Mixed Nash equilibrium I have game table posted below:

$$\begin{matrix} &amp;\#2 \\ \#1 &amp;
\begin{array}{c|c|c|c}
&amp;D &amp;E &amp;F \\ \hline
A &amp;4,4 &amp;6,6 &amp;2,6 \\ \hline
B &amp;6,4 &amp;2,2 &amp;0,4 \\ \hline
C &amp;2,6 &amp;8,4 &amp;0,0
\end{array}
\end{matrix}
$$

Are there any Nash equilibrium where the agent 1 mixes between 

(i) A &amp; B

(ii) A and C

(iii) B &amp; c?

What I did is that

For example I did first one 

I calculate its expected utilities 

$$u1(A, p2)= 4p_D+ 6 p_E+ 2+(1-p_D-p_E)$$

$$u1(B,p)=6p_D+2p_E$$

Since first payoff is higher than the second payoff, there is no such a mixed Nash equilibrium. 

I think this approach to such a question seems wrong. Therefore I ask this question. What is your idea about the solution?
(i) $A$ &amp; $B$

If player 1 play $A$ with probability $p$ and $B$ with probability $(1-p)$, where $0&lt;p&lt;1$, then player 2's expected payoff from playing 


$D$ is $4p+4(1-p) = 4$
$E$ is $6p + 2(1-p) = 4p + 2$
$F$ is $6p + 4(1-p) = 2p + 4$


Since payoff from playing $F$ is more than the payoff from playing any other strategy for player 2, he will always choose $F$ in response to any mixed strategy of player 1 where 1 play $A$ with probability $p$ and $B$ with probability $(1-p)$. Therefore, $F$ is the best response of player 2 if 1 play a mixed strategy $A$ with probability $p$ and $B$ with probability $(1-p)$. However, given the strategy $F$ of player 2, player 1's best response is to play a pure strategy $A$. Therefore, there is no Nash equilibrium in which player 1 mixes between $A$ &amp; $B$.

(ii) $A$ &amp; $C$

If player 1 play $A$ with probability $p$ and $C$ with probability $(1-p)$, where $0&lt;p&lt;1$, then player 2's expected payoff from playing 


$D$ is $4p+6(1-p) = 6 - 2p$
$E$ is $6p + 4(1-p) = 2p + 4$
$F$ is $6p + 0(1-p) = 6p$


Since payoff from playing $E$ is more than the payoff from playing $F$, he will never choose $F$ in response to any mixed strategy of player 1 where 1 play $A$ with probability $p$ and $C$ with probability $(1-p)$. Therefore, $F$ is never a best response of player 2 if 1 play a mixed strategy $A$ with probability $p$ and $C$ with probability $(1-p)$. Now let us consider the following cases for $p$ :


$0 &lt; p &lt; 0.5$ In this case player 2 has a unique best response and that is to play $D$, in response to which player 1 would like to play a pure strategy $B$, and not a mix of $A$ and $C$. 
$0.5 &lt; p &lt; 1$ In this case player 2 has a unique best response and that is to play $E$, in response to which player 1 would like to play a pure strategy $C$, , and not a mix of $A$ and $C$. 
$p = 0.5$ In this case player 2 is indifferent between playing $D$ and $E$, and if 2 mixes between them appropriately i.e. by choosing to play $D$ with probability $0.5$ and $E$ with probability $0.5$, it will leave player 1 with two best responses $A$ and $C$ and therefore, 1 can now choose to play $A$ and $C$ with equal probability. So, we've found a mixed strategy Nash Equilibrium in this case: Player 1 choose to play $A$ with probability 0.5 and $C$ with probability 0.5, and player 2 choose to play $D$ with probability 0.5 and $E$ with probability 0.5.


(iii) $B$ &amp; $C$

If player 1 play $B$ with probability $p$ and $C$ with probability $(1-p)$, where $0&lt;p&lt;1$, then player 2's expected payoff from playing 


$D$ is $4p+6(1-p) = 6 - 2p$
$E$ is $2p + 4(1-p) = 4 - 2p$
$F$ is $4p + 0(1-p) = 4p$


Since payoff from playing $D$ is more than the payoff from playing any other strategy for player 2, he will always choose $D$ in response to any mixed strategy of player 1 where 1 play $B$ with probability $p$ and $C$ with probability $(1-p)$. Therefore, $D$ is the best response of player 2 if 1 play a mixed strategy $B$ with probability $p$ and $C$ with probability $(1-p)$. However, given the strategy $D$ of player 2, player 1's best response is to play a pure strategy $B$. Therefore, there is no Nash equilibrium in which player 1 mixes between $B$ &amp; $C$.