit guilt, then it'll be easier to convict your friend if he claims innocence. He'll get five years and we'll let you go. If, however, you both plead guilty, you'll both get four years."

In short ...... (-2,-2) ( -5,0) (0,-5) (-4,-4) Prisoner II
Don’t Confess Confess Don’t Confess (-2,-2) ( -5,0) (0,-5) (-4,-4) Prisoner I Confess

What is the Dilemma? If they both apply the “security driven” approach, the stable solution will be (-4.-4). The problem is that this solution is dominated by (-2,-2). This suggests Cooperative Games Non- cooperative Games

1.7.1 Example (page 45) (8,0) (3,3) (3,3) (0,8) A1 A2 a1 a2
(8,0) (3,3) (3,3) (0,8) A1 A2 a1 a2 Security levels = 3 , (a1,A2) If they can communicate, they may agree to alternate between (a1,A1) and (a2,A2). The average payoff will be equal to 4 for each. What if first player changes his mind after the first game?

1.7.2 Example (page 46) Player I prefers a1.
(2,5) (5,2) (1,-3) (1,-5) A1 A2 a1 a2 Player I prefers a1. Knowing this, Player II would prefer A1. But, Player I can send the following message to Player II: You better use A2, or else I’ll use a2 !! Since Player II has much more to lose she will probably agree to use A2.

Example (page 46) (100,10) (5,20) (10,10) (5,20) A1 A2 a1 a2 Player II can always “win” the game by playing A2. However, the players may agree to play (a1, A1) and share the total sum 110 according to some agreed-upon formula.

NON - cooperative 2-person non-zero sum games

1.8 Non-Cooperative 2-person Non-zero sum Games
Least departure from the classic zero-sum game. Try and use concepts from zero-sum games. Let A be the matrix of payoffs to Player I Let B be the matrix of payoffs to Player II. E.g.

EQUILIBRIUM REVISITED

Equilibrium 1.8.1 Definition:
A pair of strategies (x*,y*) in S5T is said to be in equilibrium if xAy* ≤ x*Ay* for all x in S and x*By ≤ x*By* for all y in T Note that zero sum games correspond to the case where A = –B.

Security levels can be computed as in zero-sum games, but each player will use her payoff matrix, and each player will try to maximize her payoff. We define mixed strategies as for zero sum games, as though the game is played repeatedly, except that: Under this arrangement Player II is also maximizing her payoff, but she operates on the columns of matrix B. So it is easier to use Bt rather than B and work out the security level solution as for A. In this case each player ignores the other player and works out the least they should be willing to settle for.

Example Consider the two person, non-cooperative, non-zero-sum game
This gives A has saddle: X* = (1, 0), value = 2 Bt has saddle: Y* = (0, 1), value = 3. Notice that if both play these strategies the payoff is actually (3, 3) - different to the maximin values. Also we can clearly see this is not the best solution.

Check for equilibrium xAy* ≤ x*Ay* for all x in S and
x*By ≤ x*By* for all y in T See lecture for details. The strategy pair (X*, Y*) = ((1, 0), (0, 1)) is NOT in equilibrium. Note that there is no reason why it should be, the two matrices may have no relation, so why should it turn out to be equilibrium? Check the pair ((0, 1), (0, 1)). This IS an equilibrium pair and gives better payoff than security level.

1.8.2 Example (page 48) (8,13) (1,5) (6,4) (13,8) A1 A2 a1 a2 There are three equilibrium pairs (don’t worry about where these come from at this stage, we will see methods later): ((1,0),(1,0)) ; ((0,1),(0,1)) ; ((1/3,2/3),(6/7,1/7)) The respective payoffs are all quite different: v = (8, 13), v = (13, 8), v = (7, 7). See lecture for details of where these payoffs come from. Note that the third is dominated by the the first two. None of these are security level pairs .

Some differences between zero-sum and non-zero sum games
A maximin security level pair is not necessarily an equilibrium pair (but it may be) or vice versa. Not all equilibrium pairs have the same payoffs. Even when there is only one equilibrium pair, it is not clear that this equilibrium is exactly what we want, e.g. Prisoners’ dilemma has one equilibrium pair, but it is not the best solution - see detail later.

Theorem (page 49) Let v1 and v2 be the security levels for Player I and Player II respectively (I.e. the payoffs when each player solves their own matrix (or the transpose) as though it is a zero sum game) in a non-zero sum game, and assume that (x*,y*) is an equilibrium pair for this game. Then, v1 ≤ x*Ay* and v2 ≤ x*By* In other words, equilibrium pairs are as good or better than security level pairs.

Proof Let x be any element of S. By definition of min,
min {xAy: y in T} ≤ xAy* Since (x*,y*) is in equilibrium, xAy* ≤ x*Ay* for all x in S (***) Thus, v1:=max{min{xAy: y in T}: x in S} ≤ max{xAy*: x in S} ≤ x*Ay*, using (***). Similarly for v

Theorem (page 50) Any 2-person game has at least one equilibrium pair (allowing mixed strategies). Proof: Appendix A

Finding equilibrium pairs
It is quite hard to construct algorithms for finding all equilibrium pairs in a 2-person non- zero-sum game. However there are some techniques which are useful in some situations. For a 2x2 game we can use a graphical method. Example See lecture for details of this method. (It is different to the method used for zero sum games.)

Now we go to Appendix B. See lecture discussion and examples Then we'll go to Appendix A. Read for yourself and also see the example and discussion in lectures.