while going over wiki page on Battle of the Sexes game I found something funny.

This game has two pure strategy Nash equilibria, one where both go to the opera and another where both go to the football game. For the first game, there is also a Nash equilibrium in mixed strategies, where the players go to their preferred event more often than the other. For the payoffs listed above, each player attends their preferred event with probability 3/5.

http://en.wikipedia.org/wiki/Battle_of_the_sexes_%28game_theory%29#Equilibrium_analysis Also from wiki :

A mixed strategy is an assignment of a probability to each pure strategy. This allows for a player to randomly select a pure strategy. Since probabilities are continuous, there are infinitely many mixed strategies available to a player, even if their strategy set is finite.

Of course, one can regard a pure strategy as a degenerate case of a mixed strategy, in which that particular pure strategy is selected with probability 1 and every other strategy with probability 0.

So my Q is why arent pure strategies also considered Nash equilibrium for the BotS game? I mean if players play lets say uper left corner of the matrix(always, p1=1.0,p2=1.0) then either players expected utility will drop if it lowers its px without other player chaning his px. So shouldnt this be Nash eq also?

  • 3
    $\begingroup$ They are Nash equilibria, and the article says so ("two pure strategy Nash equilibria"). $\endgroup$
    – Charles
    Commented Jan 23, 2013 at 15:11
  • $\begingroup$ my question maybe isnt clear, but my question is why article doesnt mention them when talking about mixed strategies NE... same thing with recent lectures on Coursera... lecturer only mentioned fractional probabilities mixed strategies, not the ones with (1.0,0.0) probs. basically my question is if all pure str are subset of the mixed strategies set, why arent all pure str subset of the mixed nash eq set? $\endgroup$ Commented Jan 23, 2013 at 15:17
  • $\begingroup$ math.stackexchange.com/questions/742332/… $\endgroup$
    – GuyEternal
    Commented Apr 24 at 19:45

2 Answers 2


It's like tonic, which isn't considered a mixed drink ($0$ parts gin and $1$ part tonic). That's degenerate for you: "In mathematics [as in mixology], a degenerate case is a limiting case in which a class of object changes its nature so as to belong to another, usually simpler, class."


Yes. All pure Nash equilibria is a subset of mixed Nash equilibria. Pure strategy equilibrium is just a special case of mixed ones. However, some professors (or wikipedia entries) use "mixed equilibrium" to refer completely mixed equilibrium. Some professors distinguish between those cases : 1) completely mixed 2) semi mixed 3) pure equilibrium.

  1. completely mixed: all pure strategies should be played with a probability >0
  2. semi mixed: some pure strategies are not used (used with probability 0)
  3. pure: a pure strategy is used with probability 1.

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