The Emergence of Conventions- The Crossroads Game
Game theory can explain the emergence of road rules, as the outcome of two drivers passing each other depends on their spontaneous actions and not on a pre-designed system of rules.
Consider the following example (after a model by Dr Dieter Schmidtchen (30) and a model by Dr. Bertrand Lemennicier (31).
Suppose that a random driver, Driver 1 (D 1) is approaching an intersection. Another car driving in the opposite direction by Driver 2 (D 2) gets at the intersection at the same time.
Since both drivers have to cross the intersection, D 1 has two options: he can either slow down (strategy S) and let the other car pass or he can maintain speed (strategy M). Suppose D 1 has an expectation of ? that D 2 will slow down and an expectation of 1 - ? that D 2 will maintain speed, where ? ? (0.1) Suppose D 2 has an expectation of ? that D 1 will slow down and an expectation of 1 - ? that D 1 will maintain speed, where ? ? (0.1).
Below are the payoffs measured in the time drivers lose at the intersection. Suppose that if one of the drivers slows down and the other one maintains speed, the one who slows down will lose one minute and that if both drivers slow down, they will each lose 10 minutes, as they have to decide who will gain priority in crossing the road. Also, suppose that if both drivers maintain speed (and thus collide), they will each lose 21 minutes (assess and agree on the damages, exchange insurance information, etc.), in addition to later on having to incur some cost for repairing their cars.
Assume that the game is played by a very large number of drivers, who have to encounter this situation very often. The players cannot predict each other’s actions (i.e. the game is not valid if one of the drivers maintains speed as he knows that the other driver will slow down, due to the fact that he has previously encountered him). The only thing known to the players is the game’s payoffs.
Expected Losses for D 1
The expected loss for D 1 from adopting strategy M is:
EM 1 = (1- ?)(21) + (?)(0) , ( D 1's expected loss for maintaining speed when the other driver maintains speed as well + D 1's expected loss for maintaining speed when the other driver slows down)
The expected loss for D 1 from adopting strategy S is:
ES 1 = (1- ?)(1) + (?)(10) , ( D 1's expected loss for slowing down when the other driver maintains speed + D 1's expected loss for slowing down when the other driver slows down as well)
Expected Losses for D 2
The expected loss for D 2 from adopting strategy M is:
EM 2 = (1- ?)(21) + (?)(0), ( D 2's expected loss for maintaining speed when the other driver maintains speed as well + D 2's expected loss for maintaining speed when the other driver slows down)
The expected loss for D 2 from adopting strategy S is:
ES 2 = (1- ?)(1) + (?)(10), ( D 2's expected loss for slowing down when the other driver maintains speed + D 2's expected loss for slowing down when the other driver slows down as well)
Analysis for D 1
The decision on which strategy to adopt depends on each driver’s expectations on what the other driver will do.
a. if ? = 1, which means that D 1 completely expects D 2 to slow down, D 1 will maintain speed, as his incurred losses will equal 0.
b. if ? = 0, which means that D 1 completely expects D 2 to maintain speed, D 1 will slow down as the losses of him slowing down (1 minute) are smaller than the losses of him maintaining speed (21 minutes).
c. if D 1 expects D 2 to have an equal chance of maintaining speed or slowing down (which means that ? = 1/2):
EM 1 = (1- 1/2)(21) + (1/2)(0) EM 1 = 21/2 + 0/2 = 10.5
ES 1 = (1- 1/2)(1) + (1/2)(10)
ES 1 = 1/2 = 5.5
Since EM 1 > ES 2, D 1 will always maintain speed when he expects D 2 to have an equal chance of maintaining speed or slowing down
d. if D 1 is indifferent between slowing down and maintaining speed as the expected loss for D 1 from adopting strategy M equals the expected loss for D 1 from adopting strategy S
EM 1 = ES 1
1-?)(21) + (?)(0) = (1- ?)(1) + (?)(10)
?= 2/3= 0.67
Thus, when ?< 0.67 D 1 should maintain speed and when ?> 0.67 D 1 should slow down.
Analysis for D 2
Same as analysis for D 1. When ?< 0.67 D 2 should maintain speed and when ?> 0.67 D 2 should slow down.
Nash Equilibria of the Game
There are three combinations (?,?) in which no tendency of change in probabilities arise: (0,1), (1,0) and (0.67, 0.67). While (0,1) and (1,0)are stable equilibria (both drivers clearly know what to expect from each other), (0.67, 0.67) is un unstable equilibrium as drivers don't clearly know what to expect from each other since they are indifferent between speeding and slowing down.
Let x1 = probability that D 1 maintains speed and x2 = probability that D 1 slows down
Let y1 = probability of D 2 maintains speed and y2 = probability of D 2 slows down
Both drivers want to minimize their expected losses
D 1’s expected losses = x1 (EM 1) + X2 (ES 1)
D 1’s expected losses = x1 [(1- ?) (21) + (?)(0)] + x2 [(1- ?) (1) + (?)(10)]
D2’s expected losses = y1 (EM 2) + y2 (ES 2)
D2’s expected losses = y1 [(1- ?) (21) + (?)(0)] + y2 [(1- ?) (1) + (?)(10)]
A Nash equilibria will consist of probability beliefs (1-?, ?, 1-?, ?), probability of chosing probabilities (x1, x2, y1, y2), such that x1 = ?, y1 = ? and x1 = ? , x2= 1- ?
For Driver 1 :
Min(x1, x2) = x1 [(1- ?) (21) + (?)(0)] + x2 [(1- ?) (1) + (?)(10)], where x1 + x2 = 1
The Lagrangian takes the form:
L = x1 [(1- ?) (21) + (?)(0)] + x2 [(1- ?) (1) + (?)(10)] - £ (x1 +x2 -1) - £2x1 - £3x2
When differentiating with respect to x1 and x2 we get the following results:
(1-?) (21) + (?)(0) = £1 + £2
(1-?) (1) + (?)(10) = £1 + £3
x1 + x2 = 1
£2 = £3 = 0 which results in
(1- ?) (21) + (?)(0) = £1
(1- ?) (1) + (?)(10) = £1
Thus,
(1- ?) (21) + (?)(0) = (1- ?) (1) + (?)(10),
20 ? = 30
? = 2/3 (thus the driver maintains speed two out of three times and slows down one out of three times)
since x1 = ? , x2= 1- ?; x1 = 2/3, x2 = 1/3
Min(x1, x2) = x1 [(1- ?)(21) + (?)(0)] + x2 [(1- ?) (1) + (?)(10)] = 2/3 (1-2/3) (21) + 1/3 (1-2/3) = 7
Thus D 1 has an expected loss of 7
For Driver 2
same method as for Driver 1 leads to
y1 [(1- ?) (21) + (?)(0)] + y2 [(1- ?) (1) + (?)(10)] = 7
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