Mixing
The value of 2 person zero sum game described by square matrix.
$begingroup$
I am trying to solve the following problem.
Find the value of G, where G is a 2-person zero sum game described by an $n$ x $n $ square matrix A such that:
$ -a_{ii} = sum_{j neq i} a_{ij} \ a_{ij} geq 0 $
and $[A]_{ij} = a_{ij} space space i neq j $
and $ [A]_{ii} = - a_{ii}$
That is, the sum of all diagonal entries are non-positive, the sum of all non-diagonal entries are non-negative and the sum of the entries in each row is zero where the diagonal entry is equal to the negative of the sum of all other entries in its row.
I believe that value is zero. I have verified this for $2$ x $2$ examples. Also, by giving each column equal weight player-2 can guarantee not paying more than $0$.
any help is welcome, thanks.
combinatorics game-theory combinatorial-game-theory
asked Jan 28 at 22:24
AdiAdi
1107
$endgroup$
add a comment |
$begingroup$
I am trying to solve the following problem.
Find the value of G, where G is a 2-person zero sum game described by an $n$ x $n $ square matrix A such that:
$ -a_{ii} = sum_{j neq i} a_{ij} \ a_{ij} geq 0 $
and $[A]_{ij} = a_{ij} space space i neq j $
and $ [A]_{ii} = - a_{ii}$
That is, the sum of all diagonal entries are non-positive, the sum of all non-diagonal entries are non-negative and the sum of the entries in each row is zero where the diagonal entry is equal to the negative of the sum of all other entries in its row.
I believe that value is zero. I have verified this for $2$ x $2$ examples. Also, by giving each column equal weight player-2 can guarantee not paying more than $0$.
any help is welcome, thanks.
combinatorics game-theory combinatorial-game-theory
asked Jan 28 at 22:24
AdiAdi
1107
$endgroup$
add a comment |
$begingroup$
I am trying to solve the following problem.
Find the value of G, where G is a 2-person zero sum game described by an $n$ x $n $ square matrix A such that:
$ -a_{ii} = sum_{j neq i} a_{ij} \ a_{ij} geq 0 $
and $[A]_{ij} = a_{ij} space space i neq j $
and $ [A]_{ii} = - a_{ii}$
That is, the sum of all diagonal entries are non-positive, the sum of all non-diagonal entries are non-negative and the sum of the entries in each row is zero where the diagonal entry is equal to the negative of the sum of all other entries in its row.
I believe that value is zero. I have verified this for $2$ x $2$ examples. Also, by giving each column equal weight player-2 can guarantee not paying more than $0$.
any help is welcome, thanks.
combinatorics game-theory combinatorial-game-theory
asked Jan 28 at 22:24
AdiAdi
1107
$endgroup$
I am trying to solve the following problem.
Find the value of G, where G is a 2-person zero sum game described by an $n$ x $n $ square matrix A such that:
$ -a_{ii} = sum_{j neq i} a_{ij} \ a_{ij} geq 0 $
and $[A]_{ij} = a_{ij} space space i neq j $
and $ [A]_{ii} = - a_{ii}$
That is, the sum of all diagonal entries are non-positive, the sum of all non-diagonal entries are non-negative and the sum of the entries in each row is zero where the diagonal entry is equal to the negative of the sum of all other entries in its row.
I believe that value is zero. I have verified this for $2$ x $2$ examples. Also, by giving each column equal weight player-2 can guarantee not paying more than $0$.
any help is welcome, thanks.
combinatorics game-theory combinatorial-game-theory
combinatorics game-theory combinatorial-game-theory
asked Jan 28 at 22:24
AdiAdi
1107
asked Jan 28 at 22:24
AdiAdi
1107
edited Jan 28 at 22:32
Adi
asked Jan 28 at 22:24
AdiAdi
1107
asked Jan 28 at 22:24
AdiAdi
1107
asked Jan 28 at 22:24
AdiAdi
1107
1107
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
by giving each column equal weight player-2 can guarantee not paying more than $0$.
Thus, in order to show that the value of the game is $0$, by the well-known theorem, it remains to show that Player 2 cannot guarantee not to pay less than $0$. Let in a mixed strategy Player 2 chooses column $i$ with probability $q_i$. Since $sum q_i=1$, there exists $i$ with $q_ile 1/n$. Then it is easy to check that when Player 1 chooses $i$-th row his expected gain is at least $0$.
answered Feb 2 at 17:19
Alex RavskyAlex Ravsky
42.7k32483
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3091479%2fthe-value-of-2-person-zero-sum-game-described-by-square-matrix%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
by giving each column equal weight player-2 can guarantee not paying more than $0$.
Thus, in order to show that the value of the game is $0$, by the well-known theorem, it remains to show that Player 2 cannot guarantee not to pay less than $0$. Let in a mixed strategy Player 2 chooses column $i$ with probability $q_i$. Since $sum q_i=1$, there exists $i$ with $q_ile 1/n$. Then it is easy to check that when Player 1 chooses $i$-th row his expected gain is at least $0$.
answered Feb 2 at 17:19
Alex RavskyAlex Ravsky
42.7k32483
$endgroup$
add a comment |
$begingroup$
by giving each column equal weight player-2 can guarantee not paying more than $0$.
Thus, in order to show that the value of the game is $0$, by the well-known theorem, it remains to show that Player 2 cannot guarantee not to pay less than $0$. Let in a mixed strategy Player 2 chooses column $i$ with probability $q_i$. Since $sum q_i=1$, there exists $i$ with $q_ile 1/n$. Then it is easy to check that when Player 1 chooses $i$-th row his expected gain is at least $0$.
answered Feb 2 at 17:19
Alex RavskyAlex Ravsky
42.7k32483
$endgroup$
add a comment |
$begingroup$
by giving each column equal weight player-2 can guarantee not paying more than $0$.
Thus, in order to show that the value of the game is $0$, by the well-known theorem, it remains to show that Player 2 cannot guarantee not to pay less than $0$. Let in a mixed strategy Player 2 chooses column $i$ with probability $q_i$. Since $sum q_i=1$, there exists $i$ with $q_ile 1/n$. Then it is easy to check that when Player 1 chooses $i$-th row his expected gain is at least $0$.
answered Feb 2 at 17:19
Alex RavskyAlex Ravsky
42.7k32483
$endgroup$
by giving each column equal weight player-2 can guarantee not paying more than $0$.
Thus, in order to show that the value of the game is $0$, by the well-known theorem, it remains to show that Player 2 cannot guarantee not to pay less than $0$. Let in a mixed strategy Player 2 chooses column $i$ with probability $q_i$. Since $sum q_i=1$, there exists $i$ with $q_ile 1/n$. Then it is easy to check that when Player 1 chooses $i$-th row his expected gain is at least $0$.
answered Feb 2 at 17:19
Alex RavskyAlex Ravsky
42.7k32483
answered Feb 2 at 17:19
Alex RavskyAlex Ravsky
42.7k32483
answered Feb 2 at 17:19
Alex RavskyAlex Ravsky
42.7k32483
answered Feb 2 at 17:19
Alex RavskyAlex Ravsky
42.7k32483
42.7k32483
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3091479%2fthe-value-of-2-person-zero-sum-game-described-by-square-matrix%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
