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Stat 155 Spring 2019Homework 2: SolutionProblem 1(2.3 from Karlin and Peres)Find the value of the following zero-sum game and determine some optimal strategies for each of the players:834147160385Solution.•The solution itself:Let’s prove that the following strategies are optimal:x=31/9636/9629/96,y=4/1203/125/12.Let’s computexTAandAy:xTA=39296,43296,39296,39296PI gets at least39296=4912on average, no matter what PII does.Ay=4912,4912,4912TPII pays at most4912on average, no matter what PI doesThus,4912is the value of the game, and thex, yabove are optimal.•How to get the solution:We have to use proposition 2.5.3:First, one may find out that there are no saddle points or domination. Moreover, there is no fully equalizing strategyof the first player (which is not surprizing, as PI only has 3 strategies, but needs to equalize 4, so we get a system of 4linear equations with only 3 variables). That implies that PII can not use all 4 columns.Thus, we need to use proposition 2.5.3 (or, equivalently, the fact that optimal strategies are exactly Nash equilibria),which means that we need to understand which strategies are used with positive probabilities.Optimal strategies are not pure:Note that all the values in each row, and all the values in each column are different. Thus, if some player uses a purestrategy, then an optimal response of another player will also use only one strategy. So, if an optimal strategy of oneplayer is pure, then the optimal strategy of another player is also pure, which means that there exists a saddle point.Thus, since there are no saddle points, the optimal strategies can not be pure.PI must use all 3 strategies:Now let’s get a lower bound on the value of the game. Note that if PI plays all 3 rows equiprobably (uses the mixedstrategy(1/3,1/3,1/3)T, then the payments of PII arexTA=(4,413,413,4).We see that the value of the game is atleast4. Moreover, it is not hard to see that there is no strategy of PII of the form(y1,0,0, y4)Tthat equalizes al 3options of PI. Thus,x= (1/3,1/3,1/3)Tis not optimal, soV >4(the inequality is actually strict.)Using the bound from the previous paragraph, we can rule out all cases in which PI only uses 2 options. Indeed, if