# 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.

Hint 1: Suppose a mixed strategy NE $(\sigma_1,\sigma_2)$ exists, where $\sigma_i$ is a distribution over $[0,1]$. By the property of MSNE, $$u_i(x_i,\sigma_j)=u_i(x_i',\sigma_j)$$ for all pure strategies $x_i,x_i'$ in the support of $\sigma_i$.
Hint 2: Note that a segment of $u_i$ is linear in strategies, and thus $$x_i+\frac12(1-x_i-\mathbb E[x_j])=x_i'+\frac12(1-x_i'-\mathbb E[x_j])$$
• @LaurenSolis: Hint 1 is essentially the definition of MSNE. Note that Hint 2 implies that, in a MSNE where $j$ uses a non-degenerate mixed strategy $\sigma_j$, $i$'s best response is a pure strategy, i.e. $x_i=x_i'$ and hence $\sigma_i$ is a degenerate distribution. By symmetry, the same holds for $j$. So no MSNE exists in the proper sense that players randomize over more than one pure strategy. Of course, you still need to fill in the details by discussing cases where $x_i+x_j>1$ etc. – Herr K. Feb 8 '17 at 23:19