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Q: game theory in microeconomics ( Answered ,   0 Comments )
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 Subject: game theory in microeconomics Category: Business and Money > Economics Asked by: vitaminc-ga List Price: \$10.00 Posted: 10 Oct 2002 20:43 PDT Expires: 09 Nov 2002 19:43 PST Question ID: 75124
 ```1) Consider the game (1,1 0,0 0,0) (0,0 2,2 2,2) (0,0 2,2 2,2) For what mixed strategies of Bob is the first row of Alice a best reply? For what mixed strategies of Alice is the second column a best reply for Bob? 2)Write a two-person game where each player has 4 pure strategies and where six rounds of iterated elimination of strictly dominated strategies yields a unique solution. is there anybody can help me as soon as possible(in 10 hours)? exam is coming!```
 ```I will try to answer rapidly due to your time constraints. 1) For notation, we suppose Bob can choose strategies s1, s2, and s3; and, without risk of confusion, we will call Alice's choices the same, s1,s2, and s3. Note that the payoffs are the same if Alice and Bob are interchanged, and note that the second and third rows are the same (as are the second and third columns). 1a) In a mixed strategy, Bob and Alice can choose to play moves with a different probability. Suppose Bob chooses strategies s1, s2, and s3 with probabilities p1,p2 and p3. If Alice chooses strategy 1, her payoff is then p1*1 + p2*0 + p3 *0 = p1 This is the DEFINITION of payoff in a mixed strategy game, and it's also common sense. Alice only gets payoff of 1 when Bob chooses s1, which he only does with probability p1, so on average Alice gets a payoff of p1 in this case. If Alice chooses strategy s2, her payoff is then p1*0+p2*2+p3*2 = 2(p2+p3)=2(1-p1) By symmetry, her payoff is the same for strategy s3. Hence, Alice chooses strategy 1 as a best strategy if and only if: p1>=2(1-p1) 3p1>=2 p1>=2/3 Therefore, Alice choosing strategy 1, the first row, is a best strategy for Alice if and only if the probability that Bob chooses strategy 1 in his mixed strategy is greater than or equal to 2/3. 1b) By symmetry considerations (since the second and third choices are the same, and Bob and Alice have the same payoff functions), the second column is a best reply for Bob if and only if the first column is not a best reply, since the second column is a best reply if and only if the third column is also a best reply. Hence, the second column is a best reply for Bob if and only if q1<=2/3, where q1 is the probability that Alice chooses strategy 1 in her mixed strategy. (If it's clearer one can work through the math as in part a or I can clarify if necessary). 2) An appropriate normal form matrix satisfying the conditions is: Bob (1,1) (2,2) (3,3) (4,4) Alice (2,2) (3,3) (4,4) (5,5) (3,3) (4,4) (5,5) (6,6) (4,4) (5,5) (6,6) (7,7) At the first round, Alice eliminates strategy 1 (row 1), since it is strictly dominated. At the second round, Bob eliminates his first column, since it is strictly dominated - that is, Bob can always do better by playing to another columns. At the third round, Alice eliminates strategy 2, since the second row is strictly dominated by the third and fourth rows - Alice can always do better by playing in the 3rd and 4th rows. At the fourth round, Bob eliminates strategy 2, his second columns. At the fifth round, Alice eliminates strategy 3, her third row. At the sixth round, Bob eliminates column 3, his third strategy. Note that the order of elimination in a particular elimination might be different. Remaining is the strategy profiles (row4, column4) with payoff of 7,7. This is the only Nash equilibrium for the system. Search Strategy -------------- For this problem I continued to use the excellent book Game Theory Evolving, by Herbert Gintis, Princeton, 2000.``` Clarification of Answer by rbnn-ga on 11 Oct 2002 01:50 PDT ```I was thinking more about problem 2) and I believe I have come up with more interesting example for you. The example I gave was a correct solution, but aesthetically it had certain disadvantages: there were many choices of strictly dominating strategy to use at each step. A natural question to ask is: can we find a payoff matrix such that there is only a SINGLE choice of strictly dominating strategy to remove at each step in the iterative removal of dominating strategies? I found an example; however, it is of course a lot harder to see that it is correct than the example I gave. Consider the payoff matrix; 1,4 1,3 1,3 1,3 4,0 2,4 2,3 2,2 4,0 5,2 3,4 3,3 4,0 4,1 4,3 4,4 There are no strictly dominated columns and only the first row is strictly dominated. So after round 1, Alice play on the first row is dominated and we can cross it out to reach the matrix: 4,0 2,4 2,3 2,2 4,0 5,2 3,4 3,3 4,0 4,1 4,3 4,4 Here there are no dominated rows and only the first column is strictly dominated, so Bob's first column play is strictly dominated and can be removed, and we get the matrix: 2,4 2,3 2,2 5,2 3,4 3,3 4,1 4,3 4,4 Now again, there are no dominated columns and the top row is strictly dominated by the other rows so we remove it to get: 5,2 3,4 3,3 4,1 4,3 4,4 Now the first column only is strictly dominated so we remove that to get: 3,4 3,3 4,3 4,4 Now only the first row is strictly dominated, and then only the first column so we get the Nash equilibrium (4,4) That is the ONLY allowed order of removal is Alice strategy 1, Bob strategy 1, Alice strategy 2, Bob strategy 2, Alice Strategy 3, Bob Strategy 3. I found this matrix through playing around with the payoff matrix values. Again, this clarification is not strictly speaking necessary for problem 2, I just thought it was a more interesting example```
 vitaminc-ga rated this answer: ```yup, the explanation will be fine with the first question, coz i solved it in the same way by myself. no need to clarify. Otherwise, thanks for helping. u r the expert!```