# Homework3Solution.pdf - Stat 155 Spring 2019 Homework 3...

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Stat 155 Spring 2019Homework 3: SolutionLet us denote by1the vector that only contains ones. Since we do not expect this to lead to ambiguitywe will in general not specify the dimension of this vector.Solution to Problem 1LetA=834147160385.Let moreoverajdenote thejth column ofAandaidenote theith row ofA. We denote strategies for playerI, respectively player II, byxandy.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=39296,43296,39296,39296— PI gets at least39296=4912on average, no matter what PII does.Ay=4912,4912,4912T— PII pays at most4912on average, no matter what PI does.Thus,4912is the value of the game, and thex, yabove are optimal, by proposition 2.5.3.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, since the rank of thematrix is at most 3, we know that there is an optimal strategy for player II that only uses 3 columns(see piazza). Now we need to check which one of the four actions player II does not use. When playerII does not use actioni,xTaineeds to be larger than the value, whilexTajneeds to be equal to thevalue for all other actionsj. So we first try to solvexTa1VxTa2=VxTa3=VxTa4=VxT1= 1.(1)I you do the calculations, you will see that this leads to a contradiction.Now we try to do the same thing for the other actions of player II and we will see that fori= 2, thissystem of linear inequalities does not lead to a contradiction and gives ˜x=(3196,3696,2996)andV=4912.1
To check whether this is an optimal strategy we now need to solvea1y=4912a2y=4912a3y=4912yT1= 1Note that we need to have equalities here since the potential strategy for player I that we found earliergives a positive probability to each of his actions. Recall that we are trying to find an optimal strategywithy2= 0, so you can use this when solving this system of linear equations. Solving this system oflinear equations gives ˜yT= (412,0,312,512).To make sure that we did not make any mistakes on the way we compute the payoff vectors for thesestrategies. This gives:˜xA=39296432963929639296andA˜y=491249124912.