We'll now work through an extended version of the "prisoner's dilemma," as detailed on pgs. 245-246 of the textbook. To do so, we'll need to develop a little bit of game-theoretic
terminology. First, each player (a firm in an oligopolistic market) will have actions they might take. In this example, the actions are "withhold" or "flood." When a firm withholds, they keep
their output low. When a firm floods the market, output is high. So there are four separate potential outcomes, and each firm receives different payoffs depending on the outcome. We
specify profits here:
1. When firm #1 withholds and firm #2 withholds, profits are p₁ and p2, respectively. These are variables, which we do not specify now. We will solve for the pair of profits which allow
the firms to achieve this cooperative outcome.
2. When firm #1 floods and firm #2 withholds, profits are 10 and 0.
3. When firm #1 withholds and firm #2 floods, profits are 2 and 6.
4. When firm #1 floods and firm #2 floods, profits are 6 and 2.
To clean things up aesthetically, let's put this information into a table, where the first number gives firm #1's payoff and the second number gives #2's:
#2 withholds
#2 floods
2,6
P1, P2
10,0
6, 2
#1 withholds
#1 floods
We now define Nash equilibrium. A pair of actions constitutes an equilibrium when neither firm wants to change their action, given what the other is set to do. So for (withhold, withhold) to
be an equilibrium, we have two conditions. Given that #2 is set to withhold, #1 must get a higher payoff by withholding than by flooding, so that they would want to play their prescribed
action. Otherwise we would not have an "equilibrium." So it must be that p₁ is at least 10. And given that #1 is set to withhold, #2 must do better by withholding than they would be
flooding. So p2 must be at least 6.
However, this doesn't make much sense. When a firm cheats and floods, they should do better than they would have by withholding. This would be in accordance with the book's example.
So it's safe to say that for our model to accurately describe reality, p1 is NOT at least 10, nor is p2 at least 6. Is cooperation at (withhold, withhold) still possible? Let's extend the model a
little bit.
Say that mutual cooperation builds trust between firms and allows for a cooperative future. In particular, if (withhold, withhold) is played, then firm #1 gets p₁ today and p₁ tomorrow, for 2p1
total. Firm #2 similarly gets p2 today and p2 tomorrow, for 2p2 total. If any other outcome results, then firms receive payoffs today according to the table above, and they each flood the
market tomorrow since trust was not built. In this case, tomorrow's payoffs are 6 for firm #1 and 2 for firm #2. For (withhold, withhold) to be an equilibrium, p₁ must be at least
6
and p2 must be at least 6
and p2
Let's consider a final modification. In this version, say that whenever one firm withholds and the other floods, the firm that withholds gets 0 tomorrow and the firm that floods gets twice as
much as they would otherwise receive tomorrow. This captures the idea that taking advantage of the other firm today means a greater market share tomorrow. So if both withhold, payoffs
tomorrow are as in the table above. If both flood, payoffs are also as above. But if firm #1 floods and firm #2 withholds, then they receive 10 and 0 today; 12 and 0 tomorrow. Similarly, if
firm #1 withholds and firm #2 floods, then they receive 2 and 6 today; 0 and 4 tomorrow. For (withhold, withhold) to be an equilibrium, p1 must be at least
must be at least
Does the ability to capture the market make it easier or harder for firms to cooperate?