It's precisely the correct word. Prisoner's dilemma are simple, mathematically. This was one. OpenBSD defected. The joke's on the security researcher, though, since this doesn't appear to have been their first time [1][2].
Robert Axelrod outlined, in his 1984 classic The Evolution of Cooperation [3] four requirements for a successful iterative prisoner's dilemma strategy. One is retaliating. Security researchers are letting OpenBSD play an iterating game as if it's an N=1, i.e. they're not retaliating. Given the community is playing "always cooperate," OpenBSD's best move is actually "always defect".
So does the simple mathematical treatment also include language like "the joke's on ____”? Or was that more of a philosophical interpretation of yours?
Both your [1] and [2] seem to conclude that violating the embargo had no significant ill effects: "since... the underlying issue was already publicly known, OpenBSD's commits don't change things much." If "defecting" causes no problems for the other participants, does it actually count as defecting? (And if not, how is this a mathematically simple prisoner's dilemma?)
It's precisely the correct word. Prisoner's dilemma are simple, mathematically. This was one. OpenBSD defected. The joke's on the security researcher, though, since this doesn't appear to have been their first time [1][2].
Robert Axelrod outlined, in his 1984 classic The Evolution of Cooperation [3] four requirements for a successful iterative prisoner's dilemma strategy. One is retaliating. Security researchers are letting OpenBSD play an iterating game as if it's an N=1, i.e. they're not retaliating. Given the community is playing "always cooperate," OpenBSD's best move is actually "always defect".
[1] https://lwn.net/Articles/726585/ thank you 0x0 [a]
[2] https://lwn.net/Articles/726580/ thank you 0x0 [a]
[a] https://news.ycombinator.com/item?id=15481980
[1] https://en.wikipedia.org/wiki/The_Evolution_of_Cooperation