Nash equilibrium in antagonist game in a 2x5 matrix
Background
Input matrix:
$$
begin{bmatrix}
1 & 2 & 3 & 4 \
5 & 4 & 3 & 2 \
end{bmatrix}$$
We have a game with 2 players. The game is antagonistic e.g value functions P(x,y) for player 1 and Q(x,y) for player 2: $P(x,y) = -Q(x,y)$
graph
We can see that all lines merge in one point -3:
https://www.graphsketch.com/?eqn1_color=1&eqn1_eqn=1%2B4x&eqn2_color=2&eqn2_eqn=2%2B2x&eqn3_color=3&eqn3_eqn=3&eqn4_color=4&eqn4_eqn=4-2x&eqn5_color=5&eqn5_eqn=&eqn6_color=6&eqn6_eqn=&x_min=0&x_max=1&y_min=-3&y_max=5&x_tick=1&y_tick=1&x_label_freq=5&y_label_freq=5&do_grid=0&do_grid=1&bold_labeled_lines=0&bold_labeled_lines=1&line_width=4&image_w=850&image_h=525
We know that the price $v=3$, and so:
$$
begin{array}{lcl}
y_1 + 2y_2 + 3y_3 + 4y_4 & = & 3 \
5y_1 + 4y_2 + 3y_3 + 2y_4 & = & 3 \
sum y_i = 1
end{array}
$$
Solution:
The following is given as a hint but I don't understand, what is the idea behind the following?:
$$P(overline{x}, 1) = v$$
$$P(overline{x}, 3) = v$$
$$P(overline{x}, 2) > v$$
$$P(overline{x}, 4) > v$$
$$overline{y_2}=0$$
Question:
- How can I get the Nash equilibrium?
- What is the idea of the hint?
linear-algebra game-theory combinatorial-game-theory algorithmic-game-theory
edited Nov 28 '18 at 21:50
mlc
4,87931332
asked Nov 28 '18 at 19:29
Harton
857
add a comment |
Background
Input matrix:
$$
begin{bmatrix}
1 & 2 & 3 & 4 \
5 & 4 & 3 & 2 \
end{bmatrix}$$
We have a game with 2 players. The game is antagonistic e.g value functions P(x,y) for player 1 and Q(x,y) for player 2: $P(x,y) = -Q(x,y)$
graph
We can see that all lines merge in one point -3:
https://www.graphsketch.com/?eqn1_color=1&eqn1_eqn=1%2B4x&eqn2_color=2&eqn2_eqn=2%2B2x&eqn3_color=3&eqn3_eqn=3&eqn4_color=4&eqn4_eqn=4-2x&eqn5_color=5&eqn5_eqn=&eqn6_color=6&eqn6_eqn=&x_min=0&x_max=1&y_min=-3&y_max=5&x_tick=1&y_tick=1&x_label_freq=5&y_label_freq=5&do_grid=0&do_grid=1&bold_labeled_lines=0&bold_labeled_lines=1&line_width=4&image_w=850&image_h=525
We know that the price $v=3$, and so:
$$
begin{array}{lcl}
y_1 + 2y_2 + 3y_3 + 4y_4 & = & 3 \
5y_1 + 4y_2 + 3y_3 + 2y_4 & = & 3 \
sum y_i = 1
end{array}
$$
Solution:
The following is given as a hint but I don't understand, what is the idea behind the following?:
$$P(overline{x}, 1) = v$$
$$P(overline{x}, 3) = v$$
$$P(overline{x}, 2) > v$$
$$P(overline{x}, 4) > v$$
$$overline{y_2}=0$$
Question:
- How can I get the Nash equilibrium?
- What is the idea of the hint?
linear-algebra game-theory combinatorial-game-theory algorithmic-game-theory
edited Nov 28 '18 at 21:50
mlc
4,87931332
asked Nov 28 '18 at 19:29
Harton
857
1
The phrasing of your question makes it very difficult to understand. You may find useful to note that if $2$ plays the third column, she has a guaranteed payoff of 3. This implies that $1$ cannot achieve more than 3.
– mlc
Nov 28 '18 at 21:53
@mlc you would get a sure value of 3 but how does it make this a Nash Equilibrium? Should't there be a mixed strategy for PL2 to have a avg outcome bigger than 3?
– Harton
Nov 29 '18 at 14:31
in an antagonistic game, your gain is my loss. No other pure strategy for 2 has a guaranteed minimum above 3, and so no other mixed strategy for 2 does. (By the way, 3 is the Nash equilibrium payoff, but you presumably need to find the equilibrium profile.)
– mlc
Nov 29 '18 at 16:43
add a comment |
Background
Input matrix:
$$
begin{bmatrix}
1 & 2 & 3 & 4 \
5 & 4 & 3 & 2 \
end{bmatrix}$$
We have a game with 2 players. The game is antagonistic e.g value functions P(x,y) for player 1 and Q(x,y) for player 2: $P(x,y) = -Q(x,y)$
graph
We can see that all lines merge in one point -3:
https://www.graphsketch.com/?eqn1_color=1&eqn1_eqn=1%2B4x&eqn2_color=2&eqn2_eqn=2%2B2x&eqn3_color=3&eqn3_eqn=3&eqn4_color=4&eqn4_eqn=4-2x&eqn5_color=5&eqn5_eqn=&eqn6_color=6&eqn6_eqn=&x_min=0&x_max=1&y_min=-3&y_max=5&x_tick=1&y_tick=1&x_label_freq=5&y_label_freq=5&do_grid=0&do_grid=1&bold_labeled_lines=0&bold_labeled_lines=1&line_width=4&image_w=850&image_h=525
We know that the price $v=3$, and so:
$$
begin{array}{lcl}
y_1 + 2y_2 + 3y_3 + 4y_4 & = & 3 \
5y_1 + 4y_2 + 3y_3 + 2y_4 & = & 3 \
sum y_i = 1
end{array}
$$
Solution:
The following is given as a hint but I don't understand, what is the idea behind the following?:
$$P(overline{x}, 1) = v$$
$$P(overline{x}, 3) = v$$
$$P(overline{x}, 2) > v$$
$$P(overline{x}, 4) > v$$
$$overline{y_2}=0$$
Question:
- How can I get the Nash equilibrium?
- What is the idea of the hint?
linear-algebra game-theory combinatorial-game-theory algorithmic-game-theory
edited Nov 28 '18 at 21:50
mlc
4,87931332
asked Nov 28 '18 at 19:29
Harton
857
Background
Input matrix:
$$
begin{bmatrix}
1 & 2 & 3 & 4 \
5 & 4 & 3 & 2 \
end{bmatrix}$$
We have a game with 2 players. The game is antagonistic e.g value functions P(x,y) for player 1 and Q(x,y) for player 2: $P(x,y) = -Q(x,y)$
graph
We can see that all lines merge in one point -3:
https://www.graphsketch.com/?eqn1_color=1&eqn1_eqn=1%2B4x&eqn2_color=2&eqn2_eqn=2%2B2x&eqn3_color=3&eqn3_eqn=3&eqn4_color=4&eqn4_eqn=4-2x&eqn5_color=5&eqn5_eqn=&eqn6_color=6&eqn6_eqn=&x_min=0&x_max=1&y_min=-3&y_max=5&x_tick=1&y_tick=1&x_label_freq=5&y_label_freq=5&do_grid=0&do_grid=1&bold_labeled_lines=0&bold_labeled_lines=1&line_width=4&image_w=850&image_h=525
We know that the price $v=3$, and so:
$$
begin{array}{lcl}
y_1 + 2y_2 + 3y_3 + 4y_4 & = & 3 \
5y_1 + 4y_2 + 3y_3 + 2y_4 & = & 3 \
sum y_i = 1
end{array}
$$
Solution:
The following is given as a hint but I don't understand, what is the idea behind the following?:
$$P(overline{x}, 1) = v$$
$$P(overline{x}, 3) = v$$
$$P(overline{x}, 2) > v$$
$$P(overline{x}, 4) > v$$
$$overline{y_2}=0$$
Question:
- How can I get the Nash equilibrium?
- What is the idea of the hint?
linear-algebra game-theory combinatorial-game-theory algorithmic-game-theory
linear-algebra game-theory combinatorial-game-theory algorithmic-game-theory
edited Nov 28 '18 at 21:50
mlc
4,87931332
asked Nov 28 '18 at 19:29
Harton
857
edited Nov 28 '18 at 21:50
mlc
4,87931332
asked Nov 28 '18 at 19:29
Harton
857
edited Nov 28 '18 at 21:50
mlc
4,87931332
edited Nov 28 '18 at 21:50
mlc
4,87931332
edited Nov 28 '18 at 21:50
mlc
4,87931332
4,87931332
asked Nov 28 '18 at 19:29
Harton
857
asked Nov 28 '18 at 19:29
Harton
857
asked Nov 28 '18 at 19:29
Harton
857
857
1
The phrasing of your question makes it very difficult to understand. You may find useful to note that if $2$ plays the third column, she has a guaranteed payoff of 3. This implies that $1$ cannot achieve more than 3.
– mlc
Nov 28 '18 at 21:53
@mlc you would get a sure value of 3 but how does it make this a Nash Equilibrium? Should't there be a mixed strategy for PL2 to have a avg outcome bigger than 3?
– Harton
Nov 29 '18 at 14:31
in an antagonistic game, your gain is my loss. No other pure strategy for 2 has a guaranteed minimum above 3, and so no other mixed strategy for 2 does. (By the way, 3 is the Nash equilibrium payoff, but you presumably need to find the equilibrium profile.)
– mlc
Nov 29 '18 at 16:43
add a comment |
1
The phrasing of your question makes it very difficult to understand. You may find useful to note that if $2$ plays the third column, she has a guaranteed payoff of 3. This implies that $1$ cannot achieve more than 3.
– mlc
Nov 28 '18 at 21:53
@mlc you would get a sure value of 3 but how does it make this a Nash Equilibrium? Should't there be a mixed strategy for PL2 to have a avg outcome bigger than 3?
– Harton
Nov 29 '18 at 14:31
in an antagonistic game, your gain is my loss. No other pure strategy for 2 has a guaranteed minimum above 3, and so no other mixed strategy for 2 does. (By the way, 3 is the Nash equilibrium payoff, but you presumably need to find the equilibrium profile.)
– mlc
Nov 29 '18 at 16:43
1
1
The phrasing of your question makes it very difficult to understand. You may find useful to note that if $2$ plays the third column, she has a guaranteed payoff of 3. This implies that $1$ cannot achieve more than 3.
– mlc
Nov 28 '18 at 21:53
The phrasing of your question makes it very difficult to understand. You may find useful to note that if $2$ plays the third column, she has a guaranteed payoff of 3. This implies that $1$ cannot achieve more than 3.
– mlc
Nov 28 '18 at 21:53
@mlc you would get a sure value of 3 but how does it make this a Nash Equilibrium? Should't there be a mixed strategy for PL2 to have a avg outcome bigger than 3?
– Harton
Nov 29 '18 at 14:31
@mlc you would get a sure value of 3 but how does it make this a Nash Equilibrium? Should't there be a mixed strategy for PL2 to have a avg outcome bigger than 3?
– Harton
Nov 29 '18 at 14:31
in an antagonistic game, your gain is my loss. No other pure strategy for 2 has a guaranteed minimum above 3, and so no other mixed strategy for 2 does. (By the way, 3 is the Nash equilibrium payoff, but you presumably need to find the equilibrium profile.)
– mlc
Nov 29 '18 at 16:43
in an antagonistic game, your gain is my loss. No other pure strategy for 2 has a guaranteed minimum above 3, and so no other mixed strategy for 2 does. (By the way, 3 is the Nash equilibrium payoff, but you presumably need to find the equilibrium profile.)
– mlc
Nov 29 '18 at 16:43
add a comment |
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1
The phrasing of your question makes it very difficult to understand. You may find useful to note that if $2$ plays the third column, she has a guaranteed payoff of 3. This implies that $1$ cannot achieve more than 3.
– mlc
Nov 28 '18 at 21:53
@mlc you would get a sure value of 3 but how does it make this a Nash Equilibrium? Should't there be a mixed strategy for PL2 to have a avg outcome bigger than 3?
– Harton
Nov 29 '18 at 14:31
in an antagonistic game, your gain is my loss. No other pure strategy for 2 has a guaranteed minimum above 3, and so no other mixed strategy for 2 does. (By the way, 3 is the Nash equilibrium payoff, but you presumably need to find the equilibrium profile.)
– mlc
Nov 29 '18 at 16:43