Mixed strategy NE solving leading to invalid equations

Question

This is a small scenario that I have formulated and want to solve for the mixed strategy NE.

Players A and B take actions T or F, and the payoffs are as shown. When I try to find the mixed strategy NE and assume that utility of B when he takes either T or F should be same, I get a wrong equation.

Assuming A takes T with probability q and F with probability (1-q), I get for B to have equal payoff in choosing T or F -

q + 1/2(1-q) = q/2 + 0 => q/2 + 0.5 = q/2, which shouldn't be possible. Where is the mistake happening ? Is there any restrictions on the payoffs I can have or the NE doesn't exist in this case ?

Answer 1

In a Nash equilibrium, a player has to be indifferent over all the strategies they are randomizing over. So what you have shown is that there is no Nash equilibrium in which player B randomizes over their two actions properly. Indeed, T strictly dominates F for player B, so player B will always choose T.

Answer 2

Let $p \in [0,1]$ denote the probability that player $A$ plays $T$ and $1-p$ that he plays $F$. Let $q \in [0,1]$ denote the probability that player $B$ plays $T$ and $1-q$ that she plays $F$. The expected payoff of $A$ is given by: \begin{align} F_A(p,q) =& 0 \cdot p \cdot q + \frac{1}{2} \cdot p \cdot (1-q) + \frac{1}{2} \cdot (1-p) \cdot q + 1 \cdot (1-p) \cdot (1-q)\\ =&1 - \frac{p}{2} - \frac{q}{2}. \end{align} Since $\partial F_A(p,q)/\partial p = -1/2 < 0$ we get $0 = \arg\max_{p \in [0,1]} F_A(p,q)$. Idea: $F$ is a dominant strategy for $A$. Same logic for $B$.