Question

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c). Washington Huskies versus Stanford Cardinal was one of the most wonderful games in the 2019-20 season. In football, the offense can either run the ball or pass the ball, and the defense can either anticipate and prepare for) a run or anticipate and prepare for) a pass. Assume that the expected payoffs (in yards) for the two teams on any given down are as follows. This game has no pure-strategy Nash equilibrium but one mixed-strategy Nash equilibrium in which Washington plays "Run" and "Pass" with probabilities 0.75 and 0.25. respectively, and Stanford plays "Anticipate Run" and "Anticipate Pass" with probabilities 0.5 and 0.5. respectively. Stanford (Defense) Anticipate Run Anticipate Pass Run 1 5 Washington (Offense) 1, - 9,-9 ,-5 -3,3 Pass False Please choose one (3 points): True Explain (7 points):

Answer 1

Consider the given game, here there are two teams each having two possible strategies. Now, if “Washington” chooses to “Run” then the optimum strategy of “Stainford” is “Anticipate Run”, because (-1) > (-5). If “Stainford” chooses to “Anticipated run” then the optimum strategy of “Washington” is “Pass”, because “9 > 1”. So, (Run, Anticipate Run) and (Run, Anticipate Pass) are not pure strategy NE.

Similarly, if “Washington” chooses to “Pass” then the optimum strategy of “Stainford” is “Anticipate Pass”, because 3 > (-9). If “Stainford” chooses to “Anticipated pass” then the optimum strategy of “Washington” is “Run”, because “5 > (-3)”. So, (Pass, Anticipate Run) and (Pass, Anticipate Pass) are not pure strategy NE. Here the game has no pure strategy NE.

Now, let’s assume “Stainford” choose “Anticipate Run” with probability 0.5 and “Anticipate Pass” with probability 0.5. So, the expected pay off of playing “Run” by Washington is given below.

=> E(R) = 1*0.5 + 5*0.5 = 3, => E(R) = 3.

The expected pay off of playing “Pass” by Washington is given below.

=> E(P) = 9*0.5 + (-3)*0.5 = 3, => E(P) = 3. Here the “E(R) = E(P)” implied at the mixed strategy the expected payoff of both the strategy are same.

Let’s assume “Washington” chooses “Run” with probability 0.75 and “Pass” with probability 0.25. So, the expected pay off of playing “Anticipated Run” by Stainford is given below.

=> E(AR) = (-1)*0.75 + (-9)*0.25 = (-3), => E(AR) = (-3).

The expected pay off of playing “Anticipated Pass” by Stainford is given below.

=> E(AP) = (-5)*0.75 + 3*0.25 = (-3), => E(AP) = (-3). Here also the “E(AR) = E(AP)”, implied at the mixed strategy the expected payoff of both the strategy are same.

So, the statement is TRUE.