Pure strategy nash equilibrium number of sheep

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Reference no: EM131441911

Question 1: There are two herders, each of whom is profit maximizing. Let qi 0 be the number of sheep chosen by herder i. Suppose that the per-sheep profit of grazing on the commons is v1 (q1, q2) = 120 - (q1 + q2) for herder 1 and is v2 (q1, q2) = 120 + k - (q1 + q2) for herder 2, where k 0. That is, different herders have different per-sheep profits (perhaps herder 2 is just better at herding) and the per-sheep profit of grazing sheep can be negative if enough sheep are grazing.

The herders choose their pure strategies simultaneously. Solve for the pure strategy Nash equilibrium number of sheep that each herder chooses to graze (as a function of k). HINT: Please make sure of two related things: (i) you don't give an answer in which either herder is herding negative sheep and (ii) you take into consideration the possibility that there might be an equilibrium in which at least one herder is herding zero sheep.

Question 2: Consider the following three stage game. There are two teams: Team 1 and Team 2. Whereas Team 1 has only one member, Team 2 is actually a team of two people: Y and Z. In the first stage, Team 1 chooses either Top or Bottom. In the second stage, players Y and Z on Team 2 flip a (potentially unfair) coin to determine which of them gets to play in stage

3. The probability that player Y plays in stage 3 is 0 < y < 1 and the probability that player Z gets to play in stage 3 is 1 - y. In the third stage, the player that was chosen by the coin flip in stage 2 chooses either Left or Right. Since teammates Y and Z have different "abilities" playing the game, the payoffs that Team 2 earns depends on who plays for Team 2. Specifically, consider the following payoff matrix (to the simultaneous move game)

 

Left

Right

Top

1, (1, 4)

4, (3, 3)

Bottom

0, (7, 2)

8, (5, 3)

where, if the payoffs are listed as a, (b, c) within a given box, this means that Team 1's payoff is a and Team 2's payoff is either b if player Y is playing or c if player Z is playing. For what value of 0 < y < 1, if any, would Team 1 be willing to strictly mix between Top and Bottom (in a SPNE)? If there is no value of 0 < y < 1 at which Team 1 would be willing to strictly mix between Top and Bottom (in a SPNE), what must Team 1 choose (Top or Bottom?) in a   SPNE?

Question 3: Consider the following three stage game. There are two teams: Team   1 and Team 2. Whereas Team 1 has only one member, Team 2 is actually a team of two people:

Y and Z. In the first stage, Team 1 chooses either Top or Bottom. In the second stage, players Y and Z on Team 2 flip a (potentially unfair) coin to determine which of them gets to play in stage 3. The probability that player Y plays in stage 3 is 0 < y < 1 and the probability that player Z gets to play in stage 3 is 1 -y. In the third stage, the player that was chosen by the coin flip in stage 2 chooses either Left or Right. Since teammates Y and Z have different "abilities" playing the game, the payoffs that both Team 1 and Team 2 earn depend on who plays for Team 2. Specifically, consider the following payoff matrix (to the simultaneous move  game)

 

Left

Right

Top

(1, 3) , (1, 4)

(4, 1) , (3, 3)

Bottom

(0, 2) , (7, 2)

(8, 2) , (5, 3)

where, if the payoffs are listed as (a, b) , (c, d) within a given box, this means that Team 1's payoff is a and Team 2's payoff is c if player Y is playing whereas Team 1's payoff is b and Team 2's payoff is d if player Z is playing. For what value of 0 < y < 1, if any, would Team 1 be willing  to strictly mix between Top and Bottom (in a SPNE)? If there is no value of 0 < y < 1 at which Team 1 would be willing to strictly mix between Top and Bottom (in a SPNE), what must Team 1 choose (Top or Bottom?)  in a SPNE?

Reference no: EM131441911

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