Math 611: Game Theory Notes Chetan Prakash 2012

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1 Math 611: Game Theory Notes Chetan Prakash 2012 Devised in 1944 by von Neumann and Morgenstern, as a theory of economic (and therefore political) interactions. For: Decisions made in conflict situations. Applications to: tactical situations, weapons systems analysis, logistics, economics, sales strategies etc. Game theory is no panacea for existing problems, giving only tentative results. However, it often stimulates deeper analysis, and offers a good language within which to analyze conflict. ASSUMPTIONS 1. There is conflict, focused on some central issue. I.e., there are opposing interests which have the same objective. 2. Participants make simultaneous decisions on their courses of action; these decisions are irrevocable. 3. Neither participant can exert complete control on the game. 4. Each participant is rational and measures the worth of all possible outcomes on the same scale. The latter is often a problem in real-world applications. A more realistic study, building on game theory, is utility theory. CRITERION for OPTIMAL PLAY In general, we prefer conservative as against speculative decisions, because the risk of failure may be unacceptable (e.g., in nuclear war). So we use the WALD CRITERION: Maximize the minimum gain (or minimize the maximum loss). So we wish to make as large as possible the least amount we can expect to win. TERMS Person: one of the opposing players Game: The collection of rules, together with the size of bets or penalties, i.e., the payoffs. These rules are: complete, unchanging, known to all persons. Zero-Sum Game: one in which the total winnings = total losses. Play of a game: A choice of a particular alternative action by each player, and the resulting exchange of payoff. Strategy: A plan of action, complete and ready before the game starts.

2 Optimal Strategy: A strategy guaranteeing the best possible results using the Wald criterion, regardless of the other players moves. Value of the Game: The expected payoff, given that each player plays their own optimal strategy. Solution: A listing of the optimal strategy for each player and a real number for the value of the game. Saddle point for a game: a set of individual alternatives for each player that are together optimal: if any one player deviates from his action, that player will do worse than if all players play the saddle-point alternatives. Most games do not have a saddle-point solution, so some other kind of strategy needs to be devised for them. Two-person, zero-sum games: Two players only, the payoff to one being equal to the loss to the other. We will only consider two-person, zero-sum games in devising solutions below. In such games, we can pictorialize the game by means of a payoff matrix, as below: Red Player s actions Blue 1. Player s 2 P(2,3) actions 3 4 Here, e.g., P(2,3) indicates the payoff blue receives from red, when blue plays his second action and red plays his third. We fill up all the entries of this matrix with such payoffs. Example 2 in the handout: If both bids are equal, we will flip a coin to see who gets the car. In this case, the net payoff is still zero, but in a probabilistic sense of expectation: e.g., if both bid $300, Rogers makes $200 if he gets the car on the coin-flip and loses $200 otherwise. We give the payoff matrix below. We write the actions as bids in multiples of $100. Verify that each entry is correct. On the right we note the worst thing that can happen to Rogers for each of his alternatives, measured over all of Cooper s possible actions. On the bottom we note the same thing for Cooper.

3 Cooper s Actions Row minima Rogers Actions Column maxima Here the game has a saddle-point actually two of them. If Rogers bids $400 or $500, and Cooper bids $400, they will each minimize their maximum loss. In either case, the value of the game is $0, so it is fair to both. What would happen if Roger got greedy and bid, say, $200, in the hope of Cooper s being chintzy and so getting $300-worth off of him? Cooper knows game theory, so even though he really wants to low-bid, rationality wins out and he bids his optimal $400 his entire pot. As a result, and because of Rogers fantasy of riches, he looses $100 to Cooper! That s why the best thing is for both players to be rational and play the optimal strategy. Remember, each player knows the others strategies and the payoff matrix. In general, it is true that the minimax of the payoff matrix (i.e., the least of the column maxima) is at least as large as its maximin (i.e., the largest of the row minima). Can you see why? Another Game with a Saddle Point Blue alternative #3, Red alternative #2, with Value 5. Why?

4 USING DOMINANCE to simplify a game Sometimes we can eliminate a strategy for a player, because there is a different strategy such that, for each of the other player s alternatives, this other strategy is better. For the row player, this translates into a row alternative that has higher numbers, in each entry, across the dominated row. For the column player, this translates into lower numbers. Why? Here is an example: Blue s alternative #1 is at least as good as Blue s alternative #4: 2>1, 3>2, 3 3, 4>1. So we can get rid of the fourth alternative: Now a new dominance emerges (one which wasn t there before (check that!)): Column #1 dominates column #4, because 2<4, 1<3, 1<2. So get rid of column #4: A Game with No Saddle Point We both hold up one or two fingers. If it s the same number, I loose $1 to you. If both are different, I gain $1 from you: Work out the row minima and column maxima and see why there is no saddle-point, i.e., a matrix entry that is both its minimum in its row and its maximum in its column. What do we do in this instance?

5 Von Neuman s and Morgenstern s brilliant solution in this instance was to let each player consider a strategy of mixed alternatives. Each alternative might be played, but in a random manner, dictated by an optimal probability of play. Random strategies may be implemented as follows: Numbers from, say, 0000 to 9999 are broken into groups, the size of each group being in proportion with the probability assigned to each alternative. For example, if x = 2/3, the numbers from 0000 to 6666 are assigned to alternative 1, whilst those from 6667 to 9999 are assigned to alternative 2. In any given play of the game, a random number is generated by each player. Depending on the value of the random number, the corresponding alternative is chosen. There is a theorem that in any zero-sum game, an optimal mixed strategy exists for each player. In the instance where there is a saddle-point, the mixed strategy becomes a pure strategy: the probability of playing the optimal alternative is 1 and that of playing any of the sub-optimal alternatives is 0, for each player. In the current example, let x be the probability with which Blue plays alternative 1. Then 1-x is the probability with which Blue plays alternative 2 (why?). Then by definition of expectation, for the random variable Blue s Gain : Row Expectations: Expected gain to Blue if Red plays alternative 1 = (-1) (x) + (+1) (1-x) = 1-2x Expected gain to Blue if Red plays alternative 2 = (+1) (x) + (-1) (1-x) = 2x-1 Blue wants to determine a value of x such that he maximizes his minimal gain. The two lines y = 1-2x and y = 2x-1 are graphed below, for x between the probability values of 0 and 1. For 0<x< ½, the second line gives a lower expectation.: this is the worst that can happen to Blue. For ½<x<1, the first line is the lower one: this is now the worst that can happen to Blue. The x-value for which these worst possibilities is maximized is clearly where they intersect: at x = ½. At this x, the expected gain is found by simply plugging that number into either of the expressions above (the result will be the same, as the two lines are of the same height at their intersection). Here, the Value of the game is 0.

6 plot 1 K 2 x, 2 x K 1, x = 0..1, y =K y x K0.5 K1 The situation is similar for Red. Let s say that Red plays his strategy 1 with probability z and therefore his alternative 2 with probability 1-z. Then Column Expectations Expected gain to Red if Blue plays alternative 1 = (-1) (z) + (+1) (1-z) = 1-2z Expected gain to Red if Blue plays alternative 2 = (+1) (z) + (-1) (1-z) = 2z-1 Again, if we were to plot the two lines y = 1-2z and y = 2z-1, their intersection would be at z = ½ with a value of 0. We express the solution of the game as follows: Blue s optimal strategy is (½, ½); Red s optimal strategy is (½, ½); Value of the game is 0. This graphical method works for any game in which a player has exactly two alternatives available. Simply replace the payoffs given above by those in the game under consideration, and proceed to compute the intersection values of x and z, and the value of the game (the height of the appropriate straight lines at the optimal point).