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?
Best Answer
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.