# Nash Equilibria in Target Destroying-Guarding Game

Army A has a single plane which can strike one of three possible targets, A, B and C. Army B has one anti-aircraft gun that can be assigned to one of the three targets to guard it. The value of each target, $$v_k$$ is $$v_A>v_B>v_C>1$$. Army A can destroy the target only when it is unguarded and A attacks. Army A wishes to maximize the damage whilst Army B wishes to minimize the damage. Find all Nash Equilibria.

I tried formulating the game into a $$3*3$$ payoff matrix game, by giving Army A $$0$$ if the target it aimed at was guarded and the value of the target if it was unguarded. Similarly, for payoffs of Army 2, I assigned $$-v_k$$ added to the valuations of other two target which were not destroyed, if Army A succeeded in destroying the target and $$v_A+v_B+v_C$$ if Army A failed to destroy any target, i.e., by aiming at a guarded post. Given this setup, I found that there was no Pure Strategy Nash Equilibria in the game. I do not know how to proceed with the Mixed Strategy Nash Equilibria, if any.

I'm very doubtful of the approach that I tried to employ. I would really appreciate a little help!

Excuse the formatting for the game's normal form. This is a zero sum game, so I've written only the payoffs for the attacker:

$$\begin{pmatrix} A|D & a & b & c\\ \hline a & 0 & v_{A} & v_{A}\\ b & v_{B} & 0 & v_{B}\\ c & v_{C} & v_{C} & 0 \end{pmatrix}$$

Evidently, since $$v_{A} > v_{B} > v_{C}$$, there may be payoffs such that there exists some $$\lambda$$ such that $$(1-\lambda)v_{B} > v_{C}$$ and $$\lambda v_{A} > v_{C}$$.

First, suppose that the parameters are such that there exists such a $$\lambda$$ e.g. $$v_{A} = 20$$, $$v_{B} = 10$$ and $$v_{C} = 1$$. In this case, $$c$$ is strictly dominated and so we can eliminate $$c$$ from the support of the attacker's mixed strategy, and hence we may also eliminate $$c$$ from the support of the defender's mixed strategy. Write the new matrix as

$$\begin{pmatrix} A|D & a & b\\ \hline a & 0 & v_{A}\\ b & v_{B} & 0\\ \end{pmatrix}$$

Hence, the unique mixed strategy equilibrium is $$p := \Pr(a|A) = \frac{v_{B}}{v_{A}+v_{B}}$$ and $$q := \Pr(a|D) = \frac{v_{A}}{v_{A}+v_{B}}$$. Note that as the value of target $$A$$ increases, the attacker will actually attack it more infrequently.

If, $$c$$ is not dominated (strict or weak) then the algebra becomes more tedious, but the unique equilibrium will be a mixed strategy with full support over all three pure strategies. I leave this to you.

Finally, I leave to you the case in which $$c$$ is only weakly dominated. Hint: In general there can be weakly dominated strategies in support of a mixed strategy equilibrium, but if the opponent's mixed strategy has full support, there cannot be.