A Game[edit | edit source]
To begin, let's play the game of Odds or Evens.
This is a classic game to determine tiebreakers or settle contests. It only requires your fingers.
- An indefinitely repeated round-robin (i.e. play the round-robin using the schedule from the link below, and then play it again, and then play it again, and so on, until the professor says stop after some unpredictable number of repetitions).
- A payoff matrix (see below), the payoffs are in the form of [A, B], so the first number is the payoff to player A, the second to player B. Player A is the first player in the round-robin schedule.
- The game Odds or Evens
- The strategy (write it down, keep it secret for now)
- Scorekeeping (record your score … honor system)
- The prize: The awe of your peers
Use a round-robin scheduler to determine the schedule of the tournament.
|Player A||Odd||[3, 3]||[0, 5]|
|Even||[5, 0]||[1, 1]|
Discussion[edit | edit source]
What does this all mean?
System Rational vs. User Rational
Tit for Tat vs. Myopic Selfishness
See Anatol Rapoport, who developed the successful Tit-for-Tat strategy for a similar, computer based strategy.
Game Theory[edit | edit source]
Game theory is concerned with general analysis of strategic interaction of economic agents whose decisions affect each other.
Problems that can be analyzed with Game Theory:
- Bus vs. Car
who are the agents in each game?
Strategies[edit | edit source]
In game theory, dominance (also called strategic dominance) occurs when one strategy is better than another strategy for one player, no matter how that player's opponents may play. Many simple games can be solved using dominance. The opposite, intransitivity, occurs in games where one strategy may be better or worse than another strategy for one player, depending on how the player's opponents may play. (ref: Dominant strategy)
Nash Equilibrium (NE): a pair of strategies is defined as a NE if A's choice is optimal given B's and B's choice is optimal given A's choice. A NE can be interpreted as a pair of expectations about each person's choice such that once one person makes their choice neither individual wants to change their behavior.
If a strictly dominant strategy exists for one player in a game, that player will play that strategy in each of the game's Nash equilibria. If both players have a strictly dominant strategy, the game has only one unique Nash equilibrium. However, that Nash equilibrium is not necessarily Pareto optimal, meaning that there may be non-equilibrium outcomes of the game that would be better for both players. The classic game used to illustrate this is the Prisoner's Dilemma. (ref: Dominant strategy)
Payoffs for player A are represented is the first number in a cell, the payoffs for player B are given as the second number in that cell. Thus strategy pair [i,i] implies a payoff of 3 for player A and also a payoff of 3 for player B. The NE is asterisked in the above illustrations. This represents a situation in which each firm or person is making an optimal choice given the other firm or persons choice. Here both A and B clearly prefer choice i to choice j. Thus [i,i] is a NE.
Prisoner's Dilemma[edit | edit source]
Earlier, we played both a finite one-time game and an indefinitely repeated game. The game was formulated as what is referred to as a ‘prisoner’s dilemma’.
The term prisoner’s dilemma comes from the situation where two partners in crime are both arrested and interviewed separately. If they both ‘hang tough’, they get light sentences for lack of evidence (say 1 year each). If they both crumble in interrogation and confess, they both split the time for the crime (say 10 years). But if one confesses and the other doesn’t, the one who confesses turns state’s evidence (and gets parole) and helps convict the other (who does 20 years time in prison)
In the one-time or finitely repeated Prisoners' Dilemma game, to confess (toll, defect, evens) is a dominant strategy, and when both prisoners confess (states toll, defect, evens), that is a dominant strategy equilibrium.
Applications of Game Theory to Transportation[edit | edit source]
Tolling at a Frontier[edit | edit source]
' (based on Levinson, David (1999) Tolling at a Frontier: A Game Theoretic Analysis. Proceedings of the 14th International Symposium on Transportation and Traffic Theory 173-187.)
Two states (Delaware and New Jersey) are separated by a body of water. They are connected by a bridge over that body. How should they finance that bridge and the rest of their roads?
Should they toll or tax?
Let and be tolls of the two jurisdictions. Demand is a negative exponential function. (Objective is to maximize local welfare (utility of residents plus toll revenue from non-residents (toll revenue from residents is considered a transfer)).
|'||'||Jurisdiction J (New Jersey)||'|
|Jurisdiction I (Delaware)||i||[1153, 1153]*||[2322,883]|
The table is read like this: Each jurisdiction chooses one of the two strategies (Toll or Tax). In effect, Jurisdiction 1 (Delaware) chooses a row and jurisdiction 2 (New Jersey) chooses a column. The two numbers in each cell tell the outcomes for the two states when the corresponding pair of strategies is chosen. The number to the left of the comma tells the payoff to the jurisdiction who chooses the rows (Delaware) while the number to the right of the column tells the payoff to the state who chooses the columns (New Jersey). Thus (reading down the first column) if they both toll, each gets $1153/hour in welfare , but if New Jersey Tolls and Delaware Taxes, New Jersey gets $2322 and Delaware only $883.
So: how to solve this game? What strategies are "rational" if both states want to maximize welfare? New Jersey might reason as follows: "Two things can happen: Delaware can toll or Delaware can keep tax. Suppose Delaware tolls. Then I get only $883 if I don't toll, $1153 if I do, so in that case it's best to toll. On the other hand, if Delaware taxes and I toll, I get $2322, and if I tax we both get $1777. Either way, it's best if I toll. Therefore, I'll toll."
But Delaware reasons similarly. Thus they both toll, and lost $624/hour. Yet, if they had acted "irrationally," and taxed, they each could have gotten $1777/hour.
Coordination Game[edit | edit source]
In Britain, Japan, Australia, and some other island nations people drive on the left side of the road; in the US and the European continent they drive on the right. But everywhere, everyone drives on the same side as everywhere else, even if that side changes from place to place.
How is this arrangement achieved?
There are two strategies: drive on the left side and drive on the right side. There are two possible outcomes: the two cars pass one another without incident or they crash. We arbitrarily assign a value of one each to passing without problems and of -10 each to a crash. Here is the payoff table:
(Objective: Maximize payoff)
Verify that LL and RR are both Nash equilibria.
But, if we do not know which side to choose, there is some danger that we will choose LR or RL at random and crash. How can we know which side to choose? The answer is, of course, that for this coordination game we rely on social convention. Conversely, we know that in this game, social convention is very powerful and persistent, and no less so in the country where the solution is LL than in the country where it is RR.
See Driving on the left or right for historical discussion
Issues in Game Theory[edit | edit source]
- What is “rationality” ?
- What happens when the rational strategy depends on strategies of others?
- What happens if information is incomplete?
- What happens if there is uncertainty or risk?
- Under what circumstances is cooperation better than selfishness? Under what circumstances is cooperation selfish?
- How do continuing interactions differ from one-time events?
- Can morality be derived from rational selfishness?
- How does reality compare with game theory?
Thought Question[edit | edit source]
How does an infinitely or indefinitely repeated Prisoner’s Dilemma game differ from a finitely repeated or one-time game? Why?
Problem[edit | edit source]
Two airlines (United, American) each offer 1 flight from New York to Los Angeles.
Price = $/pax, Payoff = $/flight.
Each plane carries 500 passengers.
Fixed cost is $50000 per flight, total demand at $200 is 500 passengers.
At $400, total demand is 250 passengers.
Passengers choose cheapest flight.
Payoff = Revenue - Cost
Work in pairs (4 minutes):
1. Formulate the Payoff Matrix for the Game.
2. What is equilibrium ?
Solution[edit | edit source]
Zero-Sum[edit | edit source]
Zero-Sum game: If we add up the wins and losses in a game, treating losses as negatives, and we find that the sum is zero for each set of strategies chosen, then the game is a “zero-sum game."
Problem Extension[edit | edit source]
3. What happens if there is a third price $300, for which demand is 375 passengers.
Reformulate the problem.
Solution[edit | edit source]
Mixed Strategies[edit | edit source]
Mixed strategy: If a player in a game chooses among two or more strategies at random according to specific probabilities, this choice is called a "mixed strategy."
Further Applications[edit | edit source]
- Levinson, David (2005) Micro-foundations of Congestion and Pricing: A Game Theory Perspective. Transportation Research part A Volume 39, Issues 7-9 , August-November 2005, Pages 691-704.
- Levinson, David (2000) Revenue Choice on a Serial Network. Journal of Transport Economics and Policy 34,1: 69-98.