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Suppose in [link] , Robert decides to show a dime with . 20 size 12{ "." "20"} {} probability and a quarter with . 80 size 12{ "." "80"} {} probability, and Carol decides to show a dime with . 70 size 12{ "." "70"} {} probability and a quarter with . 30 size 12{ "." "30"} {} probability. What is the expected payoff for Robert?

Let R size 12{R} {} denote Robert's strategy and C size 12{C} {} denote Carol's strategy.

Since Robert is a row player and Carol is a column player, their strategies are written as follows:

R = . 20 . 80 size 12{R= left [ matrix { "." "20" {} # "." "80"{}} right ]} {} and C = . 70 . 30 size 12{C= left [ matrix { "." "70" {} ##"." "30" } right ]} {} .

To find the expected payoff, we use the following reasoning.

Since Robert chooses to play row 1 with . 20 size 12{ "." "20"} {} probability and Carol chooses to play column 1 with . 70 size 12{ "." "70"} {} probability, the move row 1, column 1 will be chosen with . 20 . 70 = . 14 size 12{ left ( "." "20" right ) left ( "." "70" right )= "." "14"} {} probability. The fact that this move has a payoff of 10 cents for Robert, Robert's expected payoff for this move is . 14 . 10 = . 014 size 12{ left ( "." "14" right ) left ( "." "10" right )= "." "014"} {} cents. Similarly, we compute Robert's expected payoffs for the other cases. The table below lists expected payoffs for all four cases.

Move Probability Payoff Expected Payoff
Row 1, Column 1 . 20 . 70 = . 14 size 12{ left ( "." "20" right ) left ( "." "70" right )= "." "14"} {} 10 cents 1.4 cents
Row 1, Column 2 . 20 . 30 = . 06 size 12{ left ( "." "20" right ) left ( "." "30" right )= "." "06"} {} -10 cents -.6 cents
Row 2, Column 1 . 80 . 70 = . 56 size 12{ left ( "." "80" right ) left ( "." "70" right )= "." "56"} {} -25 cents -14 cents
Row 2, Column 2 . 80 . 30 = . 24 size 12{ left ( "." "80" right ) left ( "." "30" right )= "." "24"} {} 25 cents 6.0 cents
Totals 1 -7.2 cents

The above table shows that if Robert plays the game with the strategy R = . 20 . 80 size 12{R= left [ matrix { "." "20" {} # "." "80"{}} right ]} {} and Carol plays with the strategy C = . 70 . 30 size 12{C= left [ matrix { "." "70" {} ##"." "30" } right ]} {} , Robert can expect to lose 7.2 cents for every game.

Alternatively, if we call the game matrix G size 12{G} {} , then the expected payoff for the row player can be determined by multiplying matrices R size 12{R} {} , G size 12{G} {} and C size 12{C} {} . Thus, the expected payoff P size 12{P} {} for Robert is as follows:

P = RCG P = . 20 . 80 10 10 25 25 . 70 . 30   = 7 . 2   cents size 12{ matrix { P= ital "RCG" {} ##P= left [ matrix { "." "20" {} # "." "80"{}} right ] left [ matrix {"10" {} # - "10" {} ## - "25" {} # "25"{}} right ] left [ matrix {"." "70" {} ## "." "30"} right ] {} ##= - 7 "." 2" cents" } } {}

Which is the same as the one obtained from the table.

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For the following game matrix G size 12{G} {} , determine the optimal strategy for both the row player and the column player, and find the value of the game.

G = 1 2 3 4 size 12{G= left [ matrix { 1 {} # - 2 {} ##- 3 {} # 4{} } right ]} {}

Let us suppose that the row player uses the strategy R = r 1 r size 12{R= left [ matrix { r {} # 1 - r{}} right ]} {} . Now if the column player plays column 1, the expected payoff P size 12{P} {} for the row player is

P r = 1 r + 3 1 r = 4r 3 size 12{P left (r right )=1 left (r right )+ left ( - 3 right ) left (1 - r right )=4r - 3} {} .

Which can also be computed as follows:

P r = r 1 r 1 3 size 12{P left (r right )= left [ matrix { r {} # 1 - r{}} right ] left [ matrix {1 {} ## - 3} right ]} {} or 4r 3 size 12{4r - 3} {} .

If the row player plays the strategy r 1 r size 12{ left [ matrix { r {} # 1 - r{}} right ]} {} and the column player plays column 2, the expected payoff P size 12{P} {} for the row player is

P r = r 1 r 2 4 = 6r + 4 size 12{P left (r right )= left [ matrix { r {} # 1 - r{}} right ] left [ matrix {- 2 {} ## 4} right ]= - 6r+4} {} .

We have two equations

P r = 4r 3 size 12{P left (r right )=4r - 3} {} and P r = 6r + 4 size 12{P left (r right )= - 6r+4} {}

The row player is trying to improve upon his worst scenario, and that only happens when the two lines intersect. Any point other than the point of intersection will not result in optimal strategy as one of the expectations will fall short.

Solving for r size 12{r} {} algebraically, we get

4r 3 = 6r + 4 size 12{4r - 3= - 6r+4} {}

r = 7 / 10 size 12{r=7/"10"} {} .

Therefore, the optimal strategy for the row player is . 7 . 3 size 12{ left [ matrix { "." 7 {} # "." 3{}} right ]} {} .

Alternatively, we can find the optimal strategy for the row player by, first, multiplying the row matrix with the game matrix as shown below.

r 1 r 1 2 3 4 = 4r 3 6r + 4 size 12{ left [ matrix { r {} # 1 - r{}} right ] left [ matrix {1 {} # - 2 {} ## - 3 {} # 4{}} right ]= left [ matrix {4r - 3 {} # - 6r+4{} } right ]} {}

And then by equating the two entries in the product matrix. Again, we get r = . 7 size 12{r= "." 7} {} , which gives us the optimal strategy . 7 . 3 size 12{ left [ matrix { "." 7 {} # "." 3{}} right ]} {} .

We use the same technique to find the optimal strategy for the column player.

Suppose the column player's optimal strategy is represented by c 1 c size 12{ left [ matrix { c {} ##1 - c } right ]} {} . We, first, multiply the game matrix by the column matrix as shown below.

1 2 3 4 c 1 c = 3c 2 7c + 4 size 12{ left [ matrix { 1 {} # - 2 {} ##- 3 {} # 4{} } right ]left [ matrix { c {} ##1 - c } right ]= left [ matrix { 3c - 2 {} ##- 7c+4 } right ]} {}

And then equate the entries in the product matrix. We get

3c 2 = 7c + 4 size 12{3c - 2= - 7c+4} {}
c = . 6 size 12{c= "." 6} {}

Therefore, the column player's optimal strategy is . 6 . 4 size 12{ left [ matrix { "." 6 {} ##"." 4 } right ]} {} .

To find the expected value, V size 12{V} {} , of the game, we find the product of the matrices R size 12{R} {} , G size 12{G} {} and C size 12{C} {} .

V = . 7 . 3 1 2 3 4 . 6 . 4 . 2 size 12{ matrix { V= left [ matrix {"." 7 {} # "." 3{} } right ]left [ matrix { 1 {} # - 2 {} ##- 3 {} # 4{} } right ]left [ matrix { "." 6 {} ##"." 4 } right ]{} ## = - "." 2} } {}

That is, if both players play their optimal strategies, the row player can expect to lose . 2 size 12{ "." 2} {} units for every game.

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Source:  OpenStax, Applied finite mathematics. OpenStax CNX. Jul 16, 2011 Download for free at http://cnx.org/content/col10613/1.5
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