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June 26, 1993
 Feruary 7, 1996

  
\section*{ 6 DYNAMIC GAMES WITH ASYMMETRIC INFORMATION }  


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     \noindent
 {\bf The Gang of Four Model}

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    One of the most important explanations of reputation is that of
Kreps, Milgrom, Roberts  \& Wilson (1982), hereafter referred to as
the Gang of Four. In their model, a few players are genuinely unable to
play any strategy but tit-for-tat, and many players pretend to be that
type.  The beauty of the model is that it requires only a little
incomplete information,  a small probability $\gamma$ that player Row
is a tit-for-tat player.  

   
 The unusual players have a
small direct influence, but they matter because other players imitate
them.  Even if Column knows that with high probability Row is just
pretending to be tit-for-tat, Column does not care what the truth is
so long as Row keeps on pretending. Hypocrisy is not only the
tribute vice pays to virtue;  it can be just as good for  deterring misbehavior. 


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 {\bf Theorem 6.2: The Gang of Four Theorem}\\
 {\it Consider a    T-stage, repeated prisoner's dilemma,  without discounting
but with a probability $\gamma$ of a tit-for-tat player. In   any
perfect Bayesian equilibrium,  the number of stages in which either
player chooses {\rm Confess} is less than some number M that depends on
$\gamma$ but not on T.} 


   The significance of the Gang of Four Theorem is that while the
players do resort to $Confess$ as the last period approaches, the number
of periods during which they $Confess$ is independent of the total
number of periods. Suppose $M=2,500$. If
$T=2,500$, there might be a $Confess$ every period.  But if $T=10,000$,
  there are 7,500 periods without  $Confess$. For reasonable probabilities  of the unusual type, the number of periods of cooperation can be much larger.   Wilson (unpublished) has set up an entry
deterrence model in which the incumbent fights entry (the equivalent
of {\it Deny} above) up to seven periods from the end, although
the probability the entrant is  of the unusual type is only 0.008.

Consider what would happen in a  10,001  period game with a   probability of 0.01 that Row is playing the grim strategy of $Deny$ until the first $Confess$,  and   $Deny$   every period thereafter. If the payoffs are as in Table 5.2a, a best response for Column to  a known grim player is  ($Confess$ only in the last period, unless Row chooses $Confess$ first, in which case respond with $Confess$).  Both players will choose $Deny$ until the last period, and Column's payoff will be     50,010 (= (10,000)(5) + 10).   Suppose for  the moment that if Row is not grim, he is highly aggressive, and will choose $Confess$   every period.  If Column  follows the strategy just described, the outcome will be ($Confess, Deny$) in the first period and {\it (Confess, Confess}) thereafter, for a payoff to Column of   $ -5 (= -5 +  (10,000)(0) )$. If   the probability of  the two outcomes is 0.01 and 0.99, then Column's  expected payoff from the strategy described is 495.15. If he instead follows a strategy of ($Confess$  every period), his expected payoff is just 1 ($=0.01 (10) + 0.99 (0)$). It is clearly in Column's advantage to take a chance by cooperating with Row, even if Row has a 0.99 probability of following a very aggressive strategy. 


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