Mixed strategy problem - game theory I have a basic doubt in a question of game theory.
Assume that in a $2$ player game the mixed strategy profile $((a,b,0),(c,d,0))$ is a mixed strategy NE. Does the indifference condition in a mixed strategy NE imply that $a = b = 1/2$ ? Why or why not?
Thanks
 A: The indifference condition in mixed strategy NE does not imply $a=b=1/2$. Here is an example:
 $\hskip1.8in$
First, note that since there are three actions for player 1, there are technically seven possible supports for his strategy: UCD, UC, UD, CD, U, C, D. Similarly, there are are seven possible supports for player 2. Therefore, there are in fact 49 different combinations of mixed strategies that we could consider in looking for mixed equilibria. Since that sounds like an unpleasant exercise, let's try to narrow it down.
First note that $M$ strictly dominates $L$ for player 2. Also, a 50/50 mix of $U$ and $C$ strictly dominates $D$ for player 1. Then, we are left with:
$\hskip1.7in$
Note that we can do this elimination because strictly dominated actions are never played with positive probability in mixed equilibria.
First, the underlining above for best responses shows that there is no pure-strategy NE. Looking at mixed equilibria, since, for each player, the best response to each of his opponents actions is unique, neither player wants to mix unless the other is mixing. Therefore, both plays must mix to make the other indifferent. Let P1 put probability $p$ on $U$, i.e. $\alpha_1(U)=p$ and probability $1-p$ on $C$. Let P2 put probability $q$ on $M$, i.e. $\alpha_2(M)=q$ and probability $1-q$ on $R$. 
\begin{align*}
\text{To make P2 indiff: } && 5p+5(1-p) &= 3p + 8(1-p) &\Leftrightarrow p=\frac{3}{5}\\
\text{To make P1 indiff: } && 3q+6(1-q) &= 5q + 4(1-q) &\Leftrightarrow q=\frac{1}{2}
\end{align*}
Therefore, the unique mixed equilibrium is $\left((\frac{3}{5},\frac{2}{5},0),(0,\frac{1}{2},\frac{1}{2}) \right) $. As you can see, depending on the game, none of the following is guaranteed: $a=1/2$, $b=1/2$, $a=b$. You often will see $1/2$ in there, just because those problems are easiest to calculate, but there's really no reason $1/2$ should appear more often than any other fraction. The key is that P1 must be mixing to make P2 indifferent between his actions. Sometimes this will involve mixing evenly across P1's actions, but often it won't.
