When I use arbitrary numbers for the payoff function for Player 2, I get 2 NEPS, but am not certain about this approach.
Consider the game with two players “Player 1” and “Player 2”. Now, “P1” have two strategies “UP” and “DOWN” and “P2” have two strategies “LEFT” and “RIGHT”. Now, we have given that “P1’s” payoff function is such that for any combination of the player’s chosen strategies, “P1” always receive payoff equal to “0” and “P2’s” payoff function is such that no two combination of the player’s chosen strategies ever give “P2” the same payoff, => the payoff matrix is given by.
Now, if “P1” choses “UP” then “P2” will chose either “LEFT” or “RIGHT” depending on “a” or “b” which one is greater. Similarly, if “P1” choses “DOWN” then “P2” will chose either “LEFT” or “RIGHT” depending on “c” or “d” which one is greater and “P1” will always get “0”, => is completely independent of any strategies. SO, there are exactly “2 pure strategy NE”
So, here “2nd” be the correct answer.
When I use arbitrary numbers for the payoff function for Player 2, I get 2 NEPS,...
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Every time I have plugged in the values, I keep getting only 1 NEPS. I have done the math twice. 4) Suppose that r is a non-zero real number. In which case does the following game have exactly two NEPS? Player 2 Player 1
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