-
7/31/2019 p321ovh8
1/6
I. Failures of Equilibrium: Multiple Equilibria
Example 1. Clash of Wills (two distinct equilibria)
Baker
Mtns. Sea
Mtns. (2, 1) (0, 0)
Able
Sea (-1, -1) (1, 2)
Without communication
- two equilibria
- mixed strategy may be best With communication
- coin flip
- change the game: threats, rewards, intransigence
Example 2. Gift of the Magi
Wife
Sell hair/buy chain Dont sell
Sell watch/buy comb (-1, -1) (2, 1)Husband
Dont sell (1, 2) (0, 0)
-
7/31/2019 p321ovh8
2/6
2
II. Failures of Equilibrium: sub-optimal equilibrium
1. Example: Prisoners Dilemma (PD)
B
Confess Dont
Confess (10, 10) (3, 15)
A
Dont (15, 3) (5,5)
Solution here is (10, 10): both use their dominant strategy.
Co-ordination problem: a profile of individually rational
strategies may not be rational for
the group.
2. What is group rationality?
Pareto optimality is a necessary condition
Pareto optimal: no other profile in which one player is better
off and all other players
are no worse off.
3. Examples of PD
a) OPEC or other embargo.
b) Gas Wars.
- cost to vendor is 30 cents/litre; vendor sets price for the
day- lowest price gets all the sales for that day
- volume of sales starts at 10,000 for 38 and goes up 10,000 for
each 2 cent drop
Total profits (in dollars):
Essos price
32 34 36 38
32 (400, 400) (800, 0) (800, 0) (800, 0)
Shells 34 (0, 800) (600, 600) (1200, 0) (1200, 0)
price
36 (0, 800) (0, 1200) (600, 600) (1200, 0)
38 (0, 800) (0, 1200) (0, 1200) (400, 400)
-
7/31/2019 p321ovh8
3/6
3
c) Honour system for exams
III. Iterated Prisoners Dilemma
Two cases: a definite and indefinite version of the game.
Case 1: A definite number of PD games, number known in
advance.
A. The surprise exam paradox.
B. The regress argument.
One game: No way to achieve the optimal solution.
Two games: I know the other player will cheat on the last game.
So I may as
well maximize outcome by cheating on 2nd last esp. since he will
do so.
n games: same argument.
C. The empirical objection: Tit-for-tat (TfT) does better.
Virtues of TfT:
1. Niceness: dont initiate cheating. (Not a virtue of strategy
D, but close)
2. Provocability: punishes cheaters.
3. Forgiveness: able to restore co-operation.
4. Stability: performs well against itself.
Whats wrong with the regress argument?
1) Rationality of individual decisions vs. rationality of a
strategy (or sequence of
decisions). (2nd order rationality)
2) Strategies actually available may change rationality of
cheating.
Idea: Construct a super game table.
a) Basic game
B
Co-operate Defect
Co-operate (3, 3) (1, 4)
A
Defect (4, 1) (2, 2)
b) Parameters for super game:
- exactly ten games
-
7/31/2019 p321ovh8
4/6
4
- just four strategies: co-operate, cheat, TfT and TfT* (TYT but
cheat on last round)
B
Co-op Tit-for-Tat TfT* Cheat
Co-operate 30 30 28 10
TfT 30 30 28 19
A
TfT* 31 31 29 19
Cheat 40 22 22 20
Observations:
1) Unconditional co-operation is a dominated strategy.
2) In reduced game table, Always cheat is much less
attractive.
3) TfT is dominated by TfT*.
4) In remaining 2 x 2 table, have two equilibria: both use TfT*,
or both Cheat.
But this game is NOT a PD! Arguably, both will opt for TfT*,
which achieves
the Pareto optimal equilibrium.
c) Restoring the regress: Adding additional available strategies
restores (Cheat,Cheat) as the only equilibrium.
- TfT** (cheat on last two rounds) dominates TfT*
- etc.: until Cheat dominates TfT*********
Upshot: what is rational depends upon what strategies are
available to you and to
other agents as the second point makes clear.
3) Actual dispositions of other agents in the environment
(proportions in which various
strategies occur): similar to 2) above.
- This is an empirical matter; goes outside pure game theory
**So: not always cheating may be conditionally rational rational
given the proportions of
different strategies actually represented in the population.
Case 2: Indefinite number of PD games.
denied the use of the strategies TfT*, TfT**, etc.
-
7/31/2019 p321ovh8
5/6
5
Can only use *Ch, **Ch, etc. (= switch to cheat after 1 round of
TfT or 2 rounds, etc.)
And these don't dominate TfT.
IV. Evolutionary PD games
a) Basic PD game:
B
Co-operate Defect
Co-operate (3, 3) (1, 4)
A
Defect (4, 1) (2, 2)
Consider just two strategies: TfT* and Cheat.
b) Assumptions for iterated PD:
1) 10 games per round.
2) Individuals from both populations pair off randomly.
3) No recognition of the others type.
4) New proportion for each group is just the ratio of its
expected utility divided by the
total expected utility. (Think of utility in terms of
meals.)
Two sorts of problems:
Problem 1): Given present proportions, find proportions in next
generation.
Step 1: Compute outcome for each possible pair. (# of meals in
10 games)
TfT* Cheat
TfT* 29 19
(just shows value for Row)
Cheat 22 20
Step 2: Compute the expected utility for each strategy (EU =
expected # of meals).
Letp be proportion of TfT* and q be proportion of Cheaters.
-
7/31/2019 p321ovh8
6/6
6
EU(TfT*) =p29 + q19
EU(Cheat) =p22 + q20
Step 3: Compute the relative utility earned by each group. This
is just Step 2 times the
proportion of the population represented by the group:
RU(TfT*) =p (p29 + q19)
RU(Cheat) = q (p22 + q20)
Step 4: Compute the new proportions in the population. For each
group, this is just its relative
utility divided by the total relative utility.
TfT2 = RU(TfT*) / RU(TfT*) + RU(Cheat)
=p (p29 + q19) / [p (p29 + q19) + q (p22 + q20)]
Cheat2 = q (p22 + q20) / [p (p29 + q19) + q (p22 + q20)]
Problem 2): Find the equilibrium proportions.
Equilibrium occurs if the new proportions are the same as the
old. So just set TfT2 = p
and Cheat2 = q in the above equations, and find all solutions. 0
and 1 will always be
solutions; there will be at most one more.
Three equations:
1)p + q = 12)p =p (p29 + q19) / [p (p29 + q19) + q (p22 +
q20)]
3) q = q (p22 + q20) / [p (p29 + q19) + q (p22 + q20)]
Now solve from 1) and 2): get p = 0 or p = 1 or p = 1/8.
What happens is: if the proportion of TfT* = 1/8, stays that
way. If it is greater, eventually the
Cheaters vanish. If it is smaller, the TfT* vanish.
