# hw3.pdf - Stat 155 Spring 2017 Homework 3(2.3 Find the...

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Stat 155, Spring 2017Homework 3(2.3) Find the value of the following zero-sum game and determine some optimal strategies for eachof the players:834147160385ab, we see that there existsSolution:Using proposition 2.5.3, supposing thatx=1-a-bV1such that:8a+ 4bV13 + 4bV18-4a-7bV15-4a+bV1whereaandbare positive numbers such thata+b1.First, let us assume that none of Player 2’s strategies are inactive. Thus, all of those inequal-ities are in fact equalities. We then get8a+ 4b=V13 + 4b=V18-4a-7b=V15-4a+b=V1If we subtract the fourth equation from the third equation, we get 3-8b= 0, sob=38.Plugging this value into the second equation,we get 3 + 438=V1,orV1= 4.5. If we now setthese two values into the first equation, we get 8a+ 438= 4.5, which meansa=38. However,this creates a contradiction, as plugging each of our values into the fourth equation gives5-1.5 +38= 4.5, which is obviously false. Therefore, it is not the case that all of Player 2’sstrategies are inactive; we conclude that at least one strategy must be inactive.Next, we suppose that the first strategy is inactive. This means 8a+ 4bV1, while the otherthree remain as equalities.3 + 4b=V18-4a-7b=V15-4a+b=V1We can subtract the third equation from the second to get 3-8b= 0, orb=38. Puttingthis into the first equation gives 3 + 438=V1, orV1= 4.5. Solving foraat last, we arrive at-4a=-78, ora=732.2/17/2017
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Stat 155, Spring 2017Homework 3We must confirm that the inequality is also satisfied; we have 8a+ 4b=13492. Therefore,this inequality does not hold, and we have reached a contradiction. It cannot be the case thatonly the first strategy is inactive. We next try a strategy where the only inactive strategy isthe second:8a+ 4b=V18-4a-7b=V15-4a+b=V1Again, subtracting the third equation from the second gives usb=38. We then have 8a+1.5 =V1and 8-4a-218=V1.Setting these two equations equal to each other gives8a+ 1.5 = 8-4a-218, or 12a= 8-218-1.5 =318. Solving foragivesa=3196. Then, wegetV1=4912.We must confirm that the inequality is also satisfied; we have 3 + 4b=924912. Therefore,this satisfies our necessary conditions, and we can conclude that the value of this game is4912, the optimal strategy for Player 1 is31/963/829/96, and that Player 2 has an inactive secondstrategy. We can now consider the following matrix:841416085Suppose Player 2’s strategy is given bycd1-c-d. Since we already know that all three ofPlayer 1’s strategies are active, and that the value of the game is4912, we get:1 + 7c+ 3d=49126-2c-5d=49125-5c+ 3d=4912We can add all three equations together to get 12 +d=494, ord=14. Then, we getc=13.These two values satisfy all three equations; thus, the optimal strategy for Player 2 in theoriginal payoff matrix is1/301/45/12.2/17/2017
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Stat 155, Spring 2017Homework 3Thus, the value of this game is4912. Player 1’s optimal strategy is31/963/829/96, and Player 2’soptimal strategy is1/301/45/12.
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