I’ve been fascinated by the Prisoner’s Dilemma ever since I first encountered it. It goes something like this.
You have two prisoners which are under interrogation. They have only two choices.
Choice A: Stay silent.
Choice B: Betray the other prisoner.
Call the two prisoners Bob and Wendy.
Now, there’s 4 permutations.
Wendy plays A and Bob plays A. They both serve a month in prison.
Wendy plays A and Bob plays B. Bob goes free Wendy does 1 year.
Wendy plays B and Bob plays A. Wendy goes free Bob does 1 year.
Wendy plays B and Bob plays B. In which case both get 3 months.
The question is, if Bob and Wendy replay the game over and over is there a way of playing that minimises the time spend in prison?
Well someone did ask this question and also asked for people to write programs to play the role of a prisoner. There were evil programs that would always choose B, nice programs that always choose A and other more sophisticated ones that did statistical analysis of the other prisoners responses and of course a whole lot of other strategies in between.
After running these programs a winner emerged called Tit for Tat. Its strategy was simple; start with A then depending on how the other prisoner responds, start the next round with that.
You don’t want a fluke, so everyone was told about Tit for Tat’s strategy and off they all went to revise their programs and the drama was played out all over again and the winner was … Tit for Tat.