ECON 201 – A
GAMES AND STRATEGY
). Bonnie and Clyde are playing a simultaneous-move stage game (a
prisoner’s dilemma to which a third strategy is added) as shown below
. They play
this game two times;
they learn the first-round strategies and then play the second-
Is there a SPE in which the players cooperate in the first round? If yes show it, if
not, explain why there is none.
1 , 1
-2 , 2
0 , 0
The Stage game has two Nash equilibria: (Don’t, don’t) and (defect, defect). Note that
the payoffs in the second Nash equilibrium are both -1, whereas those in the first equilibrium are
0. In the finitely repeated game, the first Nash equilibrium can be used as a “punishment path”.
Let us see if it is possible to have as SPE the outcome written in the question.
We know than in ANY SPE, the last round play must be one of the two Nash equilibria.
So, the strategy could be:
“Play Coop in the first round. If in round 1 the outcome is (coop, coop), play don’t coop in round
two. If any outcome other than (coop, coop) is observed in round 1, then play defect in round 2.”
If both players adopt this strategy, would any one of them deviate? If your rival is playing coop in
round one, the best deviation is to don’t cooperate; this will increase your payoff from 1 to 2. But
in the second round your rival will play defect, to which defect is the best response, and you will
get -1, instead of 0.
So the benefit from the best deviation is 2 – δ(-1), which is larger than the benefit from no
deviation, 1 + δ(0) for all δ smaller than one (the two payoffs are equal only if δ = 1, when the
payers do not discount future payoffs). We conclude that it is impossible to have an SPE in which