A deck of cards is split into $2$ halves of $26$ cards each. What is the probability of one partition containing all $4$ aces?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
0
down vote

favorite












A deck of cards is split into $2$ halves of $26$ cards each. What is the probability of one partition containing all $4$ aces?



Find the probability corresponding to this. It deals with the concepts of partitions which explicitly added twist.







share|cite|improve this question






















  • Please, instead of uploading images, type your question and add what have you tried so far, in order to encourage us to help you.
    – Taroccoesbrocco
    Aug 22 at 9:23











  • Welcome to MathSE. Please type your question rather than posting a link since links may be deleted. This MathJax tutorial explains how to typeset mathematics on this site. Also, when you pose a question here, it is expected that you include your own thoughts on the problem. Please edit your question to show what you have tried and explain where you are stuck.
    – N. F. Taussig
    Aug 22 at 9:24














up vote
0
down vote

favorite












A deck of cards is split into $2$ halves of $26$ cards each. What is the probability of one partition containing all $4$ aces?



Find the probability corresponding to this. It deals with the concepts of partitions which explicitly added twist.







share|cite|improve this question






















  • Please, instead of uploading images, type your question and add what have you tried so far, in order to encourage us to help you.
    – Taroccoesbrocco
    Aug 22 at 9:23











  • Welcome to MathSE. Please type your question rather than posting a link since links may be deleted. This MathJax tutorial explains how to typeset mathematics on this site. Also, when you pose a question here, it is expected that you include your own thoughts on the problem. Please edit your question to show what you have tried and explain where you are stuck.
    – N. F. Taussig
    Aug 22 at 9:24












up vote
0
down vote

favorite









up vote
0
down vote

favorite











A deck of cards is split into $2$ halves of $26$ cards each. What is the probability of one partition containing all $4$ aces?



Find the probability corresponding to this. It deals with the concepts of partitions which explicitly added twist.







share|cite|improve this question














A deck of cards is split into $2$ halves of $26$ cards each. What is the probability of one partition containing all $4$ aces?



Find the probability corresponding to this. It deals with the concepts of partitions which explicitly added twist.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 22 at 9:29









N. F. Taussig

38.8k93153




38.8k93153










asked Aug 22 at 9:20









Sahil Arora

1




1











  • Please, instead of uploading images, type your question and add what have you tried so far, in order to encourage us to help you.
    – Taroccoesbrocco
    Aug 22 at 9:23











  • Welcome to MathSE. Please type your question rather than posting a link since links may be deleted. This MathJax tutorial explains how to typeset mathematics on this site. Also, when you pose a question here, it is expected that you include your own thoughts on the problem. Please edit your question to show what you have tried and explain where you are stuck.
    – N. F. Taussig
    Aug 22 at 9:24
















  • Please, instead of uploading images, type your question and add what have you tried so far, in order to encourage us to help you.
    – Taroccoesbrocco
    Aug 22 at 9:23











  • Welcome to MathSE. Please type your question rather than posting a link since links may be deleted. This MathJax tutorial explains how to typeset mathematics on this site. Also, when you pose a question here, it is expected that you include your own thoughts on the problem. Please edit your question to show what you have tried and explain where you are stuck.
    – N. F. Taussig
    Aug 22 at 9:24















Please, instead of uploading images, type your question and add what have you tried so far, in order to encourage us to help you.
– Taroccoesbrocco
Aug 22 at 9:23





Please, instead of uploading images, type your question and add what have you tried so far, in order to encourage us to help you.
– Taroccoesbrocco
Aug 22 at 9:23













Welcome to MathSE. Please type your question rather than posting a link since links may be deleted. This MathJax tutorial explains how to typeset mathematics on this site. Also, when you pose a question here, it is expected that you include your own thoughts on the problem. Please edit your question to show what you have tried and explain where you are stuck.
– N. F. Taussig
Aug 22 at 9:24




Welcome to MathSE. Please type your question rather than posting a link since links may be deleted. This MathJax tutorial explains how to typeset mathematics on this site. Also, when you pose a question here, it is expected that you include your own thoughts on the problem. Please edit your question to show what you have tried and explain where you are stuck.
– N. F. Taussig
Aug 22 at 9:24










2 Answers
2






active

oldest

votes

















up vote
2
down vote













Hint:



Number the parts: $1,2$.



The probability that part $1$ has all $4$ aces is:




$$fracbinom264binom260binom524$$




It cannot happen that both parts have all $4$ aces.






share|cite|improve this answer




















  • How did you get that "Click to show Spoiler"?
    – Mohammad Zuhair Khan
    Aug 22 at 9:31










  • @MohammadZuhairKhan A spoiler is achieved by starting the line with ">!" and becomes visible if the cursor is on it.
    – drhab
    Aug 22 at 9:34










  • we wont do $binom5226$ ?How do we split it into 2 parts?
    – Damn1o1
    Aug 22 at 9:50











  • @Damn1o1 You can indeed also find it as $fracbinom44binom4822binom5226$. That gives the same outcome.
    – drhab
    Aug 22 at 10:11


















up vote
0
down vote













We can divide the four aces over the two partitions, ignoring the other cards. For the first ace it is irrelevant in which partition it is placed, but the other aces must be in the same partition as the first. So the probability that all aces are in the same partition is given by $$1cdot frac2551cdotfrac2450cdotfrac2349=frac92833approx 0.11.$$






share|cite|improve this answer




















    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',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );








     

    draft saved


    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2890838%2fa-deck-of-cards-is-split-into-2-halves-of-26-cards-each-what-is-the-probabi%23new-answer', 'question_page');

    );

    Post as a guest






























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote













    Hint:



    Number the parts: $1,2$.



    The probability that part $1$ has all $4$ aces is:




    $$fracbinom264binom260binom524$$




    It cannot happen that both parts have all $4$ aces.






    share|cite|improve this answer




















    • How did you get that "Click to show Spoiler"?
      – Mohammad Zuhair Khan
      Aug 22 at 9:31










    • @MohammadZuhairKhan A spoiler is achieved by starting the line with ">!" and becomes visible if the cursor is on it.
      – drhab
      Aug 22 at 9:34










    • we wont do $binom5226$ ?How do we split it into 2 parts?
      – Damn1o1
      Aug 22 at 9:50











    • @Damn1o1 You can indeed also find it as $fracbinom44binom4822binom5226$. That gives the same outcome.
      – drhab
      Aug 22 at 10:11















    up vote
    2
    down vote













    Hint:



    Number the parts: $1,2$.



    The probability that part $1$ has all $4$ aces is:




    $$fracbinom264binom260binom524$$




    It cannot happen that both parts have all $4$ aces.






    share|cite|improve this answer




















    • How did you get that "Click to show Spoiler"?
      – Mohammad Zuhair Khan
      Aug 22 at 9:31










    • @MohammadZuhairKhan A spoiler is achieved by starting the line with ">!" and becomes visible if the cursor is on it.
      – drhab
      Aug 22 at 9:34










    • we wont do $binom5226$ ?How do we split it into 2 parts?
      – Damn1o1
      Aug 22 at 9:50











    • @Damn1o1 You can indeed also find it as $fracbinom44binom4822binom5226$. That gives the same outcome.
      – drhab
      Aug 22 at 10:11













    up vote
    2
    down vote










    up vote
    2
    down vote









    Hint:



    Number the parts: $1,2$.



    The probability that part $1$ has all $4$ aces is:




    $$fracbinom264binom260binom524$$




    It cannot happen that both parts have all $4$ aces.






    share|cite|improve this answer












    Hint:



    Number the parts: $1,2$.



    The probability that part $1$ has all $4$ aces is:




    $$fracbinom264binom260binom524$$




    It cannot happen that both parts have all $4$ aces.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Aug 22 at 9:24









    drhab

    88k541120




    88k541120











    • How did you get that "Click to show Spoiler"?
      – Mohammad Zuhair Khan
      Aug 22 at 9:31










    • @MohammadZuhairKhan A spoiler is achieved by starting the line with ">!" and becomes visible if the cursor is on it.
      – drhab
      Aug 22 at 9:34










    • we wont do $binom5226$ ?How do we split it into 2 parts?
      – Damn1o1
      Aug 22 at 9:50











    • @Damn1o1 You can indeed also find it as $fracbinom44binom4822binom5226$. That gives the same outcome.
      – drhab
      Aug 22 at 10:11

















    • How did you get that "Click to show Spoiler"?
      – Mohammad Zuhair Khan
      Aug 22 at 9:31










    • @MohammadZuhairKhan A spoiler is achieved by starting the line with ">!" and becomes visible if the cursor is on it.
      – drhab
      Aug 22 at 9:34










    • we wont do $binom5226$ ?How do we split it into 2 parts?
      – Damn1o1
      Aug 22 at 9:50











    • @Damn1o1 You can indeed also find it as $fracbinom44binom4822binom5226$. That gives the same outcome.
      – drhab
      Aug 22 at 10:11
















    How did you get that "Click to show Spoiler"?
    – Mohammad Zuhair Khan
    Aug 22 at 9:31




    How did you get that "Click to show Spoiler"?
    – Mohammad Zuhair Khan
    Aug 22 at 9:31












    @MohammadZuhairKhan A spoiler is achieved by starting the line with ">!" and becomes visible if the cursor is on it.
    – drhab
    Aug 22 at 9:34




    @MohammadZuhairKhan A spoiler is achieved by starting the line with ">!" and becomes visible if the cursor is on it.
    – drhab
    Aug 22 at 9:34












    we wont do $binom5226$ ?How do we split it into 2 parts?
    – Damn1o1
    Aug 22 at 9:50





    we wont do $binom5226$ ?How do we split it into 2 parts?
    – Damn1o1
    Aug 22 at 9:50













    @Damn1o1 You can indeed also find it as $fracbinom44binom4822binom5226$. That gives the same outcome.
    – drhab
    Aug 22 at 10:11





    @Damn1o1 You can indeed also find it as $fracbinom44binom4822binom5226$. That gives the same outcome.
    – drhab
    Aug 22 at 10:11











    up vote
    0
    down vote













    We can divide the four aces over the two partitions, ignoring the other cards. For the first ace it is irrelevant in which partition it is placed, but the other aces must be in the same partition as the first. So the probability that all aces are in the same partition is given by $$1cdot frac2551cdotfrac2450cdotfrac2349=frac92833approx 0.11.$$






    share|cite|improve this answer
























      up vote
      0
      down vote













      We can divide the four aces over the two partitions, ignoring the other cards. For the first ace it is irrelevant in which partition it is placed, but the other aces must be in the same partition as the first. So the probability that all aces are in the same partition is given by $$1cdot frac2551cdotfrac2450cdotfrac2349=frac92833approx 0.11.$$






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        We can divide the four aces over the two partitions, ignoring the other cards. For the first ace it is irrelevant in which partition it is placed, but the other aces must be in the same partition as the first. So the probability that all aces are in the same partition is given by $$1cdot frac2551cdotfrac2450cdotfrac2349=frac92833approx 0.11.$$






        share|cite|improve this answer












        We can divide the four aces over the two partitions, ignoring the other cards. For the first ace it is irrelevant in which partition it is placed, but the other aces must be in the same partition as the first. So the probability that all aces are in the same partition is given by $$1cdot frac2551cdotfrac2450cdotfrac2349=frac92833approx 0.11.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 22 at 10:57









        molarmass

        1,327513




        1,327513






















             

            draft saved


            draft discarded


























             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2890838%2fa-deck-of-cards-is-split-into-2-halves-of-26-cards-each-what-is-the-probabi%23new-answer', 'question_page');

            );

            Post as a guest













































































            這個網誌中的熱門文章

            How to combine Bézier curves to a surface?

            Mutual Information Always Non-negative

            Why am i infinitely getting the same tweet with the Twitter Search API?