In: Economics
Consider the following situation
Thief |
|||
Steal |
Don't Steal |
||
Police |
Patrol |
-10,-20 |
20,10 |
Don’t Patrol |
-20,30 |
30,20 |
a. There is no Nash equilibrium in pure strategies as there is no dominant strategy for any player, hence no strategy can be eliminated via iterative dominance, thus Nash doesn't exist in pure strategy.
Checking for mixed strategy
Assume, Prob(patrol) = q, Prob(no patrol) = 1-q
Prob(steal)= p and Prob(not steal) = 1-p
For Police to be indifferent between patrolling and not patrolling,
Payoff from patrolling = Payoff from not patrolling
-10*prob(steal) + 20*prob(no steal) = -20*prob(steal) + 30*prob(no steal)
-10p +20(1-p) = -20p +(1-p)30
p = 1/2
Hence prob(steal) = 1/2 = prob(don't steal)
Similarly, for thief to be indifferent between stealing and no stealing,
-20q + 30(1-q) = 10q + 20(1-q)
q = 1/4
thus, prob(patrol) = 1/4 and prob( don't patrol) = 3/4
Hence, assuming police is player 1 and theif as player 2
{1/4, 3/4} ; {1/2, 1/2} is a mixed strategy nash equilibrium of the above game.
b. If the police patrols and the thief steals,
the payoff will be -10,-20
Police gets negative utility of -10 from cost of patrolling and theif gets negative utility of -20 from getting caught.
c. In equilibrium each player has a mixed strategy, where each strategy is played with a certain probability calculated in part (a), as no single strategy is unconditionally better than other given the other player's strategy.
Police has a strategy of patrolling with probability 1/4 and not patrolling with probability 3/4, while theif has the strategy of stealing with probability 1/2 and not stealing with probability 1/2.
in part (b) Police is playing it's dominant strategy given the theif steals, however given police patrols, stealing is not theif's dominant strategy, hence there is no equilibrium in pure strategies.
Hope the answers helped you :)