Mixed strategies Nash equilibrium I'm working on a game theory problem.I'm having trouble understanding what the mixed strategy nash equilibrium is exactly in this game.
The game is :Two players have to choose how distribute a piece of land ($size = 1$). Each player (1, 2) send his decision $x_i \in{[0, 1]}$ to an external agent.
If $x_1 +x_2 \leq 1$, each player gain a land portion equal to $x_i +\frac{1-x_1-x_2}{2}$
If $x_1+x_2 > 1$, they end up with nothing.
The first part of the problem is to solve the game in pure strategies. I figured out that the answer of this part is:
Player 1 should pick $1- x_2$, any higher and he gets 0. The game is symmetric so the same hold from player2's perspective. The optimal strategy is the intersection of these two functions.
The second part is to solve the game in mixed strategies (if exist). But I don't know how to aproach this part. 
If I use some values, the answer in mixed strategies would be $[\frac{40}{91}, \frac{3}{7}, \frac{12}{91}]$:
                    (X=0.2) (q1)    (Y=0.8)(1-q1)    (Z=0.5)(1-q1-q2)
 (A=0.2) (p1)        [(0.5, 0.5)      (0.2, 0.8)        (0.35, 0.65)]
 (B=0.8) (p2)        [(0.8, 0.2)        (0, 0)              (0, 0)  ]
 (C=0.5) (1-p1-p2)   [(0.65, 0.35)      (0, 0)          (0.5, 0.5)  ]

Any help would be appreciated.
 A: Your answer to the first part is not correct. When you say that $1$ should pick $1-x_{1}+\frac{1-x_{1}-x_{2}}{2}$, this does not make sense. $1$ is the one picking $x_{1}$, so what is the $x_{1}$ in your expression?
In pure strategies, think about best responses. Best response of player $i$ to player $-i$ choosing $x_{-i}$ is $1-x_{-i}$. Any intersection of best responses is a Nash equilibrium (in pure strategies), hence there is continuum of Nash equilibria where $1$ chooses $z$ and $2$ chooses $1-z$, for any $z\in[0,1]$.
For mixed strategies, my conjecture is that there is no Nash equilibrium in which one player plays strictly mixed strategy and the other pure strategy. Here is a sketch of the proof, you can fill in the details. Suppose $i$ plays strictly mixed strategy in a NE. Take $x_{i}$ in the support of the strategy and suppose, without loss of generality, that $x_{i}\neq1-x_{-i}$ (wlog since if $x_{i}=1-x_{-i}$ then we can take another $x_{i}'\neq1-x_{-i}$). If $x_{i}<1-x_{-i}$, then $i$ can strictly increase her payoff by deviating to $x_{i}=1-x_{-i}$. If $x_{i}>1-x_{-i}$, then $i$ can strictly increase her payoff by deviating to $x_{i}=1-x_{-i}$.
However, I am not sure how to prove that there does not exist a NE in which both players strictly mix.
