Interpreting the results of a 3x3 Nash Mixed Strategies Equilibrium
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I've got a 3x3 Nash game cube I'm trying to interpret. The decimals below represent the number of times a player 'wins' against the other player.
A B C
A 0.722,0.278 0.722, 0.278 0.800, 0.200
B 0.750, 0.250 0.780, 0.220 0.875, 0.125
C 0.286, 0.714 0.200, 0.800 1.000, 0.000
When I run this through the method for solving 3x3 Nash squares, this is the equation I use:
A->B: (p+q+(1-p-q) = p1 + q1 + (1-p1-q1))
B->C: (p1+q1+(1-p1-q1) = p2 + q2 + (1-p2-q2))
and I substitute p from one into the other to find q.
The result I get for the horizontal player, from the above square however, don't make sense:
Horizontal Player
A = 0.07460291360037259
B = 0.18149088118987192
C = 0.7439062052097555
This would indicate that the horizontal player should play 75% of the time. Yet, strategy B clearly produces better results for the horizontal player consistently. Am I interpreting this correctly? It is possible I made a programming error, but the Vertical Player makes sense:
Vertical Player
A = 0.14930944382232172
B = 1.1347517730496453
C = -0.28406121687196695
It's gobbledy gook. But just visually analyzing the square indicates that B is the best strategy.
Any feedback you can provide is much appreciated.
nash-equilibrium
asked Dec 25 '18 at 0:48
Jess BehrensJess Behrens
1
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add a comment |
$begingroup$
I've got a 3x3 Nash game cube I'm trying to interpret. The decimals below represent the number of times a player 'wins' against the other player.
A B C
A 0.722,0.278 0.722, 0.278 0.800, 0.200
B 0.750, 0.250 0.780, 0.220 0.875, 0.125
C 0.286, 0.714 0.200, 0.800 1.000, 0.000
When I run this through the method for solving 3x3 Nash squares, this is the equation I use:
A->B: (p+q+(1-p-q) = p1 + q1 + (1-p1-q1))
B->C: (p1+q1+(1-p1-q1) = p2 + q2 + (1-p2-q2))
and I substitute p from one into the other to find q.
The result I get for the horizontal player, from the above square however, don't make sense:
Horizontal Player
A = 0.07460291360037259
B = 0.18149088118987192
C = 0.7439062052097555
This would indicate that the horizontal player should play 75% of the time. Yet, strategy B clearly produces better results for the horizontal player consistently. Am I interpreting this correctly? It is possible I made a programming error, but the Vertical Player makes sense:
Vertical Player
A = 0.14930944382232172
B = 1.1347517730496453
C = -0.28406121687196695
It's gobbledy gook. But just visually analyzing the square indicates that B is the best strategy.
Any feedback you can provide is much appreciated.
nash-equilibrium
asked Dec 25 '18 at 0:48
Jess BehrensJess Behrens
1
$endgroup$
$begingroup$
Doing the expected value calculation can sometimes produce negative "probabilities" if the game has a pure strategy equilibrium. Your game has one at B,A since the payoff to left player is the highest in the column, and the payoff to either player is highest in the row.
$endgroup$
– dbx
Dec 25 '18 at 3:18
$begingroup$
Thanks! That makes a lot of sense, actually. These are strategies and C is the weakest. When I apply the rule you described to all of the different groupings of the data, the pure strategies most often involve C because the other two essentially take advantage of it. Is there anything to be gleaned from the fact that the row player produces a mixed strategies equilibrium? Or, does that only have meaning if the column player also has a mixed strategy?
$endgroup$
– Jess Behrens
Dec 25 '18 at 6:47
add a comment |
$begingroup$
I've got a 3x3 Nash game cube I'm trying to interpret. The decimals below represent the number of times a player 'wins' against the other player.
A B C
A 0.722,0.278 0.722, 0.278 0.800, 0.200
B 0.750, 0.250 0.780, 0.220 0.875, 0.125
C 0.286, 0.714 0.200, 0.800 1.000, 0.000
When I run this through the method for solving 3x3 Nash squares, this is the equation I use:
A->B: (p+q+(1-p-q) = p1 + q1 + (1-p1-q1))
B->C: (p1+q1+(1-p1-q1) = p2 + q2 + (1-p2-q2))
and I substitute p from one into the other to find q.
The result I get for the horizontal player, from the above square however, don't make sense:
Horizontal Player
A = 0.07460291360037259
B = 0.18149088118987192
C = 0.7439062052097555
This would indicate that the horizontal player should play 75% of the time. Yet, strategy B clearly produces better results for the horizontal player consistently. Am I interpreting this correctly? It is possible I made a programming error, but the Vertical Player makes sense:
Vertical Player
A = 0.14930944382232172
B = 1.1347517730496453
C = -0.28406121687196695
It's gobbledy gook. But just visually analyzing the square indicates that B is the best strategy.
Any feedback you can provide is much appreciated.
nash-equilibrium
asked Dec 25 '18 at 0:48
Jess BehrensJess Behrens
1
$endgroup$
I've got a 3x3 Nash game cube I'm trying to interpret. The decimals below represent the number of times a player 'wins' against the other player.
A B C
A 0.722,0.278 0.722, 0.278 0.800, 0.200
B 0.750, 0.250 0.780, 0.220 0.875, 0.125
C 0.286, 0.714 0.200, 0.800 1.000, 0.000
When I run this through the method for solving 3x3 Nash squares, this is the equation I use:
A->B: (p+q+(1-p-q) = p1 + q1 + (1-p1-q1))
B->C: (p1+q1+(1-p1-q1) = p2 + q2 + (1-p2-q2))
and I substitute p from one into the other to find q.
The result I get for the horizontal player, from the above square however, don't make sense:
Horizontal Player
A = 0.07460291360037259
B = 0.18149088118987192
C = 0.7439062052097555
This would indicate that the horizontal player should play 75% of the time. Yet, strategy B clearly produces better results for the horizontal player consistently. Am I interpreting this correctly? It is possible I made a programming error, but the Vertical Player makes sense:
Vertical Player
A = 0.14930944382232172
B = 1.1347517730496453
C = -0.28406121687196695
It's gobbledy gook. But just visually analyzing the square indicates that B is the best strategy.
Any feedback you can provide is much appreciated.
nash-equilibrium
nash-equilibrium
asked Dec 25 '18 at 0:48
Jess BehrensJess Behrens
1
asked Dec 25 '18 at 0:48
Jess BehrensJess Behrens
1
asked Dec 25 '18 at 0:48
Jess BehrensJess Behrens
1
asked Dec 25 '18 at 0:48
Jess BehrensJess Behrens
1
asked Dec 25 '18 at 0:48
Jess BehrensJess Behrens
1
1
$begingroup$
Doing the expected value calculation can sometimes produce negative "probabilities" if the game has a pure strategy equilibrium. Your game has one at B,A since the payoff to left player is the highest in the column, and the payoff to either player is highest in the row.
$endgroup$
– dbx
Dec 25 '18 at 3:18
$begingroup$
Thanks! That makes a lot of sense, actually. These are strategies and C is the weakest. When I apply the rule you described to all of the different groupings of the data, the pure strategies most often involve C because the other two essentially take advantage of it. Is there anything to be gleaned from the fact that the row player produces a mixed strategies equilibrium? Or, does that only have meaning if the column player also has a mixed strategy?
$endgroup$
– Jess Behrens
Dec 25 '18 at 6:47
add a comment |
$begingroup$
Doing the expected value calculation can sometimes produce negative "probabilities" if the game has a pure strategy equilibrium. Your game has one at B,A since the payoff to left player is the highest in the column, and the payoff to either player is highest in the row.
$endgroup$
– dbx
Dec 25 '18 at 3:18
$begingroup$
Thanks! That makes a lot of sense, actually. These are strategies and C is the weakest. When I apply the rule you described to all of the different groupings of the data, the pure strategies most often involve C because the other two essentially take advantage of it. Is there anything to be gleaned from the fact that the row player produces a mixed strategies equilibrium? Or, does that only have meaning if the column player also has a mixed strategy?
$endgroup$
– Jess Behrens
Dec 25 '18 at 6:47
$begingroup$
Doing the expected value calculation can sometimes produce negative "probabilities" if the game has a pure strategy equilibrium. Your game has one at B,A since the payoff to left player is the highest in the column, and the payoff to either player is highest in the row.
$endgroup$
– dbx
Dec 25 '18 at 3:18
$begingroup$
Doing the expected value calculation can sometimes produce negative "probabilities" if the game has a pure strategy equilibrium. Your game has one at B,A since the payoff to left player is the highest in the column, and the payoff to either player is highest in the row.
$endgroup$
– dbx
Dec 25 '18 at 3:18
$begingroup$
Thanks! That makes a lot of sense, actually. These are strategies and C is the weakest. When I apply the rule you described to all of the different groupings of the data, the pure strategies most often involve C because the other two essentially take advantage of it. Is there anything to be gleaned from the fact that the row player produces a mixed strategies equilibrium? Or, does that only have meaning if the column player also has a mixed strategy?
$endgroup$
– Jess Behrens
Dec 25 '18 at 6:47
$begingroup$
Thanks! That makes a lot of sense, actually. These are strategies and C is the weakest. When I apply the rule you described to all of the different groupings of the data, the pure strategies most often involve C because the other two essentially take advantage of it. Is there anything to be gleaned from the fact that the row player produces a mixed strategies equilibrium? Or, does that only have meaning if the column player also has a mixed strategy?
$endgroup$
– Jess Behrens
Dec 25 '18 at 6:47
add a comment |
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$begingroup$
Doing the expected value calculation can sometimes produce negative "probabilities" if the game has a pure strategy equilibrium. Your game has one at B,A since the payoff to left player is the highest in the column, and the payoff to either player is highest in the row.
$endgroup$
– dbx
Dec 25 '18 at 3:18
$begingroup$
Thanks! That makes a lot of sense, actually. These are strategies and C is the weakest. When I apply the rule you described to all of the different groupings of the data, the pure strategies most often involve C because the other two essentially take advantage of it. Is there anything to be gleaned from the fact that the row player produces a mixed strategies equilibrium? Or, does that only have meaning if the column player also has a mixed strategy?
$endgroup$
– Jess Behrens
Dec 25 '18 at 6:47