Pareto optimality in game theory

Question

Consider the classic Game Theory game Prisoner Dilemma:

  ______C_____________D______
 C|   -1,-1    |   -4, 0
  |------------|-----------
 D|   0, -4    |   -3,-3

Now in this game there are 3 Pareto optimals: (-4,0), (0,-4) and (-1,-1).

I have had some trouble about considering (-1,-1) as a Pareto optimal, maybe because there is a better outcome for both players (not at the same time) in other outcomes.

Is the following reasoning correct?

• initialize the set of Pareto optimals with all the outcomes (a,b);
• if there is another outcome (a',b') such that a'>a ANDb'>bthen remove from the Pareto optimals set(a,b)`.

Is this correct?

Another question: is the outcome where a player has the maximum of his/her utility always a Pareto optimal?

Answer 1

I have had some trouble about considering (-1,-1) as a Pareto optimal, maybe because there is a better outcome for both players (not at the same time) in other outcomes.

No. It does have to be at the same time. that is, since none of the outcomes have the feature that it is better than $(-1,-1)$ for both players at the same time. So, it is Pareto optimal.

You say this yourself: for some outcome $(a,b)$ to not be Pareto Optimal, you must have some other outcome $(a',b')$ where both $a'>a$ AND $b'>b$. This is not true for $(-4,0)$ (worse for player 1), $(0,-4)$ (worse for player 2), and $(-3,-3)$ (worse for both players).

p.s. in explaining your reasoning you say:

• assume every outcome (a,b) is Pareto Optimal.

I don't understand why you would be assuming that. It's typically not true.

Another question: is the outcome where a player has the maximum of his/her utility always a Pareto optimal?

Yes, because for some outcome to not be a Pareto Optimal, there must be some other outcome where that player can do better, but that is not the case if the player is already at his/her maximum.