The Truly Iterated Prisoner’s Dilemma

Fol­lowup to: The True Pri­soner’s Dilemma

For ev­ery­one who thought that the ra­tio­nal choice in yes­ter­day’s True Pri­soner’s Dilemma was to defect, a fol­low-up dilemma:

Sup­pose that the dilemma was not one-shot, but was rather to be re­peated ex­actly 100 times, where for each round, the pay­off ma­trix looks like this:

Hu­mans: C Hu­mans: D
Paper­clip­per: C (2 mil­lion hu­man lives saved, 2 pa­per­clips gained) (+3 mil­lion lives, +0 pa­per­clips)
Paper­clip­per: D (+0 lives, +3 pa­per­clips) (+1 mil­lion lives, +1 pa­per­clip)

As most of you prob­a­bly know, the king of the clas­si­cal iter­ated Pri­soner’s Dilemma is Tit for Tat, which co­op­er­ates on the first round, and on suc­ceed­ing rounds does what­ever its op­po­nent did last time. But what most of you may not re­al­ize, is that, if you know when the iter­a­tion will stop, Tit for Tat is—ac­cord­ing to clas­si­cal game the­ory—ir­ra­tional.

Why? Con­sider the 100th round. On the 100th round, there will be no fu­ture iter­a­tions, no chance to re­tal­i­ate against the other player for defec­tion. Both of you know this, so the game re­duces to the one-shot Pri­soner’s Dilemma. Since you are both clas­si­cal game the­o­rists, you both defect.

Now con­sider the 99th round. Both of you know that you will both defect in the 100th round, re­gard­less of what ei­ther of you do in the 99th round. So you both know that your fu­ture pay­off doesn’t de­pend on your cur­rent ac­tion, only your cur­rent pay­off. You are both clas­si­cal game the­o­rists. So you both defect.

Now con­sider the 98th round...

With hu­man­ity and the Paper­clip­per fac­ing 100 rounds of the iter­ated Pri­soner’s Dilemma, do you re­ally truly think that the ra­tio­nal thing for both par­ties to do, is steadily defect against each other for the next 100 rounds?