proving if player 1 has a pure optimal strategy, player 2 should as well
$begingroup$
Problem:
Prove that if in a matrix game 2x2 if the player 1 has a pure optimal strategy, so has player 2
Attempt:
Given:
We know that player 1 has pure optimal strategy, meaning:
$$P(x, overline{y}) leq P(overline{x}, overline{y}) leq P(overline{x}, y) $$
Do we actually know that?
I thought about expressing $P(x,y)=-Q(x,y)$ but nowhere it's said that it's an antagonistic game, but then I found this theorem:
If we knew that there is a saddle point, we would know that, there is a price and that:
$$P(x, overline{y}) leq v leq P(overline{x},y)$$
Meaning that there is dual solution:
$$begin{cases}
v -> min\
P(i,y) leq v \
y in Y
end{cases}
$$
$$begin{cases}
v -> max\
P(x,j) geq v \
x in X
end{cases}
$$
Question:
The Problmem is ... do we actually know that we have antagonistic game or a seddle point, just by knowing that Player 1 has pure strategy? How can we prove this ?
linear-algebra optimization game-theory algorithmic-game-theory
asked Jan 19 at 13:57
HartonHarton
858
$endgroup$
add a comment |
$begingroup$
Problem:
Prove that if in a matrix game 2x2 if the player 1 has a pure optimal strategy, so has player 2
Attempt:
Given:
We know that player 1 has pure optimal strategy, meaning:
$$P(x, overline{y}) leq P(overline{x}, overline{y}) leq P(overline{x}, y) $$
Do we actually know that?
I thought about expressing $P(x,y)=-Q(x,y)$ but nowhere it's said that it's an antagonistic game, but then I found this theorem:
If we knew that there is a saddle point, we would know that, there is a price and that:
$$P(x, overline{y}) leq v leq P(overline{x},y)$$
Meaning that there is dual solution:
$$begin{cases}
v -> min\
P(i,y) leq v \
y in Y
end{cases}
$$
$$begin{cases}
v -> max\
P(x,j) geq v \
x in X
end{cases}
$$
Question:
The Problmem is ... do we actually know that we have antagonistic game or a seddle point, just by knowing that Player 1 has pure strategy? How can we prove this ?
linear-algebra optimization game-theory algorithmic-game-theory
asked Jan 19 at 13:57
HartonHarton
858
$endgroup$
$begingroup$
Maybe it's just too long since I've looked at game theory, but I thought a matrix game was zero-sum by definition. If not, how can you describe the payoff functions with one matrix?
$endgroup$
– saulspatz
Jan 19 at 14:05
$begingroup$
@saulspatz I think so as well, it should be, I just want to validate this statement as it's nowhere to be said
$endgroup$
– Harton
Jan 19 at 14:18
$begingroup$
encyclopediaofmath.org/index.php/Matrix_game
$endgroup$
– saulspatz
Jan 19 at 14:21
$begingroup$
ok ... so I have to prove the von Neumann's minimax theorem ?
$endgroup$
– Harton
Jan 19 at 14:47
add a comment |
$begingroup$
Problem:
Prove that if in a matrix game 2x2 if the player 1 has a pure optimal strategy, so has player 2
Attempt:
Given:
We know that player 1 has pure optimal strategy, meaning:
$$P(x, overline{y}) leq P(overline{x}, overline{y}) leq P(overline{x}, y) $$
Do we actually know that?
I thought about expressing $P(x,y)=-Q(x,y)$ but nowhere it's said that it's an antagonistic game, but then I found this theorem:
If we knew that there is a saddle point, we would know that, there is a price and that:
$$P(x, overline{y}) leq v leq P(overline{x},y)$$
Meaning that there is dual solution:
$$begin{cases}
v -> min\
P(i,y) leq v \
y in Y
end{cases}
$$
$$begin{cases}
v -> max\
P(x,j) geq v \
x in X
end{cases}
$$
Question:
The Problmem is ... do we actually know that we have antagonistic game or a seddle point, just by knowing that Player 1 has pure strategy? How can we prove this ?
linear-algebra optimization game-theory algorithmic-game-theory
asked Jan 19 at 13:57
HartonHarton
858
$endgroup$
Problem:
Prove that if in a matrix game 2x2 if the player 1 has a pure optimal strategy, so has player 2
Attempt:
Given:
We know that player 1 has pure optimal strategy, meaning:
$$P(x, overline{y}) leq P(overline{x}, overline{y}) leq P(overline{x}, y) $$
Do we actually know that?
I thought about expressing $P(x,y)=-Q(x,y)$ but nowhere it's said that it's an antagonistic game, but then I found this theorem:
If we knew that there is a saddle point, we would know that, there is a price and that:
$$P(x, overline{y}) leq v leq P(overline{x},y)$$
Meaning that there is dual solution:
$$begin{cases}
v -> min\
P(i,y) leq v \
y in Y
end{cases}
$$
$$begin{cases}
v -> max\
P(x,j) geq v \
x in X
end{cases}
$$
Question:
The Problmem is ... do we actually know that we have antagonistic game or a seddle point, just by knowing that Player 1 has pure strategy? How can we prove this ?
linear-algebra optimization game-theory algorithmic-game-theory
linear-algebra optimization game-theory algorithmic-game-theory
asked Jan 19 at 13:57
HartonHarton
858
asked Jan 19 at 13:57
HartonHarton
858
asked Jan 19 at 13:57
HartonHarton
858
asked Jan 19 at 13:57
HartonHarton
858
asked Jan 19 at 13:57
HartonHarton
858
858
$begingroup$
Maybe it's just too long since I've looked at game theory, but I thought a matrix game was zero-sum by definition. If not, how can you describe the payoff functions with one matrix?
$endgroup$
– saulspatz
Jan 19 at 14:05
$begingroup$
@saulspatz I think so as well, it should be, I just want to validate this statement as it's nowhere to be said
$endgroup$
– Harton
Jan 19 at 14:18
$begingroup$
encyclopediaofmath.org/index.php/Matrix_game
$endgroup$
– saulspatz
Jan 19 at 14:21
$begingroup$
ok ... so I have to prove the von Neumann's minimax theorem ?
$endgroup$
– Harton
Jan 19 at 14:47
add a comment |
$begingroup$
Maybe it's just too long since I've looked at game theory, but I thought a matrix game was zero-sum by definition. If not, how can you describe the payoff functions with one matrix?
$endgroup$
– saulspatz
Jan 19 at 14:05
$begingroup$
@saulspatz I think so as well, it should be, I just want to validate this statement as it's nowhere to be said
$endgroup$
– Harton
Jan 19 at 14:18
$begingroup$
encyclopediaofmath.org/index.php/Matrix_game
$endgroup$
– saulspatz
Jan 19 at 14:21
$begingroup$
ok ... so I have to prove the von Neumann's minimax theorem ?
$endgroup$
– Harton
Jan 19 at 14:47
$begingroup$
Maybe it's just too long since I've looked at game theory, but I thought a matrix game was zero-sum by definition. If not, how can you describe the payoff functions with one matrix?
$endgroup$
– saulspatz
Jan 19 at 14:05
$begingroup$
Maybe it's just too long since I've looked at game theory, but I thought a matrix game was zero-sum by definition. If not, how can you describe the payoff functions with one matrix?
$endgroup$
– saulspatz
Jan 19 at 14:05
$begingroup$
@saulspatz I think so as well, it should be, I just want to validate this statement as it's nowhere to be said
$endgroup$
– Harton
Jan 19 at 14:18
$begingroup$
@saulspatz I think so as well, it should be, I just want to validate this statement as it's nowhere to be said
$endgroup$
– Harton
Jan 19 at 14:18
$begingroup$
encyclopediaofmath.org/index.php/Matrix_game
$endgroup$
– saulspatz
Jan 19 at 14:21
$begingroup$
encyclopediaofmath.org/index.php/Matrix_game
$endgroup$
– saulspatz
Jan 19 at 14:21
$begingroup$
ok ... so I have to prove the von Neumann's minimax theorem ?
$endgroup$
– Harton
Jan 19 at 14:47
$begingroup$
ok ... so I have to prove the von Neumann's minimax theorem ?
$endgroup$
– Harton
Jan 19 at 14:47
add a comment |
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$begingroup$
Maybe it's just too long since I've looked at game theory, but I thought a matrix game was zero-sum by definition. If not, how can you describe the payoff functions with one matrix?
$endgroup$
– saulspatz
Jan 19 at 14:05
$begingroup$
@saulspatz I think so as well, it should be, I just want to validate this statement as it's nowhere to be said
$endgroup$
– Harton
Jan 19 at 14:18
$begingroup$
encyclopediaofmath.org/index.php/Matrix_game
$endgroup$
– saulspatz
Jan 19 at 14:21
$begingroup$
ok ... so I have to prove the von Neumann's minimax theorem ?
$endgroup$
– Harton
Jan 19 at 14:47