# Definition of Pareto efficiency and prisoner's dilemma

Pareto efficiency is defined in Wikipedia as:

Pareto efficiency or Pareto optimality is a state of allocation of resources from which it is impossible to reallocate so as to make any one individual or preference criterion better off without making at least one individual or preference criterion worse off.

And it has similar definitions in other sources.

Let's say in a Prisoner's dilemma game, A is confession, B is tie.

The strategies with results are:

(A, A) = -6, -6
(A, B) = 0, -10
(B, A) = -10, 0
(B, B) = -1, -1


Nash equilibrium is (A, A), also both confessing, and it's not Pareto efficient because by moving from (A, A) to (B, B), both could improve the result (form -6 to -1, also by 5).

But why is (A, B) and (B, A) not Pareto efficient? In both cases, no player can be made better off without making another player worse off.

Moving from (A, B) to (B, B) makes "player 2" better off (from -10 to -1) but makes "player 1" worse off (from 0 to -1).

Moving from (A, B) to (A, A) makes "player 2" better off (from -10 to -6) but makes "player 1" worse off (from 0 to -6).

Moving from (A, B) to (B, A) makes "player 2" better off (from -10 to 0) but makes "player 1" worse off (from 0 to -10).

So moving from (A, B) to any other combination makes one player worse off. That's the case with (B, A) too.

So why is (A, B) and (B, A) not Pareto efficient?

• Who says they aren't? – Kenny LJ Mar 7 at 10:17
• "A is confession, B is tie." Did you mean to write "tie"? – Giskard Mar 7 at 13:13