Create a table
A/BA1A2A3B1−555B210−10−20B320−10−20
p1 - probability of choosing A1
p2 - probability of choosing A2
q1 - probability of choosing B1
q2 - probability of choosing B2
−5p1+5p2=10p1−10p2 or
p1=p2
p1+p2=1
Therefore, p1=0.5 & p2=0.5
The same we do with q1 & q2
−5q1+10q2=5q1−10q2 or
q1=2q2
q1+q2=1
Therefore, q2=1/3 & q1=2/3
First value of rows is the choice of player A. The first value of columns is the choice of player B. All others elements are values of winner(and the money he won).
Now we see that the best strategy for A is to choose 5p, thus he wins in 2 of 3 cases ( win - 0.6(6), win10 - 0,3(3), win20 - 0,3(3))
For player B the best strategy is to choose 10p, as when he chooses 10 or 20 he wins twice. But at the same time, if he loses it is better to lose 10p instead of 20p (win - 0.6(6), win10 - 0,3(3), win20 - 0,3(3))
Value of the game = [a11a22−a21a12]/[(a11+a22−(a12+a21))]=
=[(−5)∗(−10)−(5)∗(10)]/[−5+(−10)−10+5]=0
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