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Arturiano [62]
4 years ago
5

Player II is moving an important item in one of three cars, labeled 1, 2, and 3. Player I will drop a bomb on one of the cars of

her choosing. She no chance of destroying the item bombs the wrong car. If she chooses the right car, then her probability of destroying the item depends on that car. The probabilities for cars 1, 2, and 3 are equal to 3/4, 1/4, and 1/2. write the 3 times 3 payoff matrix for the game, and find an optimal strategy for each player.
Mathematics
1 answer:
Serhud [2]4 years ago
5 0

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.

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