The Nash Demand Game I am having troubles to figure out how to find all pure strategies and two mixed strategies (one in which the shares received by players are always $0$, and one in which $\omega$ is often - but may be not always distributed to players) of following game.


*

*The set of players is $N=\left \{ 1,\dots,n \right \}$. There is a sum of money $\omega > 0$ to divide among the $n$ players. 

*Each player's action set is $A_i=[0,\omega]$. 

*Given action profile $a=(a_1,\dots,a_n) \in A$, if $\sum_{j \in N}a_j \leq \omega$, then each player $i$ gets the share of $\omega$ he asked for and a utility of $u_i(a_i,a_{-i})=a_i$. Given action profile $a\in A $, if $\sum_{i \in N} a_i > \omega$, then each player gets a share of $0, \omega$ is destroyed and $u_i(a_i,a_i)=0$ for each $i \in N$. In sum $$u_i(a_i,a_{-i})=\begin{cases}a_i, &{\rm if}\; \sum_{j=1}^na_j\le\omega\\0, &{\rm if}\; \sum_{j=1}^na_j>\omega\end{cases}$$

 A: A mixed-strategy Nash equilibrium in addition to the ones given by Jimmy R. occurs when all players bid $\frac1{n+1}$ with probability $p$ and $\frac2{n+1}$ with probability $1-p$. In this case, the only sensible bids are multiples of $\frac1{n+1}$, a zero bid is certain to pay off zero, and any bid above $\frac2{n+1}$ is certain to pay off zero, so we only have to choose $p$ to create equilibrium between the two bids $\frac1{n+1}$ and $\frac2{n+1}$.
The equilibrium condition is
$$
\frac1{n+1}\left(p^{n-1}+(n-1)p^{n-2}(1-p)\right)=\frac2{n+1}p^{n-1}
$$
with solution $p=1-\frac1n$.
A: In games with $n$ players it is useful to "guess" an equilibrium $a^*=(a_1^*,\dots,a_n^*)$ and then to show that it is indeed an equilibrium by showing that given $a^*_{-i}$ the proposed $a^*_i$ is indeed a best reply to $a^*_{-i}$. Symmetry should also be exploited when possible. So here,


*

*Consider the strategy profile $a^*$ with $a^*_j=\frac{w}{n}$ for all $j=1,\dots, n$. Then, given $a^*_{-i}$ you have that $$\sum_{j=1,j\neq i}^na^*_j=\frac{n-1}{n}w=w-\frac1nw$$ Hence player $i$ can get at most $$w-\left(w-\frac1nw\right)=\frac1nw$$ which he can indeed achieve if he bids $a^*_{i}=\frac1nw$. By symmetry this shows that $a^*$ is a Nash equilibrium in pure strategies, at which everybody receives $u_i(a^*)=\frac1nw$.

*Consider any strategy profile $a^*$ with $a^*_j\ge\frac1{n-1}w$ for all $j=1,\dots, n$. Then, player $i$ is indifferent between all of his strategies - she will receive $0$ for sure, since $$\sum_{j=1,j\neq i}^{n}a^*_j\ge \frac{n-1}{n-1}w=w$$ hence any $a^*_i\ge \frac{1}{n-1}w$ is a best reply to $a^*_{-i}$. By symmetry, this shows that any such $a^*$ is a Nash equilibrium at which everybody receives $u_i(a^*)=0$. Actually this is a continuum of Nash equilibria, containing pure and mixed strategies. For example a player may mix between strategies $>\frac{1}{n-1}w$ and not choose a pure one.


I do not know if there is a third component of Nash equilibria (equilibrium) but if I come up with one, I will edit my answer. (If you do please let me know).
