Biased dice probability question Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Probability of dice thrownDice and probabilityDetermine whether the dice is biased based on 10 rollsProbability of events with biased diceProbability of biased diceProbability on biased diceProbability of rolling 2 and 3 numbers in a sequence when rolling 3, 6 sided diceDice probability helpProbability of an “at least” QuestionProbability of biased die.
New Order #5: where Fibonacci and Beatty meet at Wythoff
Is there a service that would inform me whenever a new direct route is scheduled from a given airport?
Simulating Exploding Dice
Biased dice probability question
Why is "Captain Marvel" translated as male in Portugal?
Geometric mean and geometric standard deviation
Two different pronunciation of "понял"
Autumning in love
Antler Helmet: Can it work?
Cold is to Refrigerator as warm is to?
What computer would be fastest for Mathematica Home Edition?
Why is there no army of Iron-Mans in the MCU?
Estimate capacitor parameters
What LEGO pieces have "real-world" functionality?
Area of a 2D convex hull
Problem when applying foreach loop
Can smartphones with the same camera sensor have different image quality?
How to rotate it perfectly?
How is simplicity better than precision and clarity in prose?
Can a zero nonce be safely used with AES-GCM if the key is random and never used again?
Is above average number of years spent on PhD considered a red flag in future academia or industry positions?
When is phishing education going too far?
Why does tar appear to skip file contents when output file is /dev/null?
Mortgage adviser recommends a longer term than necessary combined with overpayments
Biased dice probability question
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Probability of dice thrownDice and probabilityDetermine whether the dice is biased based on 10 rollsProbability of events with biased diceProbability of biased diceProbability on biased diceProbability of rolling 2 and 3 numbers in a sequence when rolling 3, 6 sided diceDice probability helpProbability of an “at least” QuestionProbability of biased die.
$begingroup$
A biased six sided dice is rolled twice. Show that the probability that the two results are the same is at least $frac16$.
(Hint: $(p_1 − a)^2 + . . . + (p_6 − a)^2 ≥ 0$ and choose suitable
$p_1, . . . , p_6$, a.)
probability
edited 20 mins ago
mathpadawan
2,019422
asked 23 mins ago
mandymandy
211
New contributor
mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
A biased six sided dice is rolled twice. Show that the probability that the two results are the same is at least $frac16$.
(Hint: $(p_1 − a)^2 + . . . + (p_6 − a)^2 ≥ 0$ and choose suitable
$p_1, . . . , p_6$, a.)
probability
edited 20 mins ago
mathpadawan
2,019422
asked 23 mins ago
mandymandy
211
New contributor
mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Hint: First try to show that if a coin is flipped twice, the probability that the two results are the same is at least $1/2$. This will help you figure out what to choose as $a$.
$endgroup$
– Lorenzo
10 mins ago
add a comment |
$begingroup$
A biased six sided dice is rolled twice. Show that the probability that the two results are the same is at least $frac16$.
(Hint: $(p_1 − a)^2 + . . . + (p_6 − a)^2 ≥ 0$ and choose suitable
$p_1, . . . , p_6$, a.)
probability
edited 20 mins ago
mathpadawan
2,019422
asked 23 mins ago
mandymandy
211
New contributor
mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
A biased six sided dice is rolled twice. Show that the probability that the two results are the same is at least $frac16$.
(Hint: $(p_1 − a)^2 + . . . + (p_6 − a)^2 ≥ 0$ and choose suitable
$p_1, . . . , p_6$, a.)
probability
probability
edited 20 mins ago
mathpadawan
2,019422
asked 23 mins ago
mandymandy
211
New contributor
mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 20 mins ago
mathpadawan
2,019422
asked 23 mins ago
mandymandy
211
New contributor
mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 20 mins ago
mathpadawan
2,019422
edited 20 mins ago
mathpadawan
2,019422
edited 20 mins ago
mathpadawan
2,019422
2,019422
asked 23 mins ago
mandymandy
211
New contributor
mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 23 mins ago
mandymandy
211
asked 23 mins ago
mandymandy
211
211
New contributor
mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
mandy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Hint: First try to show that if a coin is flipped twice, the probability that the two results are the same is at least $1/2$. This will help you figure out what to choose as $a$.
$endgroup$
– Lorenzo
10 mins ago
add a comment |
$begingroup$
Hint: First try to show that if a coin is flipped twice, the probability that the two results are the same is at least $1/2$. This will help you figure out what to choose as $a$.
$endgroup$
– Lorenzo
10 mins ago
$begingroup$
Hint: First try to show that if a coin is flipped twice, the probability that the two results are the same is at least $1/2$. This will help you figure out what to choose as $a$.
$endgroup$
– Lorenzo
10 mins ago
$begingroup$
Hint: First try to show that if a coin is flipped twice, the probability that the two results are the same is at least $1/2$. This will help you figure out what to choose as $a$.
$endgroup$
– Lorenzo
10 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $p_i$ be the probability of rolling $i$. Then $sum_i=1^6 p_i = 1$.
By Cauchy-Schwarz inequality,
$$beginalign*
left(sum_i=1^6 1^2right) left(sum_i=1^6 p_i^2right) &ge
left(sum_i=1^6 1p_iright)^2\
6left(sum_i=1^6 p_i^2right) &ge 1\
sum_i=1^6 p_i^2 &ge frac16endalign*$$
Equality holds when all the $p_i$ are the same, i.e. when the die is unbiased.
answered 9 mins ago
peterwhypeterwhy
12.3k21229
$endgroup$
add a comment |
Your Answer
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
);
);
mandy is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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%2f3188165%2fbiased-dice-probability-question%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
$begingroup$
Let $p_i$ be the probability of rolling $i$. Then $sum_i=1^6 p_i = 1$.
By Cauchy-Schwarz inequality,
$$beginalign*
left(sum_i=1^6 1^2right) left(sum_i=1^6 p_i^2right) &ge
left(sum_i=1^6 1p_iright)^2\
6left(sum_i=1^6 p_i^2right) &ge 1\
sum_i=1^6 p_i^2 &ge frac16endalign*$$
Equality holds when all the $p_i$ are the same, i.e. when the die is unbiased.
answered 9 mins ago
peterwhypeterwhy
12.3k21229
$endgroup$
add a comment |
$begingroup$
Let $p_i$ be the probability of rolling $i$. Then $sum_i=1^6 p_i = 1$.
By Cauchy-Schwarz inequality,
$$beginalign*
left(sum_i=1^6 1^2right) left(sum_i=1^6 p_i^2right) &ge
left(sum_i=1^6 1p_iright)^2\
6left(sum_i=1^6 p_i^2right) &ge 1\
sum_i=1^6 p_i^2 &ge frac16endalign*$$
Equality holds when all the $p_i$ are the same, i.e. when the die is unbiased.
answered 9 mins ago
peterwhypeterwhy
12.3k21229
$endgroup$
add a comment |
$begingroup$
Let $p_i$ be the probability of rolling $i$. Then $sum_i=1^6 p_i = 1$.
By Cauchy-Schwarz inequality,
$$beginalign*
left(sum_i=1^6 1^2right) left(sum_i=1^6 p_i^2right) &ge
left(sum_i=1^6 1p_iright)^2\
6left(sum_i=1^6 p_i^2right) &ge 1\
sum_i=1^6 p_i^2 &ge frac16endalign*$$
Equality holds when all the $p_i$ are the same, i.e. when the die is unbiased.
answered 9 mins ago
peterwhypeterwhy
12.3k21229
$endgroup$
Let $p_i$ be the probability of rolling $i$. Then $sum_i=1^6 p_i = 1$.
By Cauchy-Schwarz inequality,
$$beginalign*
left(sum_i=1^6 1^2right) left(sum_i=1^6 p_i^2right) &ge
left(sum_i=1^6 1p_iright)^2\
6left(sum_i=1^6 p_i^2right) &ge 1\
sum_i=1^6 p_i^2 &ge frac16endalign*$$
Equality holds when all the $p_i$ are the same, i.e. when the die is unbiased.
answered 9 mins ago
peterwhypeterwhy
12.3k21229
answered 9 mins ago
peterwhypeterwhy
12.3k21229
answered 9 mins ago
peterwhypeterwhy
12.3k21229
answered 9 mins ago
peterwhypeterwhy
12.3k21229
12.3k21229
add a comment |
add a comment |
mandy is a new contributor. Be nice, and check out our Code of Conduct.
mandy is a new contributor. Be nice, and check out our Code of Conduct.
mandy is a new contributor. Be nice, and check out our Code of Conduct.
mandy is a new contributor. Be nice, and check out our Code of Conduct.
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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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%2f3188165%2fbiased-dice-probability-question%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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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');
var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);
;
$window.on('scroll', onScroll);
);
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
$begingroup$
Hint: First try to show that if a coin is flipped twice, the probability that the two results are the same is at least $1/2$. This will help you figure out what to choose as $a$.
$endgroup$
– Lorenzo
10 mins ago