Answer:
There are 3 strategies for Player I - i) throw bomb on car 1 ii) i) throw bomb on car 2 iii) i) throw bomb on car 3
Similarly, there are 3 strategies for Player II - i) put item on car 1 ii) put item on car 2 iii) put item on car 3
The payoff matrix is \begin{pmatrix} 3/4 &0 &0 \\ 0&1/4 &0 \\ 0&0 &1/2 \end{pmatrix}
Row 1, Column 1 : If the player I drops bomb on car 1 and the item is in the car 1 then the probability of destroying the item is 3/4.
Row 1, Column 2 or Row 1, Column 3: If the player I drops bomb on car 1 and the item is in the car 2 or 3 then the probability of destroying the item is 0.
Row 2, Column 2 : If the player I drops bomb on car 2 and the item is in the car 2 then the probability of destroying the item is 1/4.
Row 2, Column 1 or Row 2, Column 3: If the player I drops bomb on car 2 and the item is in the car 1 or 3 then the probability of destroying the item is 0.
Row 3, Column 3 : If the player I drops bomb on car 3 and the item is in the car 3 then the probability of destroying the item is 1/2.
Row 3, Column 1 or Row 3, Column 2: If the player I drops bomb on car 3 and the item is in the car 1 or 2 then the probability of destroying the item is 0.
Finding the optimal strategy for both the players.
Let p = (p1, p2, p3) be the optimal strategy for the player I and q = (q1, q2, q3) be the optimal strategy for the player II where p1,p2,p3 are the probabilities of player I dropping the bomb on car 1,2 or 3. And q1,q1,q3 are the probabilities of player II putting items in car 1,2 or 3.
Let V be the value of the game.
Solving the equations derived from the pay off matrix , 3/4p1 = V,
1/4p2 = V,
1/2p3=V
and p1+p2+p3=1
p1 = 4/3V, p2=4V, p3=2V
4/3V + 4V + 2V = 1
or, V = 3/22
so, p1 = 4/3V = 2/11, p2 = 4V = 6/11, p3=2V = 3/11
Similarly, we solve for q = (q1, q2, q3), 3/4q1 = V,
1/4q2 = V,
1/2q3=V
and q1+q2+q3=1
we get q1 = 4/3V = 2/11, q2 = 4V = 6/11, q3=2V = 3/11
So, the optimal strategies for both the players are
p = (2/11,6/11,3/11) and q= (2/11,6/11,3/11 and
the value of the game is 3/22.