Probability - Mathematical Expectation

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This is a homework question but I’m not allowed to tag it.



The question is “find the expected number of aces in a poker hand of 5 cards”



Considering this is the chapter on expectancy, I attempted to find e(x) using 1/13 as the probability for getting an ace but im used to doing so using a probability function. The answer in the back of the book is 85/221 but I can’t wrap my head around how to do this problem.



Thanks!




probability




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  • 1




    Let $X$ be the number of aces in a 5 card hand. Then $Xin0,1,2,3,4$. What's the probability of each event occurring?
    – user170231
    Aug 9 at 15:11














up vote
1
down vote

favorite












This is a homework question but I’m not allowed to tag it.



The question is “find the expected number of aces in a poker hand of 5 cards”



Considering this is the chapter on expectancy, I attempted to find e(x) using 1/13 as the probability for getting an ace but im used to doing so using a probability function. The answer in the back of the book is 85/221 but I can’t wrap my head around how to do this problem.



Thanks!




probability




share|cite|improve this question
















  • 1




    Let $X$ be the number of aces in a 5 card hand. Then $Xin0,1,2,3,4$. What's the probability of each event occurring?
    – user170231
    Aug 9 at 15:11












up vote
1
down vote

favorite









up vote
1
down vote

favorite











This is a homework question but I’m not allowed to tag it.



The question is “find the expected number of aces in a poker hand of 5 cards”



Considering this is the chapter on expectancy, I attempted to find e(x) using 1/13 as the probability for getting an ace but im used to doing so using a probability function. The answer in the back of the book is 85/221 but I can’t wrap my head around how to do this problem.



Thanks!




probability




share|cite|improve this question












This is a homework question but I’m not allowed to tag it.



The question is “find the expected number of aces in a poker hand of 5 cards”



Considering this is the chapter on expectancy, I attempted to find e(x) using 1/13 as the probability for getting an ace but im used to doing so using a probability function. The answer in the back of the book is 85/221 but I can’t wrap my head around how to do this problem.



Thanks!





probability





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 9 at 15:07









PSqc

82




82







  • 1




    Let $X$ be the number of aces in a 5 card hand. Then $Xin0,1,2,3,4$. What's the probability of each event occurring?
    – user170231
    Aug 9 at 15:11












  • 1




    Let $X$ be the number of aces in a 5 card hand. Then $Xin0,1,2,3,4$. What's the probability of each event occurring?
    – user170231
    Aug 9 at 15:11







1




1




Let $X$ be the number of aces in a 5 card hand. Then $Xin0,1,2,3,4$. What's the probability of each event occurring?
– user170231
Aug 9 at 15:11




Let $X$ be the number of aces in a 5 card hand. Then $Xin0,1,2,3,4$. What's the probability of each event occurring?
– user170231
Aug 9 at 15:11










3 Answers
3






active

oldest

votes

















up vote
3
down vote



accepted










Give the cards the numbers $1,2,3,4,5$ and for $i=1,dots,5$ let $X_i$ take value $1$ if card $i$ is an ace and value $0$ otherwise.



Then the number of aces is:$$X_1+cdots+X_5$$



Applying linearity of expectation and symmetry we find:$$mathbb E[X_1+cdots+X_5]=mathbb EX_1+cdots+mathbb EX_5=5mathbb EX_1=5mathsf P(X_1=1)=frac513$$






share|cite|improve this answer






















  • This is the best way to think about the problem, and the only answer here that gives an answer; that's important in this case, because the answer that OP states is in the book is incorrect.
    – Marcus M
    Aug 9 at 15:18






  • 1




    @MarcusM Thank you. However $frac85221=frac513$ so the answers match. No idea how factor $17$ got involved.
    – drhab
    Aug 9 at 15:20










  • Oh true, that's quite odd.
    – Marcus M
    Aug 9 at 15:20






  • 1




    As a (minor) variant, let $Y_i$ be the indicator for each ace. The $i^th$ ace is in the hand with probability $frac 552$ so the answer is $frac 2052=frac 513$. I, too, have no idea how they got the extra factor of $17$.
    – lulu
    Aug 9 at 15:59






  • 1




    @lulu Indeed. In my variant a card in hand wonders: "what is my chance to be an ace?" In yours an ace wonders: "what is my chance to be in the hand?"
    – drhab
    Aug 9 at 16:59

















up vote
1
down vote













Hint.



1) What is the total number of hands?



2) What is the number of hands containing



  • one ace?

  • two aces?

  • ...





share|cite|improve this answer



























    up vote
    1
    down vote













    HINT



    You can get 0,1,2,3, or 4 aces. Find the probability and multiply by the number of aces for each of those cases



    For example, let's do 2 aces. This means you get 2 aces and 3 non-aces. The probability of that is:



    $$frac4 choose 2 cdot 48 choose 352 choose 5$$



    And multiply by this by $2$.



    Same for 1,3, and 4 aces (no need to do 0 aces), and add them all up






    share|cite|improve this answer


















    • 1




      I assume you mean $binom483$ rather than $binom393$?
      – Marcus M
      Aug 9 at 15:18










    • @MarcusM Good catch.
      – callculus
      Aug 9 at 15:32










    • @MarcusM Thanks! :)
      – Bram28
      Aug 9 at 15:35










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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    Give the cards the numbers $1,2,3,4,5$ and for $i=1,dots,5$ let $X_i$ take value $1$ if card $i$ is an ace and value $0$ otherwise.



    Then the number of aces is:$$X_1+cdots+X_5$$



    Applying linearity of expectation and symmetry we find:$$mathbb E[X_1+cdots+X_5]=mathbb EX_1+cdots+mathbb EX_5=5mathbb EX_1=5mathsf P(X_1=1)=frac513$$






    share|cite|improve this answer






















    • This is the best way to think about the problem, and the only answer here that gives an answer; that's important in this case, because the answer that OP states is in the book is incorrect.
      – Marcus M
      Aug 9 at 15:18






    • 1




      @MarcusM Thank you. However $frac85221=frac513$ so the answers match. No idea how factor $17$ got involved.
      – drhab
      Aug 9 at 15:20










    • Oh true, that's quite odd.
      – Marcus M
      Aug 9 at 15:20






    • 1




      As a (minor) variant, let $Y_i$ be the indicator for each ace. The $i^th$ ace is in the hand with probability $frac 552$ so the answer is $frac 2052=frac 513$. I, too, have no idea how they got the extra factor of $17$.
      – lulu
      Aug 9 at 15:59






    • 1




      @lulu Indeed. In my variant a card in hand wonders: "what is my chance to be an ace?" In yours an ace wonders: "what is my chance to be in the hand?"
      – drhab
      Aug 9 at 16:59














    up vote
    3
    down vote



    accepted










    Give the cards the numbers $1,2,3,4,5$ and for $i=1,dots,5$ let $X_i$ take value $1$ if card $i$ is an ace and value $0$ otherwise.



    Then the number of aces is:$$X_1+cdots+X_5$$



    Applying linearity of expectation and symmetry we find:$$mathbb E[X_1+cdots+X_5]=mathbb EX_1+cdots+mathbb EX_5=5mathbb EX_1=5mathsf P(X_1=1)=frac513$$






    share|cite|improve this answer






















    • This is the best way to think about the problem, and the only answer here that gives an answer; that's important in this case, because the answer that OP states is in the book is incorrect.
      – Marcus M
      Aug 9 at 15:18






    • 1




      @MarcusM Thank you. However $frac85221=frac513$ so the answers match. No idea how factor $17$ got involved.
      – drhab
      Aug 9 at 15:20










    • Oh true, that's quite odd.
      – Marcus M
      Aug 9 at 15:20






    • 1




      As a (minor) variant, let $Y_i$ be the indicator for each ace. The $i^th$ ace is in the hand with probability $frac 552$ so the answer is $frac 2052=frac 513$. I, too, have no idea how they got the extra factor of $17$.
      – lulu
      Aug 9 at 15:59






    • 1




      @lulu Indeed. In my variant a card in hand wonders: "what is my chance to be an ace?" In yours an ace wonders: "what is my chance to be in the hand?"
      – drhab
      Aug 9 at 16:59












    up vote
    3
    down vote



    accepted







    up vote
    3
    down vote



    accepted






    Give the cards the numbers $1,2,3,4,5$ and for $i=1,dots,5$ let $X_i$ take value $1$ if card $i$ is an ace and value $0$ otherwise.



    Then the number of aces is:$$X_1+cdots+X_5$$



    Applying linearity of expectation and symmetry we find:$$mathbb E[X_1+cdots+X_5]=mathbb EX_1+cdots+mathbb EX_5=5mathbb EX_1=5mathsf P(X_1=1)=frac513$$






    share|cite|improve this answer














    Give the cards the numbers $1,2,3,4,5$ and for $i=1,dots,5$ let $X_i$ take value $1$ if card $i$ is an ace and value $0$ otherwise.



    Then the number of aces is:$$X_1+cdots+X_5$$



    Applying linearity of expectation and symmetry we find:$$mathbb E[X_1+cdots+X_5]=mathbb EX_1+cdots+mathbb EX_5=5mathbb EX_1=5mathsf P(X_1=1)=frac513$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Aug 9 at 15:18

























    answered Aug 9 at 15:14









    drhab

    87.1k541118




    87.1k541118











    • This is the best way to think about the problem, and the only answer here that gives an answer; that's important in this case, because the answer that OP states is in the book is incorrect.
      – Marcus M
      Aug 9 at 15:18






    • 1




      @MarcusM Thank you. However $frac85221=frac513$ so the answers match. No idea how factor $17$ got involved.
      – drhab
      Aug 9 at 15:20










    • Oh true, that's quite odd.
      – Marcus M
      Aug 9 at 15:20






    • 1




      As a (minor) variant, let $Y_i$ be the indicator for each ace. The $i^th$ ace is in the hand with probability $frac 552$ so the answer is $frac 2052=frac 513$. I, too, have no idea how they got the extra factor of $17$.
      – lulu
      Aug 9 at 15:59






    • 1




      @lulu Indeed. In my variant a card in hand wonders: "what is my chance to be an ace?" In yours an ace wonders: "what is my chance to be in the hand?"
      – drhab
      Aug 9 at 16:59
















    • This is the best way to think about the problem, and the only answer here that gives an answer; that's important in this case, because the answer that OP states is in the book is incorrect.
      – Marcus M
      Aug 9 at 15:18






    • 1




      @MarcusM Thank you. However $frac85221=frac513$ so the answers match. No idea how factor $17$ got involved.
      – drhab
      Aug 9 at 15:20










    • Oh true, that's quite odd.
      – Marcus M
      Aug 9 at 15:20






    • 1




      As a (minor) variant, let $Y_i$ be the indicator for each ace. The $i^th$ ace is in the hand with probability $frac 552$ so the answer is $frac 2052=frac 513$. I, too, have no idea how they got the extra factor of $17$.
      – lulu
      Aug 9 at 15:59






    • 1




      @lulu Indeed. In my variant a card in hand wonders: "what is my chance to be an ace?" In yours an ace wonders: "what is my chance to be in the hand?"
      – drhab
      Aug 9 at 16:59















    This is the best way to think about the problem, and the only answer here that gives an answer; that's important in this case, because the answer that OP states is in the book is incorrect.
    – Marcus M
    Aug 9 at 15:18




    This is the best way to think about the problem, and the only answer here that gives an answer; that's important in this case, because the answer that OP states is in the book is incorrect.
    – Marcus M
    Aug 9 at 15:18




    1




    1




    @MarcusM Thank you. However $frac85221=frac513$ so the answers match. No idea how factor $17$ got involved.
    – drhab
    Aug 9 at 15:20




    @MarcusM Thank you. However $frac85221=frac513$ so the answers match. No idea how factor $17$ got involved.
    – drhab
    Aug 9 at 15:20












    Oh true, that's quite odd.
    – Marcus M
    Aug 9 at 15:20




    Oh true, that's quite odd.
    – Marcus M
    Aug 9 at 15:20




    1




    1




    As a (minor) variant, let $Y_i$ be the indicator for each ace. The $i^th$ ace is in the hand with probability $frac 552$ so the answer is $frac 2052=frac 513$. I, too, have no idea how they got the extra factor of $17$.
    – lulu
    Aug 9 at 15:59




    As a (minor) variant, let $Y_i$ be the indicator for each ace. The $i^th$ ace is in the hand with probability $frac 552$ so the answer is $frac 2052=frac 513$. I, too, have no idea how they got the extra factor of $17$.
    – lulu
    Aug 9 at 15:59




    1




    1




    @lulu Indeed. In my variant a card in hand wonders: "what is my chance to be an ace?" In yours an ace wonders: "what is my chance to be in the hand?"
    – drhab
    Aug 9 at 16:59




    @lulu Indeed. In my variant a card in hand wonders: "what is my chance to be an ace?" In yours an ace wonders: "what is my chance to be in the hand?"
    – drhab
    Aug 9 at 16:59










    up vote
    1
    down vote













    Hint.



    1) What is the total number of hands?



    2) What is the number of hands containing



    • one ace?

    • two aces?

    • ...





    share|cite|improve this answer
























      up vote
      1
      down vote













      Hint.



      1) What is the total number of hands?



      2) What is the number of hands containing



      • one ace?

      • two aces?

      • ...





      share|cite|improve this answer






















        up vote
        1
        down vote










        up vote
        1
        down vote









        Hint.



        1) What is the total number of hands?



        2) What is the number of hands containing



        • one ace?

        • two aces?

        • ...





        share|cite|improve this answer












        Hint.



        1) What is the total number of hands?



        2) What is the number of hands containing



        • one ace?

        • two aces?

        • ...






        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 9 at 15:11









        Arnaud Mortier

        19.2k22159




        19.2k22159




















            up vote
            1
            down vote













            HINT



            You can get 0,1,2,3, or 4 aces. Find the probability and multiply by the number of aces for each of those cases



            For example, let's do 2 aces. This means you get 2 aces and 3 non-aces. The probability of that is:



            $$frac4 choose 2 cdot 48 choose 352 choose 5$$



            And multiply by this by $2$.



            Same for 1,3, and 4 aces (no need to do 0 aces), and add them all up






            share|cite|improve this answer


















            • 1




              I assume you mean $binom483$ rather than $binom393$?
              – Marcus M
              Aug 9 at 15:18










            • @MarcusM Good catch.
              – callculus
              Aug 9 at 15:32










            • @MarcusM Thanks! :)
              – Bram28
              Aug 9 at 15:35














            up vote
            1
            down vote













            HINT



            You can get 0,1,2,3, or 4 aces. Find the probability and multiply by the number of aces for each of those cases



            For example, let's do 2 aces. This means you get 2 aces and 3 non-aces. The probability of that is:



            $$frac4 choose 2 cdot 48 choose 352 choose 5$$



            And multiply by this by $2$.



            Same for 1,3, and 4 aces (no need to do 0 aces), and add them all up






            share|cite|improve this answer


















            • 1




              I assume you mean $binom483$ rather than $binom393$?
              – Marcus M
              Aug 9 at 15:18










            • @MarcusM Good catch.
              – callculus
              Aug 9 at 15:32










            • @MarcusM Thanks! :)
              – Bram28
              Aug 9 at 15:35












            up vote
            1
            down vote










            up vote
            1
            down vote









            HINT



            You can get 0,1,2,3, or 4 aces. Find the probability and multiply by the number of aces for each of those cases



            For example, let's do 2 aces. This means you get 2 aces and 3 non-aces. The probability of that is:



            $$frac4 choose 2 cdot 48 choose 352 choose 5$$



            And multiply by this by $2$.



            Same for 1,3, and 4 aces (no need to do 0 aces), and add them all up






            share|cite|improve this answer














            HINT



            You can get 0,1,2,3, or 4 aces. Find the probability and multiply by the number of aces for each of those cases



            For example, let's do 2 aces. This means you get 2 aces and 3 non-aces. The probability of that is:



            $$frac4 choose 2 cdot 48 choose 352 choose 5$$



            And multiply by this by $2$.



            Same for 1,3, and 4 aces (no need to do 0 aces), and add them all up







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 9 at 15:21









            InterstellarProbe

            2,542619




            2,542619










            answered Aug 9 at 15:13









            Bram28

            55.2k33982




            55.2k33982







            • 1




              I assume you mean $binom483$ rather than $binom393$?
              – Marcus M
              Aug 9 at 15:18










            • @MarcusM Good catch.
              – callculus
              Aug 9 at 15:32










            • @MarcusM Thanks! :)
              – Bram28
              Aug 9 at 15:35












            • 1




              I assume you mean $binom483$ rather than $binom393$?
              – Marcus M
              Aug 9 at 15:18










            • @MarcusM Good catch.
              – callculus
              Aug 9 at 15:32










            • @MarcusM Thanks! :)
              – Bram28
              Aug 9 at 15:35







            1




            1




            I assume you mean $binom483$ rather than $binom393$?
            – Marcus M
            Aug 9 at 15:18




            I assume you mean $binom483$ rather than $binom393$?
            – Marcus M
            Aug 9 at 15:18












            @MarcusM Good catch.
            – callculus
            Aug 9 at 15:32




            @MarcusM Good catch.
            – callculus
            Aug 9 at 15:32












            @MarcusM Thanks! :)
            – Bram28
            Aug 9 at 15:35




            @MarcusM Thanks! :)
            – Bram28
            Aug 9 at 15:35












             

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