Estimate number of cards needed to be drawn from deck before full house












3














Imagine we have a well-shuffled deck of cards and we keep drawing cards until there is at least one full house in the drawn cards. How many cards will we draw on average? I would be interested in both the exact solution (which is something around 12) but, more importantly, in a good quick way of estimating this value.










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  • If this is too hard, I would be interested in the same question but for other simple poker hands like a single pair or a three-of-a-kind
    – user132290
    Dec 14 '18 at 23:48










  • And the value of 12 quoted above comes from my computer simulations.
    – user132290
    Dec 15 '18 at 12:13










  • My computer simulation gives an average of $13.9$ with a standard deviation of $3.7$.
    – Jens
    Dec 15 '18 at 13:13










  • @Jens Indeed. The exact expectation I got is $13.893157dots$, which is amazingly close to your simulation result. The interesting question is how to get it with a back of the envelope calculation (I just honestly wrote the combinatorial sum and told my PC to compute it)..
    – fedja
    Dec 15 '18 at 17:39










  • How do you guys simulate this? My simulation gives 13.55 after 1 million runs. Do you also actually simulate a deck of cards, or do you calculate the probability somehow?
    – user132290
    Dec 17 '18 at 21:51
















3














Imagine we have a well-shuffled deck of cards and we keep drawing cards until there is at least one full house in the drawn cards. How many cards will we draw on average? I would be interested in both the exact solution (which is something around 12) but, more importantly, in a good quick way of estimating this value.










share|cite|improve this question






















  • If this is too hard, I would be interested in the same question but for other simple poker hands like a single pair or a three-of-a-kind
    – user132290
    Dec 14 '18 at 23:48










  • And the value of 12 quoted above comes from my computer simulations.
    – user132290
    Dec 15 '18 at 12:13










  • My computer simulation gives an average of $13.9$ with a standard deviation of $3.7$.
    – Jens
    Dec 15 '18 at 13:13










  • @Jens Indeed. The exact expectation I got is $13.893157dots$, which is amazingly close to your simulation result. The interesting question is how to get it with a back of the envelope calculation (I just honestly wrote the combinatorial sum and told my PC to compute it)..
    – fedja
    Dec 15 '18 at 17:39










  • How do you guys simulate this? My simulation gives 13.55 after 1 million runs. Do you also actually simulate a deck of cards, or do you calculate the probability somehow?
    – user132290
    Dec 17 '18 at 21:51














3












3








3


2





Imagine we have a well-shuffled deck of cards and we keep drawing cards until there is at least one full house in the drawn cards. How many cards will we draw on average? I would be interested in both the exact solution (which is something around 12) but, more importantly, in a good quick way of estimating this value.










share|cite|improve this question













Imagine we have a well-shuffled deck of cards and we keep drawing cards until there is at least one full house in the drawn cards. How many cards will we draw on average? I would be interested in both the exact solution (which is something around 12) but, more importantly, in a good quick way of estimating this value.







probability puzzle card-games






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share|cite|improve this question




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asked Dec 14 '18 at 23:46









user132290user132290

536




536












  • If this is too hard, I would be interested in the same question but for other simple poker hands like a single pair or a three-of-a-kind
    – user132290
    Dec 14 '18 at 23:48










  • And the value of 12 quoted above comes from my computer simulations.
    – user132290
    Dec 15 '18 at 12:13










  • My computer simulation gives an average of $13.9$ with a standard deviation of $3.7$.
    – Jens
    Dec 15 '18 at 13:13










  • @Jens Indeed. The exact expectation I got is $13.893157dots$, which is amazingly close to your simulation result. The interesting question is how to get it with a back of the envelope calculation (I just honestly wrote the combinatorial sum and told my PC to compute it)..
    – fedja
    Dec 15 '18 at 17:39










  • How do you guys simulate this? My simulation gives 13.55 after 1 million runs. Do you also actually simulate a deck of cards, or do you calculate the probability somehow?
    – user132290
    Dec 17 '18 at 21:51


















  • If this is too hard, I would be interested in the same question but for other simple poker hands like a single pair or a three-of-a-kind
    – user132290
    Dec 14 '18 at 23:48










  • And the value of 12 quoted above comes from my computer simulations.
    – user132290
    Dec 15 '18 at 12:13










  • My computer simulation gives an average of $13.9$ with a standard deviation of $3.7$.
    – Jens
    Dec 15 '18 at 13:13










  • @Jens Indeed. The exact expectation I got is $13.893157dots$, which is amazingly close to your simulation result. The interesting question is how to get it with a back of the envelope calculation (I just honestly wrote the combinatorial sum and told my PC to compute it)..
    – fedja
    Dec 15 '18 at 17:39










  • How do you guys simulate this? My simulation gives 13.55 after 1 million runs. Do you also actually simulate a deck of cards, or do you calculate the probability somehow?
    – user132290
    Dec 17 '18 at 21:51
















If this is too hard, I would be interested in the same question but for other simple poker hands like a single pair or a three-of-a-kind
– user132290
Dec 14 '18 at 23:48




If this is too hard, I would be interested in the same question but for other simple poker hands like a single pair or a three-of-a-kind
– user132290
Dec 14 '18 at 23:48












And the value of 12 quoted above comes from my computer simulations.
– user132290
Dec 15 '18 at 12:13




And the value of 12 quoted above comes from my computer simulations.
– user132290
Dec 15 '18 at 12:13












My computer simulation gives an average of $13.9$ with a standard deviation of $3.7$.
– Jens
Dec 15 '18 at 13:13




My computer simulation gives an average of $13.9$ with a standard deviation of $3.7$.
– Jens
Dec 15 '18 at 13:13












@Jens Indeed. The exact expectation I got is $13.893157dots$, which is amazingly close to your simulation result. The interesting question is how to get it with a back of the envelope calculation (I just honestly wrote the combinatorial sum and told my PC to compute it)..
– fedja
Dec 15 '18 at 17:39




@Jens Indeed. The exact expectation I got is $13.893157dots$, which is amazingly close to your simulation result. The interesting question is how to get it with a back of the envelope calculation (I just honestly wrote the combinatorial sum and told my PC to compute it)..
– fedja
Dec 15 '18 at 17:39












How do you guys simulate this? My simulation gives 13.55 after 1 million runs. Do you also actually simulate a deck of cards, or do you calculate the probability somehow?
– user132290
Dec 17 '18 at 21:51




How do you guys simulate this? My simulation gives 13.55 after 1 million runs. Do you also actually simulate a deck of cards, or do you calculate the probability somehow?
– user132290
Dec 17 '18 at 21:51










1 Answer
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A simpler question to address is, having drawn n cards, what is the probability of having a full house. Calculating this is equivalent to counting the number of possible hands with a full house, and then dividing my total possible hands.



To get this, we can count the invalid hands through casework, and then subtract this from the total possible hands:



Case 1 - no doubles/ triples:



$$ 4^n cdot binom{13}{n} $$



Case 2 - a triple/quadruple:



$$ 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) $$



Case 3 - having k doubles:



$$ binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} $$



Thus, the chance that we get a straight flush drawing n cards is:



$$ frac{binom{52}{n}- left( 4^n binom{13}{n} + 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) + sum_{k=1}^n binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} right)}{binom{52}{n}}$$



Putting this into WolframAlpha, we get these values:



enter image description here



enter image description here



Now, this doesn't account from when it happens, only how likely it is to happen after a certain number of draws. To get a closer number on that, you'd have to do some stricter case work, involving ordering.



Essentially, just calculate the number of ordered decks which will has a royal flush at exactly the $n$-th card.



Let $f(n)$ be the number of orderings of $n$ cards containing a royal flush. Using the methods of before, but not diving by $binom{52}{n}$ we get $f(n)$. From here, we can calculate exactly how many decks get a royal flush in exactly the $n$-th draw. We shall denote this number as g(n).



$$ g(n) = f(n)(52-n)! - g(n-1) $$



Then, simply take the sum:



$$ sum_{n=1}^{52} frac{n cdot g(n)}{52!} $$



Which, unfortunately, I am too lazy to evaluate... consider it an exercise :)






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    1 Answer
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    1 Answer
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    A simpler question to address is, having drawn n cards, what is the probability of having a full house. Calculating this is equivalent to counting the number of possible hands with a full house, and then dividing my total possible hands.



    To get this, we can count the invalid hands through casework, and then subtract this from the total possible hands:



    Case 1 - no doubles/ triples:



    $$ 4^n cdot binom{13}{n} $$



    Case 2 - a triple/quadruple:



    $$ 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) $$



    Case 3 - having k doubles:



    $$ binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} $$



    Thus, the chance that we get a straight flush drawing n cards is:



    $$ frac{binom{52}{n}- left( 4^n binom{13}{n} + 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) + sum_{k=1}^n binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} right)}{binom{52}{n}}$$



    Putting this into WolframAlpha, we get these values:



    enter image description here



    enter image description here



    Now, this doesn't account from when it happens, only how likely it is to happen after a certain number of draws. To get a closer number on that, you'd have to do some stricter case work, involving ordering.



    Essentially, just calculate the number of ordered decks which will has a royal flush at exactly the $n$-th card.



    Let $f(n)$ be the number of orderings of $n$ cards containing a royal flush. Using the methods of before, but not diving by $binom{52}{n}$ we get $f(n)$. From here, we can calculate exactly how many decks get a royal flush in exactly the $n$-th draw. We shall denote this number as g(n).



    $$ g(n) = f(n)(52-n)! - g(n-1) $$



    Then, simply take the sum:



    $$ sum_{n=1}^{52} frac{n cdot g(n)}{52!} $$



    Which, unfortunately, I am too lazy to evaluate... consider it an exercise :)






    share|cite|improve this answer








    New contributor




    Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      0














      A simpler question to address is, having drawn n cards, what is the probability of having a full house. Calculating this is equivalent to counting the number of possible hands with a full house, and then dividing my total possible hands.



      To get this, we can count the invalid hands through casework, and then subtract this from the total possible hands:



      Case 1 - no doubles/ triples:



      $$ 4^n cdot binom{13}{n} $$



      Case 2 - a triple/quadruple:



      $$ 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) $$



      Case 3 - having k doubles:



      $$ binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} $$



      Thus, the chance that we get a straight flush drawing n cards is:



      $$ frac{binom{52}{n}- left( 4^n binom{13}{n} + 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) + sum_{k=1}^n binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} right)}{binom{52}{n}}$$



      Putting this into WolframAlpha, we get these values:



      enter image description here



      enter image description here



      Now, this doesn't account from when it happens, only how likely it is to happen after a certain number of draws. To get a closer number on that, you'd have to do some stricter case work, involving ordering.



      Essentially, just calculate the number of ordered decks which will has a royal flush at exactly the $n$-th card.



      Let $f(n)$ be the number of orderings of $n$ cards containing a royal flush. Using the methods of before, but not diving by $binom{52}{n}$ we get $f(n)$. From here, we can calculate exactly how many decks get a royal flush in exactly the $n$-th draw. We shall denote this number as g(n).



      $$ g(n) = f(n)(52-n)! - g(n-1) $$



      Then, simply take the sum:



      $$ sum_{n=1}^{52} frac{n cdot g(n)}{52!} $$



      Which, unfortunately, I am too lazy to evaluate... consider it an exercise :)






      share|cite|improve this answer








      New contributor




      Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





















        0












        0








        0






        A simpler question to address is, having drawn n cards, what is the probability of having a full house. Calculating this is equivalent to counting the number of possible hands with a full house, and then dividing my total possible hands.



        To get this, we can count the invalid hands through casework, and then subtract this from the total possible hands:



        Case 1 - no doubles/ triples:



        $$ 4^n cdot binom{13}{n} $$



        Case 2 - a triple/quadruple:



        $$ 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) $$



        Case 3 - having k doubles:



        $$ binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} $$



        Thus, the chance that we get a straight flush drawing n cards is:



        $$ frac{binom{52}{n}- left( 4^n binom{13}{n} + 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) + sum_{k=1}^n binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} right)}{binom{52}{n}}$$



        Putting this into WolframAlpha, we get these values:



        enter image description here



        enter image description here



        Now, this doesn't account from when it happens, only how likely it is to happen after a certain number of draws. To get a closer number on that, you'd have to do some stricter case work, involving ordering.



        Essentially, just calculate the number of ordered decks which will has a royal flush at exactly the $n$-th card.



        Let $f(n)$ be the number of orderings of $n$ cards containing a royal flush. Using the methods of before, but not diving by $binom{52}{n}$ we get $f(n)$. From here, we can calculate exactly how many decks get a royal flush in exactly the $n$-th draw. We shall denote this number as g(n).



        $$ g(n) = f(n)(52-n)! - g(n-1) $$



        Then, simply take the sum:



        $$ sum_{n=1}^{52} frac{n cdot g(n)}{52!} $$



        Which, unfortunately, I am too lazy to evaluate... consider it an exercise :)






        share|cite|improve this answer








        New contributor




        Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        A simpler question to address is, having drawn n cards, what is the probability of having a full house. Calculating this is equivalent to counting the number of possible hands with a full house, and then dividing my total possible hands.



        To get this, we can count the invalid hands through casework, and then subtract this from the total possible hands:



        Case 1 - no doubles/ triples:



        $$ 4^n cdot binom{13}{n} $$



        Case 2 - a triple/quadruple:



        $$ 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) $$



        Case 3 - having k doubles:



        $$ binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} $$



        Thus, the chance that we get a straight flush drawing n cards is:



        $$ frac{binom{52}{n}- left( 4^n binom{13}{n} + 13 cdot left(4 cdot 4^{n-3} binom{12}{n-3} + 4^{n-4} binom{12}{n-4}right) + sum_{k=1}^n binom{13}{k} binom{13-k}{n-2k} binom{4}{2}^k 4^{n-2k} right)}{binom{52}{n}}$$



        Putting this into WolframAlpha, we get these values:



        enter image description here



        enter image description here



        Now, this doesn't account from when it happens, only how likely it is to happen after a certain number of draws. To get a closer number on that, you'd have to do some stricter case work, involving ordering.



        Essentially, just calculate the number of ordered decks which will has a royal flush at exactly the $n$-th card.



        Let $f(n)$ be the number of orderings of $n$ cards containing a royal flush. Using the methods of before, but not diving by $binom{52}{n}$ we get $f(n)$. From here, we can calculate exactly how many decks get a royal flush in exactly the $n$-th draw. We shall denote this number as g(n).



        $$ g(n) = f(n)(52-n)! - g(n-1) $$



        Then, simply take the sum:



        $$ sum_{n=1}^{52} frac{n cdot g(n)}{52!} $$



        Which, unfortunately, I am too lazy to evaluate... consider it an exercise :)







        share|cite|improve this answer








        New contributor




        Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        share|cite|improve this answer



        share|cite|improve this answer






        New contributor




        Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        answered Jan 4 at 4:29









        Zachary HunterZachary Hunter

        53110




        53110




        New contributor




        Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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        New contributor





        Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        Zachary Hunter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






























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