Stochastic Iterated Prisoner’s Dilemma

Algorithms provide several solutions to gain more insight to the workings of Prisoner’s Dilemma and the most optimal strategy for the players involved. We’ve previously tackled the defection strategy and the deterministic strategy as two of the optimal solutions in looking at the iterated Prisoner’s Dilemma. This time around, let’s talk about the stochastic iterated Prisoner’s Dilemma.

In the stochastic iterated Prisoner’s Dilemma, actions and options are defined according to the probability of cooperation. All in all, each action is characterized by four cooperation probabilities. The first one is the instance when A defected in the previous round and B cooperated. The second probability is the one when A cooperated and B defected. The third is when both cooperated and the fourth probability is when both defected.

These probabilities define the action that A or B will act in the present round. If the probabilities equate to one then, it means that they are using the deterministic strategy. If A defects a while ago then, B will defect this time around. The same thing happens if the probabilities equate to zero.

But the algorithm for stochastic iterated Prisoner’s Dilemma also provides for the win-stay strategy. It studies the transition of a player from one strategy to another. Accordingly, it believes that a player will respond in the same way if his actions in the previous round results to a win. If it’s a loss then, the player will switch strategies. The algorithm accounts for the memory of the player which defines his present behavior.

Since the probabilities make up a matrix, the iterated Prisoner’s Dilemma becomes a stochastic process wherein the present action is defined by a set of random probabilities. It’s constantly evolving. There’s no deterministic way of ensuring that a player will respond as such.


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