Combinatorics game problem
$begingroup$
Two people, Alice and Ben, take turns removing marbles from a bag. The bag contains $1$ purple, $1$ orange, $4$ green, $6$ red and $2^8$ blue marbles. If Alice starts, at least one marble must be taken out on each turn and no more than one marble of the same color can be taken out on each turn, who removes the last marble? (whoever removes the last marble wins)
combinatorics
$endgroup$
add a comment |
$begingroup$
Two people, Alice and Ben, take turns removing marbles from a bag. The bag contains $1$ purple, $1$ orange, $4$ green, $6$ red and $2^8$ blue marbles. If Alice starts, at least one marble must be taken out on each turn and no more than one marble of the same color can be taken out on each turn, who removes the last marble? (whoever removes the last marble wins)
combinatorics
$endgroup$
add a comment |
$begingroup$
Two people, Alice and Ben, take turns removing marbles from a bag. The bag contains $1$ purple, $1$ orange, $4$ green, $6$ red and $2^8$ blue marbles. If Alice starts, at least one marble must be taken out on each turn and no more than one marble of the same color can be taken out on each turn, who removes the last marble? (whoever removes the last marble wins)
combinatorics
$endgroup$
Two people, Alice and Ben, take turns removing marbles from a bag. The bag contains $1$ purple, $1$ orange, $4$ green, $6$ red and $2^8$ blue marbles. If Alice starts, at least one marble must be taken out on each turn and no more than one marble of the same color can be taken out on each turn, who removes the last marble? (whoever removes the last marble wins)
combinatorics
combinatorics
edited Jan 30 at 23:17
abc...
3,237739
3,237739
asked Jan 30 at 22:42
IrinaIrina
11
11
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Alice removes the last marble. Take the purple and the orange out first, then apply copycat strategy: take out whatever marble(s) Ben took on his last turn. This guarantees Alice to take the last marble since the parity of all colours marbles are even.
$endgroup$
$begingroup$
So Alice takes two at a time each time?
$endgroup$
– Irina
Jan 30 at 23:35
$begingroup$
Not necessarily. She takes whatever amount Ben takes, and of the same color.
$endgroup$
– abc...
Jan 31 at 0:39
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%2f3094219%2fcombinatorics-game-problem%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
$begingroup$
Alice removes the last marble. Take the purple and the orange out first, then apply copycat strategy: take out whatever marble(s) Ben took on his last turn. This guarantees Alice to take the last marble since the parity of all colours marbles are even.
$endgroup$
$begingroup$
So Alice takes two at a time each time?
$endgroup$
– Irina
Jan 30 at 23:35
$begingroup$
Not necessarily. She takes whatever amount Ben takes, and of the same color.
$endgroup$
– abc...
Jan 31 at 0:39
add a comment |
$begingroup$
Alice removes the last marble. Take the purple and the orange out first, then apply copycat strategy: take out whatever marble(s) Ben took on his last turn. This guarantees Alice to take the last marble since the parity of all colours marbles are even.
$endgroup$
$begingroup$
So Alice takes two at a time each time?
$endgroup$
– Irina
Jan 30 at 23:35
$begingroup$
Not necessarily. She takes whatever amount Ben takes, and of the same color.
$endgroup$
– abc...
Jan 31 at 0:39
add a comment |
$begingroup$
Alice removes the last marble. Take the purple and the orange out first, then apply copycat strategy: take out whatever marble(s) Ben took on his last turn. This guarantees Alice to take the last marble since the parity of all colours marbles are even.
$endgroup$
Alice removes the last marble. Take the purple and the orange out first, then apply copycat strategy: take out whatever marble(s) Ben took on his last turn. This guarantees Alice to take the last marble since the parity of all colours marbles are even.
answered Jan 30 at 23:16
abc...abc...
3,237739
3,237739
$begingroup$
So Alice takes two at a time each time?
$endgroup$
– Irina
Jan 30 at 23:35
$begingroup$
Not necessarily. She takes whatever amount Ben takes, and of the same color.
$endgroup$
– abc...
Jan 31 at 0:39
add a comment |
$begingroup$
So Alice takes two at a time each time?
$endgroup$
– Irina
Jan 30 at 23:35
$begingroup$
Not necessarily. She takes whatever amount Ben takes, and of the same color.
$endgroup$
– abc...
Jan 31 at 0:39
$begingroup$
So Alice takes two at a time each time?
$endgroup$
– Irina
Jan 30 at 23:35
$begingroup$
So Alice takes two at a time each time?
$endgroup$
– Irina
Jan 30 at 23:35
$begingroup$
Not necessarily. She takes whatever amount Ben takes, and of the same color.
$endgroup$
– abc...
Jan 31 at 0:39
$begingroup$
Not necessarily. She takes whatever amount Ben takes, and of the same color.
$endgroup$
– abc...
Jan 31 at 0:39
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%2f3094219%2fcombinatorics-game-problem%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