$n$ players are each dealt two cards — what's the probability that $k$ of them have a pair?












3












$begingroup$


Each of $nleq26$ players is dealt $2$ cards from a standard $52$-card poker deck. What is $textrm{P}left(n,kright)$, the probability that exactly $k$ of the $n$ players have a pair?



(A pair is a hand like $8 clubsuit, 8 heartsuit$ or $K clubsuit, K diamondsuit$.)




This question was previously asked at Poker.SE without a satisfactory answer: https://poker.stackexchange.com/questions/4087/probability-of-x-pocket-pairs-at-a-table-of-n-people-nlhe






probability combinatorics







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$endgroup$












  • $begingroup$
    What is a pair?
    $endgroup$
    – Haran
    Dec 27 '18 at 9:03










  • $begingroup$
    So any $2$ of the cards in all the $4$ cards of ace for example are a pair?
    $endgroup$
    – Haran
    Dec 27 '18 at 9:05










  • $begingroup$
    @Haran: Affirmative.
    $endgroup$
    – user1180576
    Dec 27 '18 at 9:10










  • $begingroup$
    Right now I do not believe in the existence of an elegant solution and a closed formula. It has happpened before though that I was wrong in a sortlike disbelief.
    $endgroup$
    – drhab
    Dec 27 '18 at 9:27


















3












$begingroup$


Each of $nleq26$ players is dealt $2$ cards from a standard $52$-card poker deck. What is $textrm{P}left(n,kright)$, the probability that exactly $k$ of the $n$ players have a pair?



(A pair is a hand like $8 clubsuit, 8 heartsuit$ or $K clubsuit, K diamondsuit$.)




This question was previously asked at Poker.SE without a satisfactory answer: https://poker.stackexchange.com/questions/4087/probability-of-x-pocket-pairs-at-a-table-of-n-people-nlhe






probability combinatorics







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is a pair?
    $endgroup$
    – Haran
    Dec 27 '18 at 9:03










  • $begingroup$
    So any $2$ of the cards in all the $4$ cards of ace for example are a pair?
    $endgroup$
    – Haran
    Dec 27 '18 at 9:05










  • $begingroup$
    @Haran: Affirmative.
    $endgroup$
    – user1180576
    Dec 27 '18 at 9:10










  • $begingroup$
    Right now I do not believe in the existence of an elegant solution and a closed formula. It has happpened before though that I was wrong in a sortlike disbelief.
    $endgroup$
    – drhab
    Dec 27 '18 at 9:27
















3












3








3


1



$begingroup$


Each of $nleq26$ players is dealt $2$ cards from a standard $52$-card poker deck. What is $textrm{P}left(n,kright)$, the probability that exactly $k$ of the $n$ players have a pair?



(A pair is a hand like $8 clubsuit, 8 heartsuit$ or $K clubsuit, K diamondsuit$.)




This question was previously asked at Poker.SE without a satisfactory answer: https://poker.stackexchange.com/questions/4087/probability-of-x-pocket-pairs-at-a-table-of-n-people-nlhe






probability combinatorics







share|cite|improve this question











$endgroup$




Each of $nleq26$ players is dealt $2$ cards from a standard $52$-card poker deck. What is $textrm{P}left(n,kright)$, the probability that exactly $k$ of the $n$ players have a pair?



(A pair is a hand like $8 clubsuit, 8 heartsuit$ or $K clubsuit, K diamondsuit$.)




This question was previously asked at Poker.SE without a satisfactory answer: https://poker.stackexchange.com/questions/4087/probability-of-x-pocket-pairs-at-a-table-of-n-people-nlhe






probability combinatorics




probability combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 9:06







user1180576

















asked Dec 27 '18 at 8:58









user1180576user1180576

244




244












  • $begingroup$
    What is a pair?
    $endgroup$
    – Haran
    Dec 27 '18 at 9:03










  • $begingroup$
    So any $2$ of the cards in all the $4$ cards of ace for example are a pair?
    $endgroup$
    – Haran
    Dec 27 '18 at 9:05










  • $begingroup$
    @Haran: Affirmative.
    $endgroup$
    – user1180576
    Dec 27 '18 at 9:10










  • $begingroup$
    Right now I do not believe in the existence of an elegant solution and a closed formula. It has happpened before though that I was wrong in a sortlike disbelief.
    $endgroup$
    – drhab
    Dec 27 '18 at 9:27




















  • $begingroup$
    What is a pair?
    $endgroup$
    – Haran
    Dec 27 '18 at 9:03










  • $begingroup$
    So any $2$ of the cards in all the $4$ cards of ace for example are a pair?
    $endgroup$
    – Haran
    Dec 27 '18 at 9:05










  • $begingroup$
    @Haran: Affirmative.
    $endgroup$
    – user1180576
    Dec 27 '18 at 9:10










  • $begingroup$
    Right now I do not believe in the existence of an elegant solution and a closed formula. It has happpened before though that I was wrong in a sortlike disbelief.
    $endgroup$
    – drhab
    Dec 27 '18 at 9:27


















$begingroup$
What is a pair?
$endgroup$
– Haran
Dec 27 '18 at 9:03




$begingroup$
What is a pair?
$endgroup$
– Haran
Dec 27 '18 at 9:03












$begingroup$
So any $2$ of the cards in all the $4$ cards of ace for example are a pair?
$endgroup$
– Haran
Dec 27 '18 at 9:05




$begingroup$
So any $2$ of the cards in all the $4$ cards of ace for example are a pair?
$endgroup$
– Haran
Dec 27 '18 at 9:05












$begingroup$
@Haran: Affirmative.
$endgroup$
– user1180576
Dec 27 '18 at 9:10




$begingroup$
@Haran: Affirmative.
$endgroup$
– user1180576
Dec 27 '18 at 9:10












$begingroup$
Right now I do not believe in the existence of an elegant solution and a closed formula. It has happpened before though that I was wrong in a sortlike disbelief.
$endgroup$
– drhab
Dec 27 '18 at 9:27






$begingroup$
Right now I do not believe in the existence of an elegant solution and a closed formula. It has happpened before though that I was wrong in a sortlike disbelief.
$endgroup$
– drhab
Dec 27 '18 at 9:27












2 Answers
2






active

oldest

votes


















1












$begingroup$

First, let us compute $f(m)$, the number of ways to deal two cards to each of $m$ people which are all pairs. We can do this using a computer as follows. If we further let $f(m,r)$ be the probability that all $m$ people have pairs when dealt from a deck of $4r$ cards, then
$$
f(m,r) = rbinom42big(f(m-1,r-1)+(m-1)f(m-2,r-1)big)
$$
The first summand accounts for ways where the first player gets a pair that no one else has, and the second accounts for ways where some other player also gets that pair. The above recursive equation allows you to compute $f(m)=f(m,13)$ quickly using dynamic programming.



Next, use the generalized principle of inclusion exclusion. Letting $E_i$ be the event that the $i^{th}$ player has a pair, then
begin{align}
P(n,k)
&=sum_{i=0}^n(-1)^{i-k}binom{i}{k}binom{n}{i}mathbb P(E_1cap E_2cap dots cap E_i)\
&=sum_{i=0}^n(-1)^{i-k}binom{i}{k}binom{n}{i}frac{f(i)}{frac{52!}{2^i(52-2i)!}}
end{align}



You can see this in action here, just click the run button at the top and enter your desired value of $n$.




Also, there is the following "closed form" for $P(n,k)$:



$$
boxed{P(n,k) = binom{n}{k}sum_{i=k}^n(-1)^{i-k}binom{n-k}{i-k}frac{2^i(52-2i)!}{52!}cdot i!cdot [x^i](1+6x+3x^2)^{13}}
$$
Here, the notation $[x^i]f(x)$ means the coefficient of $x^i$ in the polynomial $f(x)$.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    The probability that a particular player is dealt a pair is $frac{4-1}{52-1}=frac{1}{17}$ so the expected number of pairs is $frac{n}{17}$



    I suspect the distribution is not going to be far away from Binomial with parameters $n$ and $p=frac{1}{17}$. This is not quite correct (for example when $n=26$ the probability of exactly $25$ pairs is $0$ rather than $4.24times 10^{−30}$) but I would be surprised if it were substantially misleading



    As an example, here are a couple of simulations of $100000$ million deals to $6$ players 



    Pairs                    0        1         2         3         4         5         6 
    
Simulation 1            69459    26065     4094      367       13         2         0
    
Simulation 2            69675    25918     4063      329       15         0         0
    
Binomial probability  0.695066 0.260650 0.040727 0.003394 0.000159 0.0000040 0.00000004


    and, given the uncertainty of simulation, these do not look noticeably different. The five pocket pairs out of six seen at the link in your link was indeed unlikely






    share|cite|improve this answer











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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

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      1












      $begingroup$

      First, let us compute $f(m)$, the number of ways to deal two cards to each of $m$ people which are all pairs. We can do this using a computer as follows. If we further let $f(m,r)$ be the probability that all $m$ people have pairs when dealt from a deck of $4r$ cards, then
      $$
      f(m,r) = rbinom42big(f(m-1,r-1)+(m-1)f(m-2,r-1)big)
      $$
      The first summand accounts for ways where the first player gets a pair that no one else has, and the second accounts for ways where some other player also gets that pair. The above recursive equation allows you to compute $f(m)=f(m,13)$ quickly using dynamic programming.



      Next, use the generalized principle of inclusion exclusion. Letting $E_i$ be the event that the $i^{th}$ player has a pair, then
      begin{align}
      P(n,k)
      &=sum_{i=0}^n(-1)^{i-k}binom{i}{k}binom{n}{i}mathbb P(E_1cap E_2cap dots cap E_i)\
      &=sum_{i=0}^n(-1)^{i-k}binom{i}{k}binom{n}{i}frac{f(i)}{frac{52!}{2^i(52-2i)!}}
      end{align}



      You can see this in action here, just click the run button at the top and enter your desired value of $n$.




      Also, there is the following "closed form" for $P(n,k)$:



      $$
      boxed{P(n,k) = binom{n}{k}sum_{i=k}^n(-1)^{i-k}binom{n-k}{i-k}frac{2^i(52-2i)!}{52!}cdot i!cdot [x^i](1+6x+3x^2)^{13}}
      $$
      Here, the notation $[x^i]f(x)$ means the coefficient of $x^i$ in the polynomial $f(x)$.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        First, let us compute $f(m)$, the number of ways to deal two cards to each of $m$ people which are all pairs. We can do this using a computer as follows. If we further let $f(m,r)$ be the probability that all $m$ people have pairs when dealt from a deck of $4r$ cards, then
        $$
        f(m,r) = rbinom42big(f(m-1,r-1)+(m-1)f(m-2,r-1)big)
        $$
        The first summand accounts for ways where the first player gets a pair that no one else has, and the second accounts for ways where some other player also gets that pair. The above recursive equation allows you to compute $f(m)=f(m,13)$ quickly using dynamic programming.



        Next, use the generalized principle of inclusion exclusion. Letting $E_i$ be the event that the $i^{th}$ player has a pair, then
        begin{align}
        P(n,k)
        &=sum_{i=0}^n(-1)^{i-k}binom{i}{k}binom{n}{i}mathbb P(E_1cap E_2cap dots cap E_i)\
        &=sum_{i=0}^n(-1)^{i-k}binom{i}{k}binom{n}{i}frac{f(i)}{frac{52!}{2^i(52-2i)!}}
        end{align}



        You can see this in action here, just click the run button at the top and enter your desired value of $n$.




        Also, there is the following "closed form" for $P(n,k)$:



        $$
        boxed{P(n,k) = binom{n}{k}sum_{i=k}^n(-1)^{i-k}binom{n-k}{i-k}frac{2^i(52-2i)!}{52!}cdot i!cdot [x^i](1+6x+3x^2)^{13}}
        $$
        Here, the notation $[x^i]f(x)$ means the coefficient of $x^i$ in the polynomial $f(x)$.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          First, let us compute $f(m)$, the number of ways to deal two cards to each of $m$ people which are all pairs. We can do this using a computer as follows. If we further let $f(m,r)$ be the probability that all $m$ people have pairs when dealt from a deck of $4r$ cards, then
          $$
          f(m,r) = rbinom42big(f(m-1,r-1)+(m-1)f(m-2,r-1)big)
          $$
          The first summand accounts for ways where the first player gets a pair that no one else has, and the second accounts for ways where some other player also gets that pair. The above recursive equation allows you to compute $f(m)=f(m,13)$ quickly using dynamic programming.



          Next, use the generalized principle of inclusion exclusion. Letting $E_i$ be the event that the $i^{th}$ player has a pair, then
          begin{align}
          P(n,k)
          &=sum_{i=0}^n(-1)^{i-k}binom{i}{k}binom{n}{i}mathbb P(E_1cap E_2cap dots cap E_i)\
          &=sum_{i=0}^n(-1)^{i-k}binom{i}{k}binom{n}{i}frac{f(i)}{frac{52!}{2^i(52-2i)!}}
          end{align}



          You can see this in action here, just click the run button at the top and enter your desired value of $n$.




          Also, there is the following "closed form" for $P(n,k)$:



          $$
          boxed{P(n,k) = binom{n}{k}sum_{i=k}^n(-1)^{i-k}binom{n-k}{i-k}frac{2^i(52-2i)!}{52!}cdot i!cdot [x^i](1+6x+3x^2)^{13}}
          $$
          Here, the notation $[x^i]f(x)$ means the coefficient of $x^i$ in the polynomial $f(x)$.






          share|cite|improve this answer











          $endgroup$



          First, let us compute $f(m)$, the number of ways to deal two cards to each of $m$ people which are all pairs. We can do this using a computer as follows. If we further let $f(m,r)$ be the probability that all $m$ people have pairs when dealt from a deck of $4r$ cards, then
          $$
          f(m,r) = rbinom42big(f(m-1,r-1)+(m-1)f(m-2,r-1)big)
          $$
          The first summand accounts for ways where the first player gets a pair that no one else has, and the second accounts for ways where some other player also gets that pair. The above recursive equation allows you to compute $f(m)=f(m,13)$ quickly using dynamic programming.



          Next, use the generalized principle of inclusion exclusion. Letting $E_i$ be the event that the $i^{th}$ player has a pair, then
          begin{align}
          P(n,k)
          &=sum_{i=0}^n(-1)^{i-k}binom{i}{k}binom{n}{i}mathbb P(E_1cap E_2cap dots cap E_i)\
          &=sum_{i=0}^n(-1)^{i-k}binom{i}{k}binom{n}{i}frac{f(i)}{frac{52!}{2^i(52-2i)!}}
          end{align}



          You can see this in action here, just click the run button at the top and enter your desired value of $n$.




          Also, there is the following "closed form" for $P(n,k)$:



          $$
          boxed{P(n,k) = binom{n}{k}sum_{i=k}^n(-1)^{i-k}binom{n-k}{i-k}frac{2^i(52-2i)!}{52!}cdot i!cdot [x^i](1+6x+3x^2)^{13}}
          $$
          Here, the notation $[x^i]f(x)$ means the coefficient of $x^i$ in the polynomial $f(x)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 29 '18 at 22:43

























          answered Dec 28 '18 at 14:33









          Mike EarnestMike Earnest

          27.8k22152




          27.8k22152























              0












              $begingroup$

              The probability that a particular player is dealt a pair is $frac{4-1}{52-1}=frac{1}{17}$ so the expected number of pairs is $frac{n}{17}$



              I suspect the distribution is not going to be far away from Binomial with parameters $n$ and $p=frac{1}{17}$. This is not quite correct (for example when $n=26$ the probability of exactly $25$ pairs is $0$ rather than $4.24times 10^{−30}$) but I would be surprised if it were substantially misleading



              As an example, here are a couple of simulations of $100000$ million deals to $6$ players 



              Pairs                    0        1         2         3         4         5         6 
              
Simulation 1            69459    26065     4094      367       13         2         0
              
Simulation 2            69675    25918     4063      329       15         0         0
              
Binomial probability  0.695066 0.260650 0.040727 0.003394 0.000159 0.0000040 0.00000004


              and, given the uncertainty of simulation, these do not look noticeably different. The five pocket pairs out of six seen at the link in your link was indeed unlikely






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                The probability that a particular player is dealt a pair is $frac{4-1}{52-1}=frac{1}{17}$ so the expected number of pairs is $frac{n}{17}$



                I suspect the distribution is not going to be far away from Binomial with parameters $n$ and $p=frac{1}{17}$. This is not quite correct (for example when $n=26$ the probability of exactly $25$ pairs is $0$ rather than $4.24times 10^{−30}$) but I would be surprised if it were substantially misleading



                As an example, here are a couple of simulations of $100000$ million deals to $6$ players 



                Pairs                    0        1         2         3         4         5         6 
                
Simulation 1            69459    26065     4094      367       13         2         0
                
Simulation 2            69675    25918     4063      329       15         0         0
                
Binomial probability  0.695066 0.260650 0.040727 0.003394 0.000159 0.0000040 0.00000004


                and, given the uncertainty of simulation, these do not look noticeably different. The five pocket pairs out of six seen at the link in your link was indeed unlikely






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The probability that a particular player is dealt a pair is $frac{4-1}{52-1}=frac{1}{17}$ so the expected number of pairs is $frac{n}{17}$



                  I suspect the distribution is not going to be far away from Binomial with parameters $n$ and $p=frac{1}{17}$. This is not quite correct (for example when $n=26$ the probability of exactly $25$ pairs is $0$ rather than $4.24times 10^{−30}$) but I would be surprised if it were substantially misleading



                  As an example, here are a couple of simulations of $100000$ million deals to $6$ players 



                  Pairs                    0        1         2         3         4         5         6 
                  
Simulation 1            69459    26065     4094      367       13         2         0
                  
Simulation 2            69675    25918     4063      329       15         0         0
                  
Binomial probability  0.695066 0.260650 0.040727 0.003394 0.000159 0.0000040 0.00000004


                  and, given the uncertainty of simulation, these do not look noticeably different. The five pocket pairs out of six seen at the link in your link was indeed unlikely






                  share|cite|improve this answer











                  $endgroup$



                  The probability that a particular player is dealt a pair is $frac{4-1}{52-1}=frac{1}{17}$ so the expected number of pairs is $frac{n}{17}$



                  I suspect the distribution is not going to be far away from Binomial with parameters $n$ and $p=frac{1}{17}$. This is not quite correct (for example when $n=26$ the probability of exactly $25$ pairs is $0$ rather than $4.24times 10^{−30}$) but I would be surprised if it were substantially misleading



                  As an example, here are a couple of simulations of $100000$ million deals to $6$ players 



                  Pairs                    0        1         2         3         4         5         6 
                  
Simulation 1            69459    26065     4094      367       13         2         0
                  
Simulation 2            69675    25918     4063      329       15         0         0
                  
Binomial probability  0.695066 0.260650 0.040727 0.003394 0.000159 0.0000040 0.00000004


                  and, given the uncertainty of simulation, these do not look noticeably different. The five pocket pairs out of six seen at the link in your link was indeed unlikely







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 27 '18 at 10:53

























                  answered Dec 27 '18 at 10:24









                  HenryHenry

                  101k482170




                  101k482170






























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                      Verónica Boquete

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                      Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
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                      Ida-Boy-Ed-Garten

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                      Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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