Air Strike Game This is an Air Strike Game with the solution, I have added some questions regarding the solution and I would appreciate if someone could answer them.
Army $A$ has a single plane with which it can strike one of three possible targets. Army $B$ has one anti-aircraft gun that can be assigned to one of the targets. The value of target $k$ is $v_k$, with $v_1 > v_2 > v_3 > 0$. Army $A$ can destroy a target only if the target is undefended and $A$ attacks it. Army $A$ wishes to maximize the expected value of the damage and army $B$ wishes to minimize it. Formulate the situation as a (strictly competitive) strategic game and find its mixed strategy Nash equilibria.
The solution is take from Solution Manual for a course of Game Theory by Martin Osborne.
Let $(p^∗ , q^∗ )$ be a mixed strategy equilibrium. 
Step 1. If $p^∗_i = 0$ then $q^*_i = 0$ (otherwise $q^*$ is not a best response to $p^*$ ); but if $q^*_i = 0$ and $i≤2$ then $p_{i+1} = 0$ (since player $i$ can achieve $v_i$ by choosing $i$). 

Q: may be there is a typo here and instead of $p_{i+1} = 0$ should be written $p^*_{i+1} = 0$, I am not sure, for me $p_{i+1} = 0$ doesn't make sense. On the other hand, defender knowning that $p^*_{i+1}=0$ has incentive to increase $q^*_i$ because chances to attack $i$ target with $p^*_{i+1} =0$ actually increased.

Thus if for $i ≤ 2$ target $i$ is not attacked then target $i + 1$ is not attacked either.
Step 2. $p^∗ \neq (1, 0, 0)$ it is not the case that only target 1 is attacked.

Q: why it is actually so, in general I think there is no pure strategies at all, because, knowing that $p^∗ = (1, 0, 0)$ second player will play $q^∗ = (1, 0, 0)$, so not much sense in pure strategies. On the other hand, the decisions are made simultaneously, therefore it might be possible to play pure strategy. Why it so?

Step 3. The remaining possibilities are that only targets 1 and 2 are attacked or all three targets are attacked.
• If only targets $1$ and $2$ are attacked the requirement that the players be indifferent between the strategies that they use with positive probability implies that $p^∗ =(v_2/(v_1+v_2), v_1/(v_1+v_2), 0)$ and $q^∗=(v_1/(v_1+v_2),v_2/(v1 + v2 ),0)$. Thus the expected payoff of Army A is $v_1 v_2 /(v_1 + v_2 )$. Hence this is an equilibrium if $v_3 ≤ v_1 v_2 /(v_1 + v_2 )$.
• If all three targets are attacked the indifference conditions imply that the probabilities of attack are in the proportions $v_2 v_3 : v_1 v_3 : v_1 v_2$ and the probabilities of defense are in the proportions $z − 2v_2 v_3 : z − 2v_3 v_1 : z − 2v_1 v_2$ where $z = v_1 v_2 + v_2 v_3 + v_3 v_1$ . For an equilibrium we need these three proportions to be nonnegative, which is equivalent to $z − 2v_1 v_2 ≥ 0$, or $v_3 ≥ v_1 v_2 /(v_1 + v_2 ).$

Q: I cannot derive proportions from the Step 2.

 A: Regarding Step 1 you are correct that it should be $p_{i+1}^*=0$. The reasoning is that if $q_i=0$ then whenever $A$ is tempted to play $i+1$ he should play $i$ instead, since whenever he plays $i$ he will certainly gain the value $v_i$, which is better than gaining $v_{i+1}$. 
Re Step 2: If $p^*=(1,0,0)$ then the best reply for $B$ is $q^*=(1,0,0)$. But with these strategies player $A$ never destroys anything and could do better by changing strategy to $(0,1,0)$. Hence we can not have that $A$ plays $(1,0,0)$ in equilibrium.
Re the last step: Let us consider if there is an equilibrium where $A$ attacks target 1 and 2 only with positive probability. Clearly $q_3^*=0$, in that case. A standard fact from game theory tells us that $A$ in equilibrium must be getting the same expected payoff from playing 1 and 2. (If you are not familiar with this I recommend you reread the chapter in your textbook where this is discussed.) The payoff from playing 1 for $A$ is 
$ v_1q(1-q_1^*) = v_1q_2^*$. The payoff from playing 2 is $v_2 q_1^*$. Thus we must have $$ v_1 (1-q_1^*) = v_2 q_1^*.$$ This equation can be solved to give $$q_1^*=\frac{v_1}{v_1+v_2}.$$ Hence $$q_2^* = 1-q_1^*= \frac{v_2}{v_1+v_2}.$$
The values of $p_1^*$ and $p_2^*$ can be found in a similar way, and the case where all three targets are attacked can also be handled in a similar way. 
