Question
Are there any NE in which player 1 randomizes (mixes) between all three actions? If yes, characterize all such equilibria. If no, explain why not.
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I have solved this question by utilizing this answer: how to find mixed Nash equilibria for 3×3?
I found that
For player 1 to be indifferent between T and B
$$4q_L+6q_M+2(1-q_L-q_M)=2+2q_L+4q_M$$
$$6q_L+2q_M+0(1-q_L-q_M)=6q_L+2q_M$$
They should be equal
So I get that $$q_M-2q_L=-1$$
For player 1 to be indifferent between C and B
$$2q_L+8q_M+0(1-q_L-q_M)=2q_L+8q_M$$
So they are equal as well
$$2q_L+8q_M=6q_L+2q_M$$
$$3q_M=2q_L$$
When I combine these two results
$$3q_M=2q_L$$
$$q_M-2q_L=1$$
I obtain $q_M=1/2$ an $q_M=3/4$ But $q_R=-1/4$
So I can say that there is no such a mixed Nash equilibrium.
What I understand that the question ask that player 1 randomizes (mixes) between all three actions.
Did I solve for player 1 or for player 2 in this way? Morespecifically, Could I choose right probability and payoff values from the given table?
Also, my solution is correct?
Best Answer
You made a sign error in deriving the equilibrium for $T$ and $B$. $2+2q_L+4q_M=6q_L+2q_M$ yields $q_M-2q_L=-1$. If you use that together with $3q_M=2q_L$ you get $q_L,q_M\in[0,1]$ but $q_R=1-q_L-q_M\lt0$, so your conclusion is nevertheless correct.
If you'd found a solution with $q_L,q_M,q_R\in[0,1]$, that would yield a strategy for Player $2$ that would lead Player $1$ to be indifferent among all her options. This is a necessary condition for a Nash equilibrium in which Player $1$ mixes all her strategies (if she weren't indifferent among all of them, she could improve her payoff by changing the mixture), but not yet sufficient. For this to correspond to a Nash equilibrium, Player $2$ must also be indifferent among all options for which the probability is in the interior $(0,1)$ of the interval $[0,1]$. For those options for which the probability is on the boundary $\{0,1\}$, Player $2$ need not be indifferent, but the sign of the derivative must be such that he could only profit by pushing the probability beyond the boundary. Only if there is a strategy for Player $1$ that creates these conditions for Player $2$ do you have a full Nash equilibrium. In particular, in the case $q_L,q_M,q_R\in(0,1)$, where all three probabilities are in the interior, you have to do the same calculation that you did again with roles reversed to show that there is a strategy for Player $1$ that makes Player $2$ indifferent among all three options.