* The [[http://xkcd.com/182/ xkcd]] comic link does point out the equilibrium presented is unstable, but it was also mentioned in the movie that no girl wants to be seen as second choice. If the guys all go for her friends, none of the guys will be tempted to switch to the blonde because she will be seen as second choice, won't she?
** Yes, but only if all the guys really take one of the blonde woman's friends. Afterwards, it is somehow stable. But, if the guys plan beforehand to only talk to the blonde's friends, each one of them would gain by not sticking to that plan. No equilibrium in the terms of game theory then, and not the famous Nash equilibrium.
** OP: Ah, OK, so the movie's example is actually PrisonersDilemma?
*** It's not quite a PrisonersDilemma. In the PrisonersDilemma, the globally best solution is for everybody to cooperate but each individual will do better defecting than cooperating no matter what the other player chooses. In the movie's scenario, defecting is only better if everybody else cooperates. If at least one other player defects, you will do worse by defecting rather than cooperating. This situation is more like the [[https://en.wikipedia.org/wiki/Platonia_dilemma platonia dilemma]], though the "consolation prize" for cooperating is a slight twist.
** The movie doesn't explain that second part but it doesn't contradict it either. In fact in the movie we are not informed of the official Nash Equilibrium, only the EurekaMoment (with diagram) when discussing the "blonde with her friends" dilemma. It is just implied that the theory he presents is directly connected to the bar scene revelation, which it is.