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?
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.