A difficult game theory riddle Suppose a master hires $n$ servants who work for him. At the end of the day, the service come to him and request a wage for their service. They are able to request a wage up to $\$1$, but no more.
With $1/2$ probability, the master is in a good mood an will give all servants their requested wages. 
With $1/2$ probability, the master is in a bad mood and only pays the the servant with the lowest request; however, he pays him wage $g(x)$ where $g$ is a continuous, differentiable, monotonically increasing $g(0)=0$ and $x$ is the wage requested by the servant who requested the highest wage. If there is a tie in this state, then nobody is payed any wage.
No servant knows if the master is in a good or bad mood and all share the half-half prior belief.
Show that there is a symmetric equilibrium (Nash) where each servant makes a wage request according to the same distribution over some interval in $[0,1]$. This means that anywhere in the support of this interval will give the same expected wage to the servant conditional on the strategy being used by the others and, of course, anywhere outside of the interval gives the same or lower expected wage. 
Show that this equilibrium is unique among symmetric equilibria.

Progress:
Let A be the good-mood state and B be the bad-mood state of the world.
Note that servants will never want to tie and so a symmetric solution will never contain atoms.
Let $F$ be the equilibrium distribution and consider the maximum of the support $\bar{w}$. Any wage offer close to the maximum of the support becomes arbitrarily unlikely to get paid in state B (this follows from no atoms). In this case, if $\bar{w}< 1$, the servant will eventually earn more in expectation by giving up on earning anything in state B and will be better of requesting $\$1$. This implies that the maximum must be one. 
Because every request in the support must earn the same expected wage, and $\$1$ is in the support, the expected wage is $\$1/2$. From this we can write an equation for the expected wage for any request in the interval:
\begin{align}
1=\frac{1}{2}w+\frac{1}{2} \left(1-F(w) \right)^{n-1}   \int_w^1 g(z) f(z) (n-1)  \frac{(F(z)-F(w))^{n-2}}{(1-F(w))^{n-1}} \;dz\\
1=\frac{1}{2}w+\frac{1}{2} (n-1)   \int_w^1 g(z) \left( F(z)-F(w)\right)^{n-2} \;dF(z)
\end{align}
Here, given a proposed wage $w$ and assuming all other $(n-1)$ servants use mixed strategy according to $F$, $(1-F(w))^{n-1}$ is the probability that all other requested wages were higher than $w$ and that the payment will be made. Given that a request was higher than $w$, it's distribution is then the truncated distribution $\frac{F(z)-F(w)}{1-F(w)}$. The distribution of the maximum of this is then $\left(\frac{F(z)-F(w)}{1-F(w)}\right)^{n-1}$. If a pdf exists, the maximum pdf is then $ f(z) (n-1)  \frac{(F(z)-F(w))^{n-2}}{(1-F(w))^{n-1}}$ which is what shows up in the first integral.
The integral term is the expected wage given request $w$.
Seems obvious to me that the $F$ should exist and be unique, just seems difficult to show this for arbitrary $n$. Maybe there should be some simple way to argue this?
 A: Intuitively, a servant requesting $1$ will expect $\frac 12$ because he will get $1$ with probability $\frac 12$ and nothing if somebody else requested something lower with probability $\frac 12$.  Nobody should request less than $\frac 12$.  Someone requesting $\frac 12$ will also expect $\frac 12$ because nobody will request less.  
Let us first consider the case of two servants and assume that they use the same probability distribution function $f(x)$ for their requests.  Let the cumulative distribution function be $F(x)=\int_{\frac 12}^xf(t)dt$.  Then the expectation on requesting $x$ is $E(x)=\frac 12x+\frac 12(1-F(x))x$ with the first term coming from the good mood and the second from winning the auction in the bad mood.  At equilibrium $E(x)$ must be constant.  As $E(\frac 12)=E(1)=\frac 12$ we have 
$$\frac 12=\frac 12x+\frac 12(1-F(x))x\\F(x)=2-\frac 1x\\f(x)=\frac 1{x^2}$$
so the servant should pull a random number $r$ on $[0,1]$ and ask for $\frac 1{2-r}$ 
For $n$ servants, the expectation on asking for $x$ is $E(x)=\frac 12x+\frac12(1-F(x))^{n-1}x$.  As $F(\frac 12)=0$ we still have $E(x)=\frac 12$.  This gives
$$\frac 12=\frac 12x+\frac12\left(1-F(x)\right)^{n-1}x\\
\frac 2x\left(\frac 12-\frac 12x\right)=(1-F(x))^{n-1}\\$$
$$\left(\frac 1x-1\right)^{\frac 1{n-1}}=1-F(x)\\
F(x)=1-\left(\frac 1x-1\right)^{\frac 1{n-1}}$$
