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.


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