2 player zero-sum-game rock paper scissors expected loss
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0
down vote
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For the Rock-Paper-Scissors game, I am trying to determine the expected loss for P1.
The following matrix displays how much P1 has lost:
A:
| 0 1 -1 |
| -1 0 1 |
| 1 -1 0 |
I am trying to find the expected loss for P1 when P1 uses mixed strategy x = (x1, x2, x3) and P2 uses mixed strategy y (y1, y2, y3)
I am not sure how to do this?
EDIT:
My attempt is thinking that if I should always have a 1/3 chance of winning, right?
linear-programming game-theory nash-equilibrium
asked Nov 15 at 2:58
Jerry M.
1011
|
show 1 more comment
up vote
0
down vote
favorite
For the Rock-Paper-Scissors game, I am trying to determine the expected loss for P1.
The following matrix displays how much P1 has lost:
A:
| 0 1 -1 |
| -1 0 1 |
| 1 -1 0 |
I am trying to find the expected loss for P1 when P1 uses mixed strategy x = (x1, x2, x3) and P2 uses mixed strategy y (y1, y2, y3)
I am not sure how to do this?
EDIT:
My attempt is thinking that if I should always have a 1/3 chance of winning, right?
linear-programming game-theory nash-equilibrium
asked Nov 15 at 2:58
Jerry M.
1011
There are only $9$ things that can happen. Figure out their probabilities and add up the expected gains and losses.
– saulspatz
Nov 15 at 3:10
the solution proposed by saulspatz can be written elegantly in terms of $A$, $x$ and $y$
– LinAlg
Nov 15 at 16:29
I will attempt this, and post an answer. Thank you for the feed back.
– Jerry M.
Nov 15 at 17:03
It's true that you can always break even by playing $x = (1/3,1/3,1/3)$. But you can possibly do better, depending on P2's strategy. If P2 always plays rock, then you can always play paper and win every time.
– Théophile
Nov 17 at 21:12
@Théophile thank you for the feedback. I know this is dumb question but I am trying so hard to understand it
– Jerry M.
Nov 17 at 21:14
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For the Rock-Paper-Scissors game, I am trying to determine the expected loss for P1.
The following matrix displays how much P1 has lost:
A:
| 0 1 -1 |
| -1 0 1 |
| 1 -1 0 |
I am trying to find the expected loss for P1 when P1 uses mixed strategy x = (x1, x2, x3) and P2 uses mixed strategy y (y1, y2, y3)
I am not sure how to do this?
EDIT:
My attempt is thinking that if I should always have a 1/3 chance of winning, right?
linear-programming game-theory nash-equilibrium
asked Nov 15 at 2:58
Jerry M.
1011
For the Rock-Paper-Scissors game, I am trying to determine the expected loss for P1.
The following matrix displays how much P1 has lost:
A:
| 0 1 -1 |
| -1 0 1 |
| 1 -1 0 |
I am trying to find the expected loss for P1 when P1 uses mixed strategy x = (x1, x2, x3) and P2 uses mixed strategy y (y1, y2, y3)
I am not sure how to do this?
EDIT:
My attempt is thinking that if I should always have a 1/3 chance of winning, right?
linear-programming game-theory nash-equilibrium
linear-programming game-theory nash-equilibrium
asked Nov 15 at 2:58
Jerry M.
1011
asked Nov 15 at 2:58
Jerry M.
1011
edited Nov 17 at 21:08
asked Nov 15 at 2:58
Jerry M.
1011
asked Nov 15 at 2:58
Jerry M.
1011
asked Nov 15 at 2:58
Jerry M.
1011
1011
There are only $9$ things that can happen. Figure out their probabilities and add up the expected gains and losses.
– saulspatz
Nov 15 at 3:10
the solution proposed by saulspatz can be written elegantly in terms of $A$, $x$ and $y$
– LinAlg
Nov 15 at 16:29
I will attempt this, and post an answer. Thank you for the feed back.
– Jerry M.
Nov 15 at 17:03
It's true that you can always break even by playing $x = (1/3,1/3,1/3)$. But you can possibly do better, depending on P2's strategy. If P2 always plays rock, then you can always play paper and win every time.
– Théophile
Nov 17 at 21:12
@Théophile thank you for the feedback. I know this is dumb question but I am trying so hard to understand it
– Jerry M.
Nov 17 at 21:14
|
show 1 more comment
There are only $9$ things that can happen. Figure out their probabilities and add up the expected gains and losses.
– saulspatz
Nov 15 at 3:10
the solution proposed by saulspatz can be written elegantly in terms of $A$, $x$ and $y$
– LinAlg
Nov 15 at 16:29
I will attempt this, and post an answer. Thank you for the feed back.
– Jerry M.
Nov 15 at 17:03
It's true that you can always break even by playing $x = (1/3,1/3,1/3)$. But you can possibly do better, depending on P2's strategy. If P2 always plays rock, then you can always play paper and win every time.
– Théophile
Nov 17 at 21:12
@Théophile thank you for the feedback. I know this is dumb question but I am trying so hard to understand it
– Jerry M.
Nov 17 at 21:14
There are only $9$ things that can happen. Figure out their probabilities and add up the expected gains and losses.
– saulspatz
Nov 15 at 3:10
There are only $9$ things that can happen. Figure out their probabilities and add up the expected gains and losses.
– saulspatz
Nov 15 at 3:10
the solution proposed by saulspatz can be written elegantly in terms of $A$, $x$ and $y$
– LinAlg
Nov 15 at 16:29
the solution proposed by saulspatz can be written elegantly in terms of $A$, $x$ and $y$
– LinAlg
Nov 15 at 16:29
I will attempt this, and post an answer. Thank you for the feed back.
– Jerry M.
Nov 15 at 17:03
I will attempt this, and post an answer. Thank you for the feed back.
– Jerry M.
Nov 15 at 17:03
It's true that you can always break even by playing $x = (1/3,1/3,1/3)$. But you can possibly do better, depending on P2's strategy. If P2 always plays rock, then you can always play paper and win every time.
– Théophile
Nov 17 at 21:12
It's true that you can always break even by playing $x = (1/3,1/3,1/3)$. But you can possibly do better, depending on P2's strategy. If P2 always plays rock, then you can always play paper and win every time.
– Théophile
Nov 17 at 21:12
@Théophile thank you for the feedback. I know this is dumb question but I am trying so hard to understand it
– Jerry M.
Nov 17 at 21:14
@Théophile thank you for the feedback. I know this is dumb question but I am trying so hard to understand it
– Jerry M.
Nov 17 at 21:14
|
show 1 more comment
1 Answer
1
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oldest
votes
up vote
0
down vote
The expected payoff for $P1$ when he plays the mixed strategy $x$ and his opponent plays the mixed strategy $y$ is $x A y^top$.
answered Nov 17 at 18:10
mlc
4,83431332
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The expected payoff for $P1$ when he plays the mixed strategy $x$ and his opponent plays the mixed strategy $y$ is $x A y^top$.
answered Nov 17 at 18:10
mlc
4,83431332
add a comment |
up vote
0
down vote
The expected payoff for $P1$ when he plays the mixed strategy $x$ and his opponent plays the mixed strategy $y$ is $x A y^top$.
answered Nov 17 at 18:10
mlc
4,83431332
add a comment |
up vote
0
down vote
up vote
0
down vote
The expected payoff for $P1$ when he plays the mixed strategy $x$ and his opponent plays the mixed strategy $y$ is $x A y^top$.
answered Nov 17 at 18:10
mlc
4,83431332
The expected payoff for $P1$ when he plays the mixed strategy $x$ and his opponent plays the mixed strategy $y$ is $x A y^top$.
answered Nov 17 at 18:10
mlc
4,83431332
answered Nov 17 at 18:10
mlc
4,83431332
answered Nov 17 at 18:10
mlc
4,83431332
answered Nov 17 at 18:10
mlc
4,83431332
4,83431332
add a comment |
add a comment |
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There are only $9$ things that can happen. Figure out their probabilities and add up the expected gains and losses.
– saulspatz
Nov 15 at 3:10
the solution proposed by saulspatz can be written elegantly in terms of $A$, $x$ and $y$
– LinAlg
Nov 15 at 16:29
I will attempt this, and post an answer. Thank you for the feed back.
– Jerry M.
Nov 15 at 17:03
It's true that you can always break even by playing $x = (1/3,1/3,1/3)$. But you can possibly do better, depending on P2's strategy. If P2 always plays rock, then you can always play paper and win every time.
– Théophile
Nov 17 at 21:12
@Théophile thank you for the feedback. I know this is dumb question but I am trying so hard to understand it
– Jerry M.
Nov 17 at 21:14