SOLVING EIGHT TREASURES OF GAME THEORY
PROBLEMS USING BI-CRITERIA METHOD
by
ZHINENG YE
Submitted in partial fulfillment of the requirements
For the degree of Master of Science
Thesis Advisor: Professor Behnam Malakooti
Department of Electrical Engineering and Computer Science
CASE WESTERN RESERVE UNIVERSITY
May, 2016
CASE WESTERN RESERVE UNIVERSITY
SCHOOL OF GRADUATE STUDIES
We hereby approve the thesis/dissertation of
Zhineng Ye
candidate for the degree of Master of Science
Committee Chair
Prof. Behnam Malakooti
Committee Member
Prof. Vira Chankong
Committee Member
Prof. Mingguo Hong
Date of Defense
Jan. 13, 2016
*We also certify that written approval has been
obtained for any proprietary material contained
therein.
Table of Contents Chapter 1 Introduction ...... 1
1.1 History of Game Theory ...... 1
1.2 Nash equilibrium ...... 2
1.3 Utility and the definition of rationality ...... 4
1.4 Bi-Criteria and its application in Game theory ...... 5
1.4.1 Concept of Bi-criteria and its application in Game theory ...... 5
1.4.2 An example using bi-criteria method in game ...... 8
1.5 Research object and thesis structure ...... 9
Chapter 2 ...... 11
Solving Eight Treasures of Game Theory Problems ...... 11
Using Bi-Criteria Method ...... 11
2.1 Traveler’s dilemma ...... 11
2.2 A Matching Pennies Game ...... 14
2.3 Coordination game with a secure option ...... 15
2.4 Minimum effort coordination game ...... 16
2.5 The Kreps Game ...... 19
2.6 Dynamic Game 1 with complete information ...... 21
2.7 Dynamic Game 2 with complete information ...... 24
2.8 Two-stage bargaining Game ...... 26
Chapter 3 ...... 29 Reasons and Discussions ...... 29
3.1 Reasons why Nash equilibrium fails ...... 29
3.2 Why use bi-criteria to solve ...... 31
3.3 Is Nash equilibrium a rational choice? ...... 32
APPENDIX ...... 34
Reference ...... 41
List of Figures
Figure 1- Claim Frequency in a Traveler's Dilemma for R=180(Dark bars) and R=5(Light bars) 12
Figure 2- Effort choice frequencies for a minimum-effort coordination game with high effort cost
(Dark bars) and low effort cost (Light bars) ...... 17
Figure 3- payoff matrix for row player in a minimum effort game ...... 18
Figure 4- payoff matrix for column player in a minimum effort game ...... 18
Figure 5- weighted payoff matrix for column player in a minimum effort game ...... 19
Figure 6- Graph solution to the basic Kreps Game ...... 20
Figure 7- Payoff tree for Dynamic Game 1, Experiment 1 ...... 22
Figure 8- Payoff tree for Dynamic Game 1, Experiment 2 ...... 23
Figure 9- Payoff tree for Dynamic Game 2, Experiment 1 ...... 24
Figure 10- Payoff tree for Dynamic Game 2, Experiment 2 ...... 25
List of Tables
Table 1-Payoff Matrix example of Pure Nash Equilibrium ...... 3
Table 2- Payoff matrix example of Mixed Nash Equilibrium ...... 3
Table 3- Payoff matrix of prisoner’s dilemma ...... 8
Table 4-Weighted payoff matrix of Prisoner’s dilemma ...... 9
Table 5- Payoff matrix of traveler’s dilemma for R=5 ...... 13
Table 6- Weighted payoff matrix of traveler’s dilemma for R=5 ...... 13
Table 7- Payoff matrix of traveler’s dilemma for R=180...... 13
Table 8- Weighted payoff matrix of traveler’s dilemma for R=180 ...... 13
Table 9- Payoff Matrix of a matching pennies game ...... 14
Table 10- Weighted payoff matrix for Matching Pennies asymmetric game ...... 15
Table 11- Weighted payoff matrix for Matching Pennies reversed game ...... 15
Table 12- Payoff Matrix of Coordination Game...... 15
Table 13- Weighted payoff matrix of coordination game ...... 16
Table 14- Payoff Matrix for the basic Kreps Game ...... 19
Table 15- Payoff Matrix for Positive payoff Kreps Game ...... 21
Acknowledgments
Throughout this research, I’m glad to have the opportunity to work with Professor
Behnam Malakooti, his excellent student Mohammad Komaki and other students as well.
I learned not only research skill from them, but also the personalities we need to go through research. I would like to thank them all, but here I would love to acknowledge some that have been particularly helpful and supportive.
I would like to express my greatest gratitude to my advisor, Prof. Behnam Malakooti, for his persistent support and tireless teaching. I appreciated his passion, patience and understanding throughout my graduate study.
I would like to extend my thanks to Mohammad Komaki, who gave a great helping hand on this project. I am impressed by his masterful skill in using different kinds of software to solve problems in a short period of time. And I always have my questions solved when I come to him for help.
Finally, I would like to thank my parents for their love and my friends for their support throughout this project.
Solving Eight Treasures of Game Theory Problems
Using Bi-Criteria Method
Abstract
by
Zhineng Ye
Game theory is a strategic mathematics model of how we make decisions. It is widely applied in fields like economics and psychology to make our decisions more competitive and favorable. Nash equilibrium, the foundation of game theory, is always the first method attempted to solve a problem, especially in a two-person game. In Goeree and Holt’s 2001 paper, “Ten little treasures of game theory and ten intuitive contradictions”, they did ten experiments on different kinds of two-person games, each game associates with a basic version and a contradictive version. From the experiment result, the basic version matches our assumption and intuition for most of the games, while the contradictive version disagrees with that, which the Nash equilibrium can’t explain the latter. But if certain weights are assigned to the payoffs of the two players and the additive weighted payoff is used as the new payoff for one player, then it may become possible to solve the “new” game and explain the result using Nash equilibrium. This is how the bi-criteria method works in game theory. Logically, the weight is assigned to the payoffs because the other player’s payoff is important to a decision maker. Using this method, almost all of the games in Goeree and Holt’s paper (2001) can be solved. Chapter 1 Introduction
1.1 History of Game Theory
Game theory was first discussed in 1713 in a letter by James Waldegrave when he was playing a two-person card game. The first formal study of game theory was published by
John von Neumann in 1928 (Neumann, 1928), when he proved the minimax theory in a two-person zero-sum game. Later, with the help of Oskar Morgenstern, they published
Theory of Games and Economic Behavior (Neumann & Morgenstern, 1944), explaining the theory of utility and expanding on the previous work of Daniel Bernoulli. The prisoner’s dilemma, known as the first game theory paradox, aroused great attention when it was introduced in 1950 and led to many experiments. Around then, John Nash developed a criterion to test whether a game is stable or undetermined, i.e. none of the players have a profitable deviation and thus they will play the same strategy at all time. This criterion, known as Nash equilibrium, was then extended to a wider application, and now it has become a basic tool to analyze games and economic behavior.
After the popularity of Nash equilibrium, a number of researchers started to further explore the meaning of Nash equilibrium and many branches of it have since been introduced, like subgame perfect equilibrium, correlated equilibrium (Aumann, 1974),
Bayesian Games and others. This interest has strengthened the fundamental position of
Nash equilibrium and thus Nash became one of the Nobel Laureates in Economics in 1994.
Game theory is still a popular research area. Scholars are constantly trying to establish a model that best describes people’s behavior. Human beings are complex creatures that they act differently when faced with different situations. They may be risk averse for gains and
1 risk prone for loss (Tversky, 1979); they may experience regret and choose differently if they know what consequence would happen under this choice (Loomse & Sugden, 1982); they may be jealous if they get much less payoff than others (Rabin, 2002); they may even be anti-social when you are “not friendly”. Explaining all these various situations using only one model would be unrealistic. But working towards it is always a good way to refine our previous model and inspire new ones.
1.2 Nash equilibrium
Nash equilibrium, proposed by John Forbes Nash (Nash, 1951), is a significant concept in solving two-person (or more) non-cooperative games (Non-cooperative games are the games which players have no communication before they make their own decisions and no commitment among them.). Loosely speaking, Nash equilibrium is a stable strategy for both players, and one player will suffer loss (or suboptimal gain) if she deviates from this strategy unilaterally. There are two kinds of Nash equilibria for a two-person finite game: pure strategy Nash equilibrium and mixed strategy Nash equilibrium. Pure strategy means the players play a certain strategy for sure and thus the game is determined, while mixed strategy means the player plays a combination of pure strategies with certain probabilities.
A game may have only one pure Nash Equilibrium, or multiple Nash equilibria, and it may have pure Nash equilibrium and mixed Nash equilibrium at the same time. And it is proved that any finite game has either one or more pure Nash equilibria or a mixed strategy
Nash equilibrium (Nash, 1951). The following example may illustrate the concept more in depth.
2 Table 1-Payoff Matrix example of Pure Nash Equilibrium
B1 B2 A1 (3,2) (0,0) A2 (0,0) (2,3) The game is played by row player A and column player B, which row player has alternative A1 and A2, and the column player has alternative B1 and B2. The payoff is shown in brackets where the first number refers to the row player and the second number refers to the column player. For instance, the payoffs are 3 units and 2 units for player A and B respectively when (A1, B1) is selected. It is obvious that there are two Nash equilibria in this game, i.e. (A1, B1) and (A2, B2). We can tell whether a strategy set is a Nash equilibrium by checking if any profitable unilateral deviation exists. In this case, if A knows B chooses
B1, A’s best response is A1, and if B knows A chooses A1, B’s best response is B1. Both A and B cannot have a higher payoff if they choose otherwise; thus (A1, B1) is a Nash equilibrium. A similar deduction can be made for another Nash equilibrium (A2, B2).
An example of mixed strategy Nash equilibrium is shown below.
Table 2- Payoff matrix example of Mixed Nash Equilibrium
B1 B2 B3 A1 (0,0) (-1,1) (1,-1) A2 (1,-1) (0,0) (-1,1) A3 (-1,1) (1,-1) (0,0) This is a symmetric payoff matrix of the game Rock-Paper-Scissors, with a payoff of 1 if winning, -1 if losing and 0 for a tie. (We may translate A1 and B1 as Rock alternatives,
A2 and B2 as Paper alternatives, A3 and B3 as Scissors alternatives.) Obviously, there is no pure strategy, and the mixed strategy Nash equilibrium is (1/3, 1/3, 1/3) of playing each alternative respectively based on our common sense. To find a mixed strategy Nash equilibrium for the row player, we should find a strategy set that makes the column player
3 indifferent between the strategies in their mix, i.e., the row player’s strategy set should yield the same expected payoff for the column player’s strategy.
We can check if (1/3, 1/3, 1/3) is a Nash equilibrium for the row player using the trick described above.
Column’s expected payoff if B1 is played: (1/3)*0+ (1/3)*(-1) + (1/3)*1=0;
Column’s expected payoff if B2 is played: (1/3)*1+ (1/3)*0+ (1/3)*(-1) =0;
Column’s expected payoff if B3 is played: (1/3)*(-1) + (1/3)*1+ (1/3)*0=0.
We can see that the column player is indifferent between each strategy, so we can say that (1/3, 1/3, 1/3) is the mixed strategy Nash equilibrium for the row player. A similar deduction can be made for the column player.
The basic method for solving a mixed Nash equilibrium problem is to assign a probability to each alternative that makes your opponent choose indifferently. For a 2×n game (a game in which one player has only two alternatives and the other player has n alternatives), the solution can be solved graphically.
1.3 Utility and the definition of rationality
Frequently, people evaluate different alternatives using an expected value and choose the alternative which associates with a higher expected value. Sometimes, however, people may realize that they may prefer a sure gain with a lower expected value to a risky higher payoff in which they may receive nothing. This is how expected utility comes into play.
Utility, meaning satisfaction or happiness, is used as a function to imply how much satisfaction a payoff can produce. The utility function is a non-decreasing function, and
4 four axioms should be satisfied according to the Von Neumann-Morgenstern utility theorem (Neumann & Morgenstern, 1944)for a decision maker to use this utility function.
The axioms are completeness, transitivity, independence and continuity.
1. Completeness. For any two alternatives A1 and A2, the preference must be one of three cases: A1 is preferred to A2; A2 is preferred to A1; or A1 and A2 are equally preferred.
2. Transitivity. For any three alternatives A1, A2 and A3, if A1 is preferred to A2 and A2 is preferred to A3, then A1 is preferred to A3.
3. Independence. For any three alternatives A1, A2 and A3, if A1 is preferred to A2, then for any positive number d with 0 4. Continuity. For any three alternatives A1, A2 and A3, in which A1 is preferred to A2, and A2 is preferred to A3, there exists a probability p such that the combinational alternative pA1+ (1-p) A3 is equally preferred with A2. What’s more, a decision maker is said to be rational if their choice satisfies the above axioms. In this case, they can use the utility function to assess the alternatives and maximize the function, and choose the outcome with the highest utility value. 1.4 Bi-Criteria and its application in Game theory 1.4.1 Concept of Bi-criteria and its application in Game theory Multiple criteria decision making (MCDM), or multiple criteria decision analysis (MCDA) is a famous analysis method in operation research. Since it was introduced in 60s, many scholars have established many advanced models based on it. And it is still a very 5 efficient tool to analyze the manufacturing problems. A great number of MCDM software are applied in large enterprises and companies. When we make a decision, many aspects (attributes) should be considered. Different alternatives associates with different set of attribute values. And how can we screen out the best choice with different attributes? It’s simple if there exists one alternative associates with all the highest attribute values, but it rarely happens in most of the time. Also, it may not be a good idea if you just simply choose the one with the highest attribute value that you care most. To make the “best” choice, we should do some trade-off among these criteria. Loosely speaking, if we assign different weights of importance to all the attributes, and then calculate the linear combination of the weights and the corresponding values of the attributes (this is also called the weighted sum model, i.e. WSM), then the alternative with the highest WSM is the best choice under this set of weights. Normalization is usually needed before applying this method. A different solution may occur if you use a different set of weights. In real world games, it makes sense that we not only consider our own payoff, but also our opponent’s. Then how can we build a connection between MCDM and game theory? We can regard two players’ payoff as two criteria. Suppose column player assigns a weight for row player’s payoff and her own, i.e. wb=(y,1-y), meaning that column player’s weight for row player’s payoff is y, and (1-y) for her own. Then we can have WSM payoffs from column player’s view. (For example, suppose the weight from column player’s view is wb=(0.2,0.8) for row player’s payoff and her own. Then the alternative that gives row player 3 units and herself 5 units is now 0.2*3+0.8*5=4.6 units to column player, instead of 5 units.) We’ve stated above that we can solve mixed strategy Nash equilibrium using 6 another player’s payoff, so we can use the WSM payoffs from column player’s view, instead of using her original payoff, to solve row player’s mixed Nash equilibrium strategy. This is reasonable because we often take both players’ payoffs into account and what the other gains affects our choice. Generally speaking, the weights assigned to the attributes should be non-negative and add up to one. But in some cases, especially when the players are competing against each other, a negative weight may be placed on an opponent’s payoff, which means that whatever the opponent gains, is felt like a loss. This behavior can be described as “selfish” or “anti-social.” It may happen in real world games and it can help us solve some paradoxes. In our solution to the problems, it may come out that the weight is negative. When the weight is negative, instead of maximizing your opponent’s payoff, you are now minimizing it. For example, the payoff is (5,2) for player A and player B, if wb=(-0.4,0.6), instead of using the negative weight to calculate the weighted sum model as a new payoff directly, we set player A’s payoff as a negative objective, i.e. we use the equation as (-5)*|- 0.4|+2*|0.6|=-0.8 instead of 5*(-0.4)+2*0.6=-0.8. The reason of this interpretation is that the negative weight is not well-defined to understand. So when the result shows that the weight is negative, we should change the corresponding payoff into a minimizing objective and calculate the weighted sum model as the new payoff to solve the Nash equilibrium. It does not change the final result in mathematics, but using the negative objective instead of negative weights better helps the understanding. Therefore strictly speaking, the weights assigned to the payoff for each player should satisfy the following equation: |w1|+|w2|+…+|wn|=1, where n is the number of players. 7 This equation can be simplified here as |w1|+|w2|=1 because we only discuss two- person games in this thesis. And that’s why we use the term bi-criteria instead of multi- criteria in our thesis. The following example may clarify this method in details. 1.4.2 An example using bi-criteria method in game Let’s consider the game of Prisoner’s dilemma. This game is first introduced by Merrill Flood and Melvin Dresher in 1950, and then formalize by Albert Tucker. It’s a two person game which players would converge to a less preferred outcome if they play it rationally. In this game, they can be better off if they both coordinate, but rational thinking or Nash equilibrium strategy suggests that they should both defect to each other which both of them get punished. One possible payoff matrix translation of Prisoner’s dilemma matrix is shown in Table 3. (A1 and B1 refer to coordination strategy, A2 and B2 refer to defection strategy.) Table 3- Payoff matrix of prisoner’s dilemma B1 B2 A1 (-1,-1) (-3,0) A2 (0,-3) (-2,-2) Table 3 shows that if the players both coordinate, i.e. if A1 and B1 are played, then both players get a payoff of -1. If they both defect, i.e. if A2 and B2 are played, the payoffs for both players are -2. In the other cases, the payoff is 0 for the defector and -3 for the coordinator in other cases. Using normal reasoning, (A2, B2) is the only Nash equilibrium play for each player, or they should both defect. However, this option is not Pareto optimal, because both players coordinating, as in (A1, B1), yields a better outcome from both players point of view. We can use bi-criteria to make this solution happen. Suppose column player B has a weight for row player A’s payoff and his own, wb= (0.1, 0.9), which makes the above 8 payoff matrix into Table 4 (in the column player’s point of view). For example, when row player A plays A1 and column player B plays B2, the new weighted payoff for B is - 3*0.1+0*0.9=-0.3. Table 4-Weighted payoff matrix of Prisoner’s dilemma A2 B2 A1 -1 -0.3 A2 -0.27 -2 Here, we use the Nash equilibrium method to solve this game. Suppose row player chooses A1 with probability p, and chooses A2 with probability (1-p). By using the trick that row’s mixed strategy makes their opponent indifferent of their choices, we have the following equation: (-1) p+ (-0.27) (1-p) = (-0.3) p+ (-2) (1-p), or p=0.712. This means the row player has a probability of 0.712 choosing A1 and a probability of 0.288 choosing A2 under the weight wb= (0.1, 0.9). Thus the problem is solved noting that here we use the column player’s payoff to solve the row player’s mixed strategy. 1.5 Research object and thesis structure In the paper “Ten little treasures of game theory and ten intuitive contradictions” by Goeree and Holt (2001), the authors proposed ten treasure games and found that Nash equilibrium failed to explain the results in the contradiction experiments. They give reasonable explanation of the Nash equilibrium failure, but do not demonstrate their explanation mathematically. In this thesis, we try to solve eight of Goeree and Holt’s proposed experiments using the bi-criteria method. We exclude the last two experiments which are games with incomplete information, and are therefore not appropriate in this model. Bi-criteria is a common tool to solve decision making problems and here we use it 9 as a weight to the payoffs. After we set appropriate weights to the payoffs, we discover that Nash equilibrium strategy still works. The thesis is divided into three chapters. In Chapter 1, we go through the history of how game theory comes into our research field and is getting progressively more focused. Also, some basic concepts in game theory, like Nash equilibrium, utility and rationality will be introduced according to the most accepted literature. Additionally, the bi-criteria method is also highlighted. Some simple examples are included to aid in understanding these concepts so that we can move forward. In Chapter 2, we are going to introduce each game in the paper by Goeree and Holt (2001), and then use the bi-criteria method to solve these games. We make slight modifications to the experiment data to fit in our model. In Chapter 3, we discuss the reasons why sometimes Nash equilibrium works as expected while sometimes does not. This discrepancy may be caused by the design of the game, the rationality of an opponent, the risk aversion of the players or other psychological reasons. Finally, we argue whether we should follow Nash equilibrium in our daily life games. 10 Chapter 2 Solving Eight Treasures of Game Theory Problems Using Bi-Criteria Method 2.1 Traveler’s dilemma I. Problem Description and Experiment Result The original version of traveler’s dilemma paradox can be described as below. Two travelers lost their identical item during their flight, and the flight manager promised to cover their lost. The range of the compensation is between 180 and 300 units. Two travelers are told to write down her value of the item separately on a piece of paper without communication. Suppose traveler 1 claims α, traveler 2 claims β. If α = β, which is reasonable to assume both travelers are telling the truth and the manager gives that amount of compensation to each traveler. If α > β, then traveler 2 is assumed to tell the truth while traveler 1 is over-claiming the value of the item. So the manager gives traveler 2 (β +K) unit of compensation, while traveler 1 gets only (β -K) unit of compensation, which K (K>1) unit can be viewed as a bonus as being honest to traveler 2 and a fine as being dishonest to traveler 1. Similar case can be easily deduced when α < β. Assume the travelers can only claim an integer number, then which number should you claim? At first glance, people may commonly believe that 300 unit is the best claim because both can get the highest compensation. But after further consideration, if you know your opponent is claiming 300, then the best claim for you is 299 instead of 300, which gives you (299+K) units payoff, which is greater than 300. (Note that K is greater than 1) Then 11 claiming 299 dominates claiming 300. And if your opponent knows you are claiming 299, then her best claim is 298 instead of 299. Then claiming 298 dominates 299. After this series of iterative deletion of dominated strategy, we come to the conclusion that 180 (the bottom line) is the right claim after rational thinking. In the experiment, two distinct values of K, 5 and 180, are applied. From the discussion above, we know that the Nash Equilibrium strategy for both players is claiming 180, which does not depend on the value of K. But the experiment result shows differently in Figure 1. 1 K=180 K=5 0.8 frequency 0.6 0.4 0.2 0 185 195 205 215 225 235 245 255 265 275 285 295 Claim Figure 1- Claim Frequency in a Traveler's Dilemma for R=180(Dark bars) and R=5(Light bars) As we see, when K=5, nearly 80% of the subjects choose 295, while nearly 80% choose 185 when K=180. The result can be easily explained. When K=5, which means the fine of over-claiming is not high, then most subjects tend to risk for higher compensation. While when K=180, which means the fine is high enough that you may get nothing, then most subjects tend to be risk-averse and stick to the conservative strategy. And Nash equilibrium fails to explain in this situation because the Nash equilibrium is independent on the value of R. 12 II. Solution To solve this paradox, we simplify the game into the following version as follow. For K=5, 80% of the subjects claim 300 as A1 (or B1) and 20% of the subjects claim 180 as A2 (or B2). For K=180, 80% of the subjects claim 180 and 20% of the subjects claim 300. The following tables show the solution of the game. For K=5, the payoff matrix is as below. Table 5- Payoff matrix of traveler’s dilemma for R=5 K=5 B1 B2 A1 (300,300) (175,185) A2 (185,175) (180,180) Suppose wb= (-0.474, 0.526), then the weighted payoff matrix for column player is shown in Table 6. Table 6- Weighted payoff matrix of traveler’s dilemma for R=5 K=5 B1 B2 A1 15.6 14.36 A2 4.36 9.36 After solving this game using the trick described above, we have 80% of the row player claim 300 and 20% claim 180, which matches the result. For K=180, the payoff table is as below. Table 7- Payoff matrix of traveler’s dilemma for R=180 K=180 B1 B2 A1 (300,300) (0,360) A2 (360,0) (180,180) Suppose wb= (0.4333, 0.5667), then the weighted payoff table for column player is shown in Table 8. Table 8- Weighted payoff matrix of traveler’s dilemma for R=180 K=180 B1 B2 13 A1 300 204.012 A2 155.988 180 After solving this game, we have 20% of the row player claim 300 (A1) and 80% claim 180 (A2), which matches the result. 2.2 A Matching Pennies Game I. Problem description and experiment result In this experiment, three types of matching pennies games, i.e. symmetric, asymmetric and reversed, are implemented. Table 9 shows the experiment payoff and the result. (The numbers in the brackets near the alternatives are the percentage of the subjects that choose the corresponding alternative.) Table 9- Payoff Matrix of a matching pennies game Left (48) Right (52) Symmetric Top (48) (80,40) (40,80) Matching pennies Bottom (52) (40,80) (80,40) Left (16) Right(84) Asymmetric Top (96) (320,40) (40,80) Matching pennies Bottom (4) (40,80) (80,40) Left (80) Right(20) Reversed Top (8) (44,40) (40,80) asymmetry Bottom (92) (40,80) (80,40) For the Symmetric case, the Nash equilibrium strategy is to play each alternative with equal probability, and this is almost displayed by the result (close to fifty-fifty). For Asymmetric case, as reminded the trick above, changing your payoff does not alter the Nash equilibrium play of your own, but your opponent’s instead. So changing row’s payoff should leave row’s strategy unchanged as what we predict. But the experiment shows differently. And so as the Reversed case. To sum up, Nash equilibrium only explains the symmetric case and fails to explain the others. II. Solution 14 Table 10 shows the solution for asymmetric case. Suppose the weight wb= (0.121, 0.879), the weighted payoff matrix for column player is shown below. Table 10- Weighted payoff matrix for Matching Pennies asymmetric game Asymmetric Left Right Top(96) 73.88 75.16 Bottom(4) 75.16 44.84 Solving the game using Nash equilibrium, it shows that row player should play Top with probability 0.96 and play Bottom with probability 0.04, which matches the experiment result. Table 11 shows the solution for the reversed case. Suppose the weight wb= (0.479, 0.521), the payoff table for column player is shown below. Table 11- Weighted payoff matrix for Matching Pennies reversed game Reversed Left Right Top(8) 41.918 60.822 Bottom(92) 60.822 59.188 Solving the game using Nash equilibrium, it shows that row player should play Top with probability 0.08 and play Bottom with probability 0.92, which matches the experiment result. 2.3 Coordination game with a secure option I. Problem description and experiment result The payoff table of this game is shown in Table 12. Table 12- Payoff Matrix of Coordination Game L H S L (90,90) (0,0) (Q,40) H (0,0) (180,180) (0,40) This game is a coordination game which they get paid if they coordinate successfully and get nothing if they fail. And the column player has one more secure option which will 15 give her the minimum payoff 40 if she doesn’t want to risk being coordinated. For row player, Q is the secure payoff which is 0 or 400 in this experiment. It is easy to tell that the Nash equilibrium for row player is (L,L), (H,H), and mixing (L,H) with probability (2/3, 1/3). As reminded the trick above, when Q varies, the strategy of row player should not vary accordingly. When Q=400, row player are more inclined to choose L. Anticipating that row player is going to choose L, column player is likely to choose L as well to coordinate or S for a safe play. When Q=0, the result approximately matches what we have expected. While Q=400, the result shows differently, only 64% of the row player chose H and 36% chose L. Here again, we use bi-criteria to solve this problem. II. Solution For Q=400, we use wb= (0.723, 0.277) to the payoff for row and column player respectively in column player’s point of view. Then the weighted payoff matrix for column player is shown in Table 13. Table 13- Weighted payoff matrix of coordination game L H S L 90 0 300.31 H 0 180 11.077 Solving this game using Nash equilibrium, we found out that row player mixes (L, H) with a probability of (0.36, 0.64) which makes column player indifferent with H and S choice. Then it is solved. 2.4 Minimum effort coordination game I. Problem description and experiment result 16 In this game, two players are going to claim their own effort value in integer within a certain range (from 110 to 170 in this experiment) and the final payoff is the minimum of these two effort values minus the product of her own effort and a constant ratio c(0 Assume a is the payoff for player A who claims α, and b is the payoff for player B who claims β, then according to the description, a= min{α,β}-cα, and b=min{α,β}-cβ. We can prove that any common pair of (α,β), i.e. α=β, is the Nash equilibrium when c<1. (You can also check the payoff table in the attachments.) Thus c does not affect the Nash equilibrium in this game. Figure 2 shows the experiment result for this game. 0.7 frequency 0.6 0.5 0.4 0.3 0.2 0.1 0 115 125 135 145 155 165 effort effort cost=0.9 effort cost=0.1 Figure 2- Effort choice frequencies for a minimum-effort coordination game with high effort cost (Dark bars) and low effort cost (Light bars) We can observe tremendously different results when c=0.1 and when c=0.9 which contradict with Nash equilibrium predictions. A loose explanation may be like this: when c is low, meaning the loss is low as well even though your opponent is claiming a lower number. While c is high, then the loss would be huge as long as you claim a slightly higher number, and you may end up losing money if you claim a much larger number. Nash equilibrium fails to explain this and we use bi-criteria to solve this problem. 17 II. Solution Since we have the payoff table for all the possible outcomes, we can simply screen out the best decision after your opponent’s choice. Figure 3 shows the screenshot of the payoff matrix for the original game for row player, Figure 4 shows the screenshot of the payoff matrix for the original game for column player. (Noted that the bolded number is the highest payoff, so the corresponding row/column is the best alternative.) Figure 3- payoff matrix for row player in a minimum effort game Figure 4- payoff matrix for column player in a minimum effort game 18 As said above, every matching payoff values are the Nash equilibriums. But after we assign a weight to the payoffs, things get changed dramatically. Figure 5 shows the screenshot of the weighted payoff matrix for the column player. Figure 5- weighted payoff matrix for column player in a minimum effort game Suppose the weight is wb= (-0.3, 0.7) for the column player, then we may figure out that the minimum effort claim (here is 110) is the best response no matter what your opponent claims. Thus we’ve explained that why people are inclined to choose the minimum effort value when c is high. The details of the solution can be found in the appendix. 2.5 The Kreps Game I. Problem description and experiment result Introduced by David M. Kreps, this game was designed to test the people’s aversion to negative payoff. The payoff table of this game is shown below. Table 14- Payoff Matrix for the basic Kreps Game Left Middle Non-Nash Right Basic Game Top (200,50) (0,45) (10,30) (20,-250) Bottom (0,-250) (10,-100) (30,30) (50,40) As we can see, this is a 2xn game which can be solved by graph. Figure 6 shows the plot to this game. 19 100 100 Payoff 50 50 0 0 0 0.2 0.4 0.6 0.8 1 -50 -50 L -100 -100 M N -150 -150 R -200 -200 -250 -250 -300 -300 Probability of choosing Top Figure 6- Graph solution to the basic Kreps Game As we can see, there are two Pure Nash equilibriums, they are (Top, Left) and (Bottom, Right), there is also a mixed Nash equilibrium which row player mixes (Top, Bottom) with probability (0.968, 0.032) and column player mixes (Left, Middle) with probability (0.0476, 0.9524). After rational thinking, (Top, Left) may be the dominant choice because it gives both players the highest payoffs among these three Nash equilibria. But this choice may lead to a huge loss to the column player when row player chooses Bottom. The Non-Nash alternative seems mediocre, slightly worse than the best outcome, but guarantees a minimum positive payoff. If column player is not willing to take a risk, then she might end up choosing the non-Nash alternative. And from the experiment result, 68% of the column player did choose the Non-Nash alternative, which disagrees with our previous deduction. But Kreps believes that it may be caused by the negative payoff, so he adds 300 to all the payoff to make it non-negative in the second experiment. And the game now looks like this. 20 Table 15- Payoff Matrix for Positive payoff Kreps Game Positive Left Middle Non-Nash Right Payoff Top (500,350) (300,345) (310,330) (320,50) Frame Bottom (300,50) (310,200) (330,330) (350,340) In this counter experiment, we find out that the Non-Nash is still the dominant choice with 64%. This shows the negative payoff may not be the only reason which makes people choose the Non-Nash alternative, but also the risk-averse attitude towards uncertainty. II. Solution For the basic game, wb=(0.3885, 0.6115) solves it. For the positive payoff game, wb=(0.1532,0.8468) solves it. The details are included in the appendix. 2.6 Dynamic Game 1 with complete information I. Problem description and experiment result for experiment 1 The difference between dynamic games and static games is that one player knows what her opponent’s move before making the choice. And we may consider each player’s decision as a stage of the game. Then the basic procedure is like player 1 moves on stage 1, then player 2 moves on stage 2, and player 1 moves on stage 3, and on and on. The common solution to dynamic games is backward induction. We analyze the game backwards, which is straight-forward to decide the best alternative in the final stage. And then we can see the best choice in the last but one stage knowing that the last-stage decision, and continue reasoning backward like this till the first stage. So the optimal strategy set is made. In this game, the payoff and the stages for the first experiment are shown in Figure7. 21 Figure 7- Payoff tree for Dynamic Game 1, Experiment 1 In the first stage, the first player chooses between S and R. If she chooses S, the game ends and the payoffs are (80, 50) for the first and second player respectively. If she chooses R, then the game moves to the final stage which the second player should choose between P and N. And the payoff for P is (20, 10) and for N is (90, 70) for the first player and second player respectively. Using backward induction works like this. In the final stage, the second player chooses N over P because 70 is much higher than 10 considering her payoff only. Reasoning back to the first stage, the first player should choose R over S because 90 is higher than 80. Thus the game is solved and the Nash equilibrium play is (R, N). In the experiment data, 16% of the first player chooses S and 84% chooses R, and all the second players choose N. Nash equilibrium fails to explain that 16% of the first player choosing S. The following shows the solution using bi-criteria. II. Solution for experiment 1 Suppose we set the weight to the payoff is wa=(1-|x|,x) for the first player, where |x|<1. And wb=(y,1-y) for the second player, where 0 When 80(1-|x|) +50x > 90(1-|x|) +70x, i.e. 1 22 over R. For the other cases, the first player should choose R. So we’ve found suitable weight that rationalized these decisions, then it is solved. III. Problem description and experiment result for experiment 2 In the contradiction experiment, the payoff for P is changed from (20,10) to (20,68), which makes the cost of switching much lower than the original one. Then the experiment result is shown in Figure 8. Figure 8- Payoff tree for Dynamic Game 1, Experiment 2 In this case, we can see a difference in the percentage of choosing S and N. Since the first player senses that the threat is somehow credible (because the cost is low.) and thus seeks the safe choice S. But the first player should also sense that the P is strictly worse than N. Even though the cost is low here, N is still dominant over P from both players’ view. Thus the threat is not credible, and the first player should stick to the R alternative, and the second player should choose N. The Nash equilibrium of this game is still (R, N) which ends up (90, 70) for the first and second player respectively. But only 36% of the second player chooses the Nash equilibrium play. The bi-criteria solution is shown in the below. IV. Solution for experiment 2 23 Suppose we set the weight to the payoff is wa=(1-|x|,x) for the first player, where |x|<1. And wb=(y,1-|y|) for the second player, where |y|<1. For the final stage, the second player chooses P over N if 20y+68(1-|y|) ≥ 90y+70(1-|y|), i.e. -1 Suppose N is chosen, then the first player should choose S over N if 80(1-|x|)+50x ≥ 90(1-|x|)+70x, i.e.-1 Suppose P is chosen, then the first player should choose S over P if 80(1-|x|)+50x ≥ 20(1-|x|)+68x, i.e. -1 To sum up, S is played when -1 -1/36 2.7 Dynamic Game 2 with complete information I. Problem description and experiment result for experiment 1 In this contradiction game 2 experiment 1, the payoffs and stages are shown in Figure 9. Figure 9- Payoff tree for Dynamic Game 2, Experiment 1 Here, N is much better than P from both players’ point of view. So the second player has no initiative to choose P over N. Even though she wants to punish the first player because it makes the second player less payoff, from 60 to 50, it is not worthwhile since the cost is 24 too high, i.e. 40. So the second player should choose N, and the first player should choose R since the payoff for N is 90, which is higher than the payoff of S, which is 70. So (R, N) is the Nash equilibrium play in the experiment. And the result shows mostly what we expect. All of the second players choose N, and no one chooses P, but there is 12% of the first player choose S which the Nash equilibrium cannot explain. We can simply assign weights to the payoff to make this result reasonable. II. Solution for experiment 1 Suppose the weight to the first player is wa=(1-|x|,x), then the first player chooses S over N if 70(1-|x|)+60x≥90(1-|x|)+50x, i.e. 2/3≤x<1. And choose N otherwise. III. Problem description and experiment result for experiment 2 For the contradiction experiment 2, the payoffs and stages are shown in Figure 10: Figure 10- Payoff tree for Dynamic Game 2, Experiment 2 In this experiment, the P alternative is changed from (60, 10) to (60, 48), meaning the cost of punishment is as low as 2. Using backward induction, we can tell that (S, N) is the Nash equilibrium play. But the experiment result shows differently. Here may be the explanation to it. The first player may consider this is a credible threat which may make her stick to the S alternative to get the 70 payoff other than risking to get 90 but may end 25 up getting 60 instead. For the second player, it makes sense to choose P over N as a punishment to the first player by not choosing S, which may give the second player the highest payoff 60. And choosing N is a rational choice made on her own payoff. The solution using bi-criteria is shown below. IV. Solution for experiment 2 Suppose we set the weight to the payoff is wa=(1-|x|,x) for the first player, where |x|<1. And wb=(y,1-|y|) for the second player, where |y|<1. For the final stage, the second player chooses P over N if 60y+48(1-|y|) ≥ 90y+50(1-|y|), i.e. -1 Suppose N is chosen, then the first player should choose S over N if 70(1-|x|)+60x ≥ 90(1-|x|)+50x, i.e.2/3 ≤x<1. And choose N otherwise. Suppose P is chosen, then the first player should choose S over P if 70(1-|x|)+60x ≥ 60(1-|x|)+48x, i.e. -5/11≤x<1. And choose P otherwise. To sum up, S is played when 2/3 ≤x<1. P is played when -1 -1/16 2.8 Two-stage bargaining Game I. Problem description and experiment result In this game, there is a total of $5 of prize and two players bargain how to divide it. The first player claim the share of it in the first stage, the game is over if the second player agrees. Otherwise, the second player claims the counter-offer share of $2 (considering it is 26 the shrink from time delay). It is either accepted by first player and they get their share of $2 as claimed or both players end up getting nothing if rejected and the game is over. Same tactic here using backward induction. In the final stage, theoretically speaking, the first player should accept any offer from the second player because a penny is still better than nothing in a rational way. Thus, the second player should claim as much as she can (here it is $1.99). So in the first stage, the first player should claim the offer such that the second player is indifferent between accepting the offer and claiming the most in the final stage, i.e. $1.99. In the first experiment, the average claim by the first player is $2.83, which is close to the expected claim of $3. In the second one, the original pie is still $5, but the pie for the second stage is shrunk to $0.5. According to our induction, the first player should claim $4.5 so that makes the second player indifferent between accepting the offer and claiming the rest of the pie. But the result shows that 93.3% of the first player claims below $4.5, and the rejection is easy to happen when the first player claims “too much”. Traditional game theory cannot explain this, but using bi-criteria method can solve it easily. And it really makes sense, because your decision is not solely based on your own payoff, but your opponent’s as well. II. Solution Suppose the first player claims $4.5 in the first stage, then the second player is faced by the choice of accepting it with the payoff ($4.5,$0.5) and rejecting it with the payoff ($0.01,$0.49). Assume the weight assigned to the payoff is wb=(y,1-|y|), where |y|<1, then the second player rejects the claim by the first player if 27 0.01y + (1-|y|)0.49 ≥ 4.5y+(1-|y|)0.5, i.e. -1 In this experiment, we can see that fair distribution is crucial to the participants, otherwise some players would make irrational choice which makes both suffer a loss. 28 Chapter 3 Reasons and Discussions 3.1 Reasons why Nash equilibrium fails Throughout all these games and experiments, we can observe that the games have two versions, the original one and the meta-game. For the original game, the Nash equilibrium seems reasonable to explain the experiment data. While for the meta-game, Nash equilibrium is insufficient to solve. Some reasons may be accounted for the void of the Nash equilibrium. I. One-shot game design Before the experiments are run, all participants are told this is a one-shot game and will not be repeated. Besides, no communication is allowed before making the decision. Any possibility of cooperation or learning is excluded. So as an individual, one should be careful about her choice and try to maximize her utility through this one-shot game. The stakes are high for this kind of game. Players either win a lot or nothing in a one-shot game, even though they may both get highly paid if they play it repeatedly. And it can be anticipated that the experiment result would be totally different if games are played repeatedly. II. Rationality of the player It is important that we go through the term “rationality” again before we go any further discussion. In game theory, a decision maker is “rational” if her preference over alternatives satisfies the following axioms, i.e. completeness, transitivity, independence and continuity. We are making our choice based on the rationality of our own, and our 29 opponent’s as well. Rationalizability, introduced by (Bernheim, 1984) and David Pearce (Pearce, 1984) (1984), is gaining more and more recognition. Simply put, rationalizability means players themselves are rational, and it is a common knowledge to them. Take the traveler’s dilemma as an example. Suppose you are an expert on game theory, and your opponent has no knowledge about this game. You know that the rational play is to claim the lower boundary within the range, but you assume that your opponent is not rational enough to play like you. So you can outguess your opponent by claiming a higher value than the lower boundary which gains you higher payoff. So some non-Nash outcome is acceptable because of some irrational players or players who believe their opponents are irrational. But is that for sure to achieve Nash equilibrium even though the rationalizability is satisfied? Probably not. Just consider a game with more than one Nash equilibria (e.g. battle of sex and Kreps Game). In a one-shot game, it’s hard to guess which equilibrium your opponent is going to play without communication beforehand. The most probable case is that players don’t achieve the same Nash equilibrium or they just play the non-Nash strategy instead. III. Risk aversion Most people are sensitive to risk and they try their best to avoid it. This kind of people are called risk averters. They like certain gains rather than a risky higher gamble. Those who like risking for higher payoff are seldom observed in these experiments. Participants are like taking a gamble while doing the experiment because they have no idea what their opponent is going to choose. They can predict their opponent’s behavior so that adjust their own to get better response, but still they can’t tell for sure. Some participants may thus 30 conform to the “best of the worst” policy to make their choice. First they list all the worst case scenario for each alternative, then pick the best among this. And this choice is not necessarily connected to Nash equilibrium. IV. Other psychological reasons People have a series of psychological activity before they come to a final decision. They may guess what kind of opponent she is facing with, a corporative one or a selfish one. They may conjecture their opponent’s move and what meaning is conveyed by its move. They may compare the payoff difference, and probably unwilling to choose the alternative which give themselves a lower payoff but pretty high to their opponents. All this train of thought would happen and it would somehow affect a player’s decision. And it has experimental basis that inequity does matter for people behavior (Schmidt, 1999). This kind of behavior shows clearly in the two-stage bargaining game, where the first player is claiming almost 80% of the pie, and the second player can be jealous and refuse this distribution. Here, they are irrational because they are not maximizing their utility. And this makes sense! 3.2 Why use bi-criteria to solve In real world problems, people with different background view things differently. A thousand dollar means a lot to a mediocre worker, but means much less to a billionaire. Some people may be cooperative that they value their joint payoff as much as their own, while some people may be self-oriented that they only focus on their own payoff. We can’t force all the participants to stick to one particular choice even it is a Nash equilibrium. Because they have different thoughts about this game, and they just choose the alternative that maximizes their interests, no matter what these interests are. To apply this idea, we 31 can just simply assign different weights to the payoff. Suppose the weight is 1 and 0 for your payoff and your opponent’s respectively, that means you make your choice based on your own payoff only and regardless or your opponent’s payoff. Suppose the weight is 0.8 and 0.2 for your payoff and your opponent’s respectively, that means what your opponent gets contributes 20% to your final payoff. For a rare case, the weight can be negative, like 0.7 for your payoff and -0.3 for your opponent. This weight distribution means that your opponent’s gain is like a loss to you because it is making negative utility. This can happen when the two players are rivalling with each other, or one is unwilling to choose the alternative which the payoff is much less than her opponent’s. The two-stage bargaining game best fits in this negative weight model. After we assign the weights from a player, we can have a weighted value as her new payoff, then we can use the trick that calculating your opponent’s mixed strategy by using your own payoff to solve the game. The only difference is that we use a weighted payoff as your new payoff value, then solve the game as usual. 3.3 Is Nash equilibrium a rational choice? There is no denying that Nash equilibrium is the most influential concept in game theory, it helps people to analyze the game in a systematical and theoretical way. But is that appropriate to play Nash equilibrium on every single game? Is that a rational choice? The answer is probably no (Risse, 2000). Nowadays, the concept of rationality is still under debate by most well-known scholars. They advocate the rationality should be recommendations that maximizes her utility or simply justifies it. Any precise or specific definition for rationality is at stake. How can we say Nash equilibrium is a rational strategy when the concept of rationality is still under discussion. 32 Prisoner’s dilemma is an example where Nash equilibrium yields a worse outcome for both players. And a lot more examples are created to demonstrate the “odd” outcome derived by Nash equilibrium. Because Nash equilibrium only limits unilateral deviation, by its definition, but not the conjoint cooperation. It may be reasonable only by a single player’s point of view, but not if two players are regarded as a whole. And that’s why this bi-criteria method comes into mind. It has weight for both players’ payoff, it can justify some of the decisions where Nash equilibrium fails. And after the weighting, Nash equilibrium successfully explain the data again! To conclude, Nash equilibrium may not be the best choice in most of the real world games, especially in a one-shot game, or a game under risk (Starmer, 2000). Nash equilibrium may work in repeated games because it may help people learn from previous lesson and start to cooperate to get higher payoff for both players. 33 Appendix Notations Ai: row player’s alternative i, where i=1,2,…,m. Bj: column player’s alternative j, where j=1,2,…, n. aij: the payoff to row player when Ai and Bj is selected. bij: the payoff to column player when Ai and Bj is selected. d: a constant number with 0 wa: the weight for both players’ payoff from row player’s view, i.e. wa=(1-|x|,x), when x is negative, the column player’s objective should be changed into minimizing and change the negative weight into a positive one. wb: the weight for both players’ payoff from column player’s view, i.e. wb=(y,1-|y|); when y is negative, the row player’s objective should be changed into minimizing and change the negative weight into a positive one. K: a parameter in the traveler’s dilemma game, with K=5 or K=180. c: effort cost ratio in the minimum effort coordination game, where c=0.1 or c=0.9 34 Q: secure payoff for row player when (L,S) is chosen in the secure payoff coordination game. Q=0 or Q=400 are applied in the experiment. 35 Part of experiment result 1. Traveler’s dilemma 1.1 When K=5 K=5 300 180 wb=(x,x+1) p(A)= 0.8 0.2 300 300,300 175,185 0.8 600x+300 360x+185 180 185,175 180,180 → 0.2 360x+175 360x+180 weighted payoff matrix x= -0.47396 0.8 15.625 14.375 0.2 4.375 9.375 U(300)= 13.375 U(180)= 13.375 1.2 When K=180 K=180 300 180 wb=(x,1-x) p(A)= 0.2 0.8 300 300,300 0,360 0.2 300 360-360x 180 360,0 180,180 → 0.8 360x 180 weighted payoff matrix x= 0.433333 0.2 300 204 0.8 156 180 U(300)= 184.8 U(180)= 184.8 2. Matching Pennies 2.1 Asymmetric for column player 36 Asymmetric for column wb= 0.121053 0.878947 pa= 0.96 0.04 LEFT RIGHT TOP 320 40 40 80 BOTTOM 40 80 80 40 weighted payoff matrix LEFT RIGHT 0.96 TOP 73.89474 75.15789 0.04 BOTTOM 75.15789 44.84211 EV(LEFT)= 73.94526 EV(RIGHT)= 73.94526 2.2 Asymmetric for row player Asymmetric for row wa= 0.708 -0.292 pa= 0.16 0.84 LEFT RIGHT TOP 320 40 40 80 BOTTOM 40 80 80 40 weighted payoff matrix 0.16 0.84 LEFT RIGHT TOP 215 5 BOTTOM 5 45 EV(TOP)= 38.6 EV(BOTTOM)= 38.6 2.3 Reversed for column player 37 Reversed for column wb= 0.479452 0.520548 pa= 0.08 0.92 LEFT RIGHT TOP 44 40 40 80 BOTTOM 40 80 80 40 weighted payoff matrix LEFT RIGHT 0.08 TOP 41.91781 60.82192 0.92 BOTTOM 60.82192 59.17808 EV(LEFT)= 59.30959 EV(RIGHT)=59.30959 2.4 Reversed for row player Reversed for row wa= 5/6 - 1/6 pb= 0.8 0.2 LEFT RIGHT TOP 44 40 40 80 BOTTOM 40 80 80 40 weighted payoff matrix 0.8 0.2 LEFT RIGHT TOP 30 20 BOTTOM 20 60 EV(TOP)= 28 EV(BOTTOM)= 28 3. Coordination game with secure option 38 L H S L 90,90 0,0 X,40 H 0,0 180,180 0,40 wb= 0.723077 0.276923 pa= 0.36 0.64 H S L 0 0 400 40 350 H 180 180 0 40 300 H S 0.36 L 0 300.3077 250 0.64 H 180 11.07692 200 U(H)= 115.2 U(S)= 115.2 150 L H S 0 0 180 11.07692 100 0.1 9 162 40 0.2 18 144 68.92308 50 0.36 32.4 115.2 115.2 0.4 36 108 126.7692 0 0.5 45 90 155.6923 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.6 54 72 184.6154 0.7 63 54 213.5385 L H S 0.8 72 36 242.4615 0.9 81 18 271.3846 1 90 0 300.3077 4. Kreps Game 4.1 Basic Kreps Game (not weighted) LEFT MIDDLE NON RIGHT 100 TOP 200,50 0,45 10,30 20,-250 Payoff BOT 0,-250 10,-100 30,30 50,40 50 LEFT 300p-250 MID 145p-100 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 NON 30 RIGHT 40-290p -50 p L M N R -100 0 -250 -100 30 40 0.1 -220 -85.5 30 11 -150 0.2 -190 -71 30 -18 0.3 -160 -56.5 30 -47 0.4 -130 -42 30 -76 -200 0.5 -100 -27.5 30 -105 0.6 -70 -13 30 -134 -250 0.7 -40 1.5 30 -163 0.8 -10 16 30 -192 -300 0.9677 40.31 40.3165 30 -240.633 L M N R 1 50 45 30 -250 Probability of choosing Top 4.1 Basic Kreps Game (weighted) 39 LEFT MIDDLE NON RIGHT wb=(x,1-x) L M N R TOP 200,50 0,45 10,30 20,-250 y T 150x+50 45-45X 30-20X 270x-250 X= 0.3885 BOT 0,-250 10,-100 30,30 50,40 1-y B 250X-250 110X-100 30 10X+40 a b LEFT 261.15 -152.875 150 MID 84.7825 -57.265 NON -7.77 30 100 RIGHT -188.99 43.885 Y L M N R 50 0 -152.875 -57.265 30 43.885 L 0.1 -126.76 -48.7868 29.223 24.986 0 0.2 -100.645 -40.3085 28.446 6.087 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 M 0.3 -74.53 -31.8303 27.669 -12.812 N -50 0.4 -48.415 -23.352 26.892 -31.711 R 0.5 -22.3 -14.8738 26.115 -50.61 0.68 24.707 0.3871 24.7164 -84.6282 -100 0.7 29.93 2.08275 24.561 -88.408 0.8 56.045 10.561 23.784 -107.307 -150 0.9 82.16 19.03925 23.007 -126.206 1 108.275 27.5175 22.23 -145.105 -200 4.2 Positive Payoff Kreps Game LEFT MIDDLE NON RIGHT wb=(x,1-x) L M N R TOP 500,350 300,345 310,330 320,50 y T 150x+350 345-45X 330-20X 270x+50 X= 0.153172 BOT 300,50 310,200 330,330 350,340 1-y B 250X+50 110X+200 330 10X+340 a b 400 LEFT 284.6828 88.293 MID 121.2583 216.8489 NON -3.06344 330 350 RIGHT -250.175 341.5317 300 Y L M N R 0 88.293 216.8489 330 341.5317 250 0.1 116.7613 228.9748 329.6937 316.5142 L 0.2 145.2296 241.1006 329.3873 291.4967 200 M 0.3 173.6978 253.2264 329.081 266.4791 N 0.4 202.1661 265.3523 328.7746 241.4616 150 R 0.5 230.6344 277.4781 328.4683 216.4441 0.6 259.1027 289.6039 328.1619 191.4266 0.7 287.571 301.7298 327.8556 166.409 100 0.84 327.4266 318.7059 327.4267 131.3845 0.9 344.5075 325.9814 327.2429 116.374 50 1 372.9758 338.1073 326.9366 91.35644 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 40 Reference Bernheim, B. 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