3 Should you compromise (Gintis/Schecter) Suppose that two friends (players 1 an

Question

3 Should you compromise (Gintis/Schecter) Suppose that two friends (players 1 and 2) must choose among three activities (A, B and C). Both players are neutral about activity B. Player 1 prefers A and dislikes C. Player 2 prefers C and dislikes A Each player independently chooses an activity with the following values (v> 0) the value of an activity the player prefers is v the value of the neutral activity B is 0. the value of the activity the player dislikes is -v In addition each activity involves a cost c >0 in the following way . if both players choose the same activity they both pay the cost c. . if a player is the only one to choose a particular activity then the cost is ac, where a > 1 The factor a represents includes a greater financial cost that may be incurred from doing the activity alone as well as a subjective cost of not having your friend with you This results in the following payoff matrix Player 2 A(-e,-v-(-ac,-ac) (v- ac, -ac) (-c,- ac, v-ac) C (-ac,--ac) -ac,-ac) -v-c,v -c) Player(-ac,--ac) 1. Suppose (a-1)c

Explanation / Answer

Accor ding to given data i can ability to solve the problems are given below

1. Given: (a-1)c<v

= ac - c < v

= ac < v + c or equivalently, -ac > -v - c or v - ac > -c ... (*)

Then, comparing the payoffs of player 1 from each strategy, by fixing player 2's strategy:

1) If player 2 chooses activity A, for player 1, since c,v,a are all positive, -v-ac< -ac ... (i)

Also, since a>1, ac>c or -ac < -c < v - c ... (ii)

From (i) and (ii), (v-c) is the highest payoff, thus, pl.1 chooses activity A.

2) If pl.2 chooses activity B, for pl.1, v-ac > -v-ac and from (*), v-ac > -c

thus, from above, again best response for pl.1 is activity A.

3) If pl. 2 chooses activity C, for pl. 1, since v>0, v-ac > - ac and from (*) -ac > -v-c

thuus, from above, again best response for pl.1 is activity A.

Hence, activity A is dominant strategy for pl. 1.

Now, since no matter what player 2 will choose, player 1 always chooses activity A. When pl. 1 chooses activity A best response for pl. 2 is:

v-ac> -ac > -v - c ... from (*) same as in case 3)

Making (A,C) as our required Nash equilibrium for (pl.1, pl.2).

(Also, C is dominant stratagy for player 2, which can be checked in the same way we checked for pl.1. So, (A,C) will be Nash equilibrium.)

2. Given: (a-1/2)c < v < (a-1)c

= (a-1)c < 2v and v < ac - c

= ac - c < 2v and.... -ac< -v - c or v - ac < -c... (**)

Again, comparing the payoffs

if pl.2 chooses A, for pl. 1, clearly v - c > -ac > -v- ac (with reasoning same as in case 1)) and pl. 1 chooses A

if pl.2 chooses B, for pl. 1, clearly v-ac > -v - ac and v-ac< -c (from (**)), thus pl.1 chooses B

if pl.2 chooses C, for pl.1, clearly v - ac > -ac and ac - c < 2v (from (**))

= v - ac > -v - c, thus, pl. 1 chooses A

Hence, pl. 1 either chooses A or B

if pl.1 chooses A, for pl. 2, v - ac > -ac and v - ac > -v-c (from (**) as above), thus pl.2 chooses C

if pl.1 chooses B, for pl. 2 -v - ac < v - ac and v - ac < -c, thus, pl.2 chooses B

Hence, our two nash equilibriums are as follows: (A,C) and (B,B) for (pl.1,pl.2).

Also, since v - ac< -c, for both player, (B,B) is best.

3. Given: v < (a-1/2)c

= 2v < (a - 1)c implying v < (a - 1)c (as v>0) ... (***)

On comparing the payoffs,

If pl.2 chooses A, pl.1 always chooses A. (for all cases above, with basic conditions given)

if pl.2 chooses B, for pl.1, v - ac > -v - ac and -c > v - ac (from (***)), thus, pl. 1 chooses B

if pl.2 chooses C, for pl.1, v - ac > -ac and v - ac < - v - c, thus pl.1 chooses C

Similarly, if pl.1 chooses A, for pl. 2, v - ac > -ac and v -ac < -v-c (from (***)), thus pl. 2 chooses A

if pl.1 chooses B, for pl.2,  v - ac > -v - ac and -c > v - ac (from (***)), thus, pl.2 chooses B

if pl.1 chooses C, pl.2 chooses C always (with above given properties as v,c>0 and a>1)

Hence, our Nash equilibriums are: (A,A), (B,B), and (C,C) for (pl.1,pl.2).

Since, v-c > -c > -v-c thus, (A,A) and (C,C) are best Nash equilbriums for pl.1 and pl.2, respectively, and (B,B) is second-best for both.

4. From above results, we can see activity A is always the best strategy for pl.1 and activity C is always the best strategy for pl.2. B is always the second best strategy for both the players. This is in accordance with the primary information provided in the question, that player 1 likes A, is neutral about B, and dislikes C while player 2 likes C, is neutral about B, and dislikes A for the given costs, and whatever the conditions are on the preferance values and costs. If you have any doubts just leave a message in comment box thank you

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