www.egwald.com Egwald Web Services

Egwald Web Services
Domain Names
Web Site Design

Egwald Website Search JOIN US AS A FACEBOOK FAN Twitter - Follow Elmer Wiens

 

Statistics Programs - Econometrics and Probability Economics - Microeconomics & Macroeconomics Operations Research - Linear Programming and Game Theory Egwald's Mathematics Egwald's Optimal Control
Egwald HomeMathematics HomeOperations Research HomeLinear ProgrammingGame Theory Home  PagePlay Two Person GamePlay Own Two Person GameCooperative Two Person GameGame Theory References
 

Operations Research - Game Theory

by

Elmer G. Wiens

Egwald's popular web pages are provided without cost to users.
Please show your support by joining Egwald Web Services as a Facebook Fan: JOIN US AS A FACEBOOK FAN
Follow Elmer Wiens on Twitter: Twitter - Follow Elmer Wiens

Game Theory - Introduction | Battle of the Sexes | Prisoner's Dilemma | Free Rider | Game of Chicken | Play a Game Online | Metagame Example

Nonnegotiable Two Person Generalized (Non-Zero) Sum Games

Game of Chicken - Rebel Without a Cause

In the movie, Rebel Without a Cause, James Dean, playing Jim, and Buzz Gunderson compete for the favours of sixteen year old Judy, played by the heartthrob Natalie Wood. Buzz's gang members steal two cars. The group of teenagers gather on a Los Angeles lookout, with a cliff that drops down to the Pacific Ocean. James and Buzz are to drive the stolen cars toward the cliff. The first person to jump from his car is declared the chicken. The last person to jump is the hero, capturing Judy's affection and the gang's respect. The driverless cars continue over the cliff and plunge to the rocks at its base.

Each driver has two strategies. Jump when he feels endangered; jump after the other driver jumps. To keep the game symmetrical, let us assume that both drivers will feel endangered at the same time.

Clearly, there are four possible outcomes of the game. The worst that can happen is that both James and Buzz chose to jump after the other driver jumps. Both cars will plunge over the cliff killing James and Buzz.

The following game tableau represents the payoffs for James and Buzz for each choice of strategy.

Game of Chicken

Tableau
Buzz
Chicken JumpJump Last
Y1Y2
James Chicken JumpX15, 5-10, 10
Jump LastX210, -10-1000, -1000

Unlike the free-rider and prisoner's dilemma games, the game of chicken does not have a dominating strategy for either player. If both James and Buzz jump at the same time, they score some brownie points with the gang (utility 5 each) for participating, but leave Judy unimpressed. If Buzz — believing James will chicken jump — jumps last, Buzz gets Judy (utility 10). If James in fact chicken jumps, he loses Judy and the gangs respect (utility -10). So James is tempted to also choose "jump last."

This game has two equilibria using just pure strategies, (James, Buzz) = (Jump Last, Chicken Jump) and (James, Buzz) = (Chicken Jump, Jump Last). If James convinces Buzz that he will jump last, then Buzz should chicken jump. If Buzz knows that James will chicken jump, then Buzz should jump last. Clearly, James prefers the first equilibrium, and Buzz prefers the second. If both James and Buzz play for their preferred equilibrium, the result is (Jump Last, Jump Last) — death for both players.

Another equilibrium using mixed strategies also exists. Here James and Buzz play each strategy with some positive probability. This does not mean that they will play "Chicken" more than once. Each player randomizes his strategy choice. (Play a few Two-Person Zero-Sum games if this concept is not clear.)

Both James and Buzz want to avoid the (Jump Last, Jump Last) result. But, both James and Buzz want to play the "Jump Last" strategy with some small probability. The equilibrium mixed strategies are (X1, X2) = (990/995, 5/995) = (Y1, Y2), with an expected payoff of 4.92 each.

In the movie, Buzz tells James before the game that he likes him, and Buzz teaches James how to play the game. They could have had an implicit agreement, based on mutual respect, to both "Chicken Jump." After all, they had already proven their metal in the knife fight at the Planetarium. Unfortunately, Buzz's leather jacket's sleeve is caught on the door handle of his car. He cannot jump, even though James jumps. Both cars and Buzz plunge over the cliff.

The game tableau above is skewed toward the "Chicken Jump" strategy for each player. To see what happens when the payoffs have different values, consider the following "Chicken" game tableau for the general case.

Game of Chicken

Tableau
Buzz
Chicken JumpJump Last
Y1Y2
James Chicken JumpX1a, a-b, b
Jump LastX2b, -b-c, -c

To be classified as a "Game of Chicken," the variables a, b, and c must satisfy

c > b > a > 0.

With these restrictions on the variables, the game has two equilibria using just pure strategies, (Jump Last, Chicken Jump) and (Chicken Jump, Jump Last). The equilibrium with mixed strategies is:

(X1, X2) = [(c - b)/(c - a), (b - a)/(c - a)] = (Y1, Y2)


with an expected payoff for each player of:

(c*a - b*b) / (c - a)

To see this, consider the expected payoff to James if he plays "Chicken Jump" with probability X1 and "Jump Last" with probability X2 = 1 - X1, and Buzz plays "Chicken Jump" with probability Y1 and "Jump Last" with probability Y2 = 1 - Y1.

The expected payoff to James is:

J = a*X1*Y1 - b*X1*(1-Y1) + b*(1-X1)*Y1 - c*(1-X1)(1-Y1)

    = [(c - b) - (c - a)*Y1]*X1 + (b + c)*Y1 - c

The expected payoff to Buzz is:

B = a*X1*Y1 + b*X1*(1-Y1) - b*(1-X1)*Y1 - c*(1-X1)(1-Y1)

    = [(c - b) - (c - a)*X1]*Y1 + (b + c)*X1 - c

Taking the derivative of J w.r.t. X1:

dJ/dX1 = (c - b) - (c - a)*Y1 = 0


Taking the derivative of B w.r.t. Y1:

dB/dY1 = (c - b) - (c - a)*X1 = 0

Solving for Y1 and X1 we get the equilibrium mixed strategies above.

For example, the game tableau for the "Game of Chicken" in Rapoport (137) is:

Game of Chicken

Tableau
Buzz
Chicken JumpJump Last
Y1Y2
James Chicken JumpX11, 1-2, 2
Jump LastX22, -2-5, -5

The equilibrium with mixed strategies is:

(X1, X2) = (3/4, 1/4) = (Y1, Y2)


with an expected payoff for each player of 1/4.

Neither player can improve this mixed strategy, equilibrium payoff. Suppose James plays (X1, X2) = (3/4, 1/4), and Buzz decides to play (Y1, Y2) = (1, 0).

James's expected payoff is:

.75 * 1 + .25 * 2 = 1.25


Buzz's expected payoff is:

.75 * 1 - .25 * 2 = .25

If Buzz decides to play (Y1, Y2) = (0, 1),

James's expected payoff is:

-.75 * 2 - .25 * 5 = -2.75


Buzz's expected payoff is:

.75 * 2 - .25 * 5 = .25

In fact, if James plays the mixed strategy (X1, X2) = (3/4, 1/4), Buzz will always get an expected payoff of .25, no matter how he plays.

By symmetry, James will always get an expected payoff of .25 if Buzz plays the mixed strategy (Y1, Y2) = (3/4, 1/4).

While neither player can improve his payoff by changing from the equilibrium mixed strategies:

(X1, X2) = (3/4, 1/4) = (Y1, Y2)


either player can hurt the other player by, say, playing the (Chicken Jump, Jump Last) = (0, 1) strategy.

If both players try to hurt the other player, the result is the (James, Buzz) = (Jump Last, Jump Last) outcome.

 

References

  • Black, Max. Perplexities. Ithaca: Cornell, 1990.
  • Howard, Nigel. Paradoxes of Rationality: Theory of Metagames and Political Behavior. Cambridge: MIT, 1971.
  • Intriligator, Michael D. Mathematical Optimization and Economic Theory. Englewood Cliffs: Prentice-Hall, 1971.
  • Nash, John F. "Noncooperative Games." Annals of Mathematics. 54 (1951): 286-295.
  • Oxford Dictionary: The Concise Oxford Dictionary of Current English. 5th ed. Ed. H.W. Fowler and F.G. Fowler. Oxford: Oxford UP, 1964
  • Owen, Guillermo. Game Theory, 2nd Edition. New York: Academic, 1982.
  • Rapoport, Anatol. Two-Person Game Theory: The Essential Ideas. Ann Arbor: U Michigan, 1966.
  • Thomas, L. C. Games, Theory and Applications. New York: John Wiley, 1984.
 
   

      Copyright © Elmer G. Wiens:   Egwald Web Services       All Rights Reserved.    Inquiries