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What Prisoner’s Dilemma?

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2 Responses

  1. Law Student says:

    This isn’t a true prisoner’s dilemma (i.e. one with a Nash equilibrium), despite the article’s invocation of it. Under a prisoner’s dilemma, a player has two options and is unambiguously better off by selecting a certain option, regardless of what the other player picks. In this scenario, the students are not unambiguously better off by always selecting cooperate no matter what the other players select. Indeed, had someone defected, everyone else would have chosen to defect as well. Therefore, no Nash equilibrium (and not a Prisoner’s Dilemma) because what any individual player chooses depends entirely on what the other players choose to do.

  2. prometheefeu says:

    The article says they found a way to cooperate without iterating, but that is not correct. They all sat outside the door for around an hour. Every second was an iteration where they could see that none of their classmates were defecting. I’m betting the first few minutes were very tense with everyone ready to dash in and that after a couple iterations, (minutes) everyone started to relax as they built mutual trust.

    More importantly, this was not (necessarily) the PD. In the PD, if everyone else is cooperating, defecting gives you an advantage. But here, taking the exam would not boost your grade if no-one else took it. And on top of that you have to go through the unpleasantness of taking the exam. Before the students use social pressure to change the game, full cooperation is already a Nash. (Though an unstable one) Given that all of your peers are cooperating, you are better off cooperating than defecting. In the PD, all out cooperation is not a Nash.

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