Inspired by the bar room scene from A Beautiful Mind (link), as an extra assignment for a Game Theory course we were asked to analyze this scene. We assume there are $n \geq 2$ men, an equal amount of brunettes and one blonde in a bar. Every man decides simultaneously and independently whether he will go for the blonde. They all agree that getting a blonde (with payoff $a$) is preferable to getting a brunette (with payoff $b < a$). If two or more men decide to go for the blonde they block each other and these men get a payoff of $0 < b < a$.

I want to show that there exists a symmetrical mixed Nash equilibrium but I'm not sure how to go about this.

My attempt so far:

Since the equilibrium should be symmetric, I define $P($player $i$ goes for the blonde$) = p = 1 - P($player i goes for the brunette$)$, where $i = 1, \ldots,n$ (so everyone man in the bar plays this mixed strategy).

Since player $i$ gets payoff $a$ if he goes for the blonde and the rest go for the brunettes, which happens with probability $p\cdot \Pi_{i=1}^{n-1}(1-p)$. Similarly, player $i$ gets payoff $b$ with probability $(1-p)$. So the expected value of playing this mixed strategy is $$a\cdot p\cdot \Pi_{i=1}^{n-1}(1-p) + (1-p)\cdot b$$ For any player. What would be the best way to show that this is a mixed Nash equilibrium? Any advice would be appreciated! (This is not homework but rather an exercise to practise with.)

[edit] I just had another idea; would it be sufficient to show that if one of the players changes his probability from $p$ to, say, $p+\epsilon$, his expected payoff would then be smaller?

  • $\begingroup$ I still think this is difficult. Suppose player 1 plays blonde with probability $q > p$ and everyone else plays blonde with probability $p$. Then the expected value of player 1 is $a\cdot q\Pi_{i=1}^{n-1}(1-p) + (1-q)b$, but to me it's not easy to see that this is less or equal to $a\cdot p\Pi_{i=1}^{n-1}(1-p) + (1-p)b$... $\endgroup$
    – Stijn
    Commented Nov 1, 2011 at 17:55

1 Answer 1


You're on the right track. Since the equilibrium is symmetric, we can assume that all the other players have the same strategy $p$ and a single player optimizes his strategy $q$ against that background. You already wrote down the right payoff for this in your comment: $aq(1-p)^{n-1}+(1-q)b$. At equilibrium, this musn't depend on $q$, since the single player isn't able to improve his strategy by changing $q$. Setting the derivative with respect to $q$ to zero yields $a(1-p)^{n-1}-b=0$, and thus $p=1-\sqrt[n-1]{b/a}$. Substituting this into the symmetric payoff (with $q=p$) gives an equilibrium payoff of $b$. That makes sense: Since each player can get payoff $b$ independent of the strategies of the other players, the probabilities are chosen such that the expected payoff for trying to get $a$ is also $b$ – if it weren't, the expected payoff could be improved by shifting the strategy.

  • $\begingroup$ Very helpful, thanks! $\endgroup$
    – Stijn
    Commented Nov 1, 2011 at 18:12
  • $\begingroup$ Just to be sure, is this reasoning correct? "If I'm the player $i$ who plays blonde with probability $q$, then if I want $q$ to be part of a mixed NE, then this should be a best response, given the probabilities of the other players. This means that $\frac{\partial}{\partial q}[aq(1−p)^{n−1}+(1-q)b]=0$, which does not depend on $q$, but which is maximized for $p = 1 - (\frac{b}{a})^{n-1}$. Therefor the expected payoff for player $i$ is maximized if it is equal to $p = 1 - (\frac{b}{a})^{n-1}$." $\endgroup$
    – Stijn
    Commented Nov 1, 2011 at 21:43
  • $\begingroup$ @Stijn: As far as I can make sense of that statement, no, it isn't correct. First, your double use of "I" in the first sentence is confusing. If the two "I"s refer first to one of the players and then to you as a mathematician, that's very confusing; if the second "I" also refers to the player like the first "I", then the sentence is wrong, since a Nash equilibrium isn't something one of the players "wants"; each player maximizes his expected payoff. Second, the condition that $\frac{\partial}{\partial q}\ldots=0$ is fulfilled, not maximized. Third, $p$ is a probability, not a payoff. $\endgroup$
    – joriki
    Commented Nov 1, 2011 at 22:46

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