What is the probability of 3 heart and 2 distinct pairs in a hand of five cards?












0












$begingroup$


Given that the hand is drawn from a standard deck of 52 cards, what is the probability that it contains three hearts and two distinct pairs?



I know that there are ${52 choose 5}$ distinct hands of five, and that each pair will contain exactly one heart, with the third heart not being in a pair.



So far, my work is that there are:



${13 choose 3}$ ways to draw three hearts



${3 choose 2}$ ways to select which of the two hearts are part of the two pairs



${3choose2}$ ways to select the suit of the other card for each pair



And so the probability would be $frac{{13 choose 3}{3 choose 2}^3}{{52 choose 5}}$, about 0.003



I can't help but feel like I am leaving something out, though. Have I forgotten to include anything in the numerator?






probability poker







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$endgroup$












  • $begingroup$
    Does a "distinct pair" rule out three of a kind (full house-one pair and then three of another)?
    $endgroup$
    – fleablood
    Dec 6 '18 at 19:00










  • $begingroup$
    "${3 choose 2}$ ways to select the suit of the other card for each pair" That's not correct. You have ${3choose 1}$ way to choose one of the suits and ${3choose 1}$ way to chose the other.
    $endgroup$
    – fleablood
    Dec 6 '18 at 19:15










  • $begingroup$
    Hmm.... no I'm mistaken. Brain fart. YOu DID square it. Sorry.
    $endgroup$
    – fleablood
    Dec 6 '18 at 19:25










  • $begingroup$
    Well, it should be ${3choose 1}$ and not ${3 choose 2}$ for the other suit of each pair but as ${3choose 1} = {3choose 2}$ .....
    $endgroup$
    – fleablood
    Dec 6 '18 at 19:27
















0












$begingroup$


Given that the hand is drawn from a standard deck of 52 cards, what is the probability that it contains three hearts and two distinct pairs?



I know that there are ${52 choose 5}$ distinct hands of five, and that each pair will contain exactly one heart, with the third heart not being in a pair.



So far, my work is that there are:



${13 choose 3}$ ways to draw three hearts



${3 choose 2}$ ways to select which of the two hearts are part of the two pairs



${3choose2}$ ways to select the suit of the other card for each pair



And so the probability would be $frac{{13 choose 3}{3 choose 2}^3}{{52 choose 5}}$, about 0.003



I can't help but feel like I am leaving something out, though. Have I forgotten to include anything in the numerator?






probability poker







share|cite|improve this question









$endgroup$












  • $begingroup$
    Does a "distinct pair" rule out three of a kind (full house-one pair and then three of another)?
    $endgroup$
    – fleablood
    Dec 6 '18 at 19:00










  • $begingroup$
    "${3 choose 2}$ ways to select the suit of the other card for each pair" That's not correct. You have ${3choose 1}$ way to choose one of the suits and ${3choose 1}$ way to chose the other.
    $endgroup$
    – fleablood
    Dec 6 '18 at 19:15










  • $begingroup$
    Hmm.... no I'm mistaken. Brain fart. YOu DID square it. Sorry.
    $endgroup$
    – fleablood
    Dec 6 '18 at 19:25










  • $begingroup$
    Well, it should be ${3choose 1}$ and not ${3 choose 2}$ for the other suit of each pair but as ${3choose 1} = {3choose 2}$ .....
    $endgroup$
    – fleablood
    Dec 6 '18 at 19:27














0












0








0





$begingroup$


Given that the hand is drawn from a standard deck of 52 cards, what is the probability that it contains three hearts and two distinct pairs?



I know that there are ${52 choose 5}$ distinct hands of five, and that each pair will contain exactly one heart, with the third heart not being in a pair.



So far, my work is that there are:



${13 choose 3}$ ways to draw three hearts



${3 choose 2}$ ways to select which of the two hearts are part of the two pairs



${3choose2}$ ways to select the suit of the other card for each pair



And so the probability would be $frac{{13 choose 3}{3 choose 2}^3}{{52 choose 5}}$, about 0.003



I can't help but feel like I am leaving something out, though. Have I forgotten to include anything in the numerator?






probability poker







share|cite|improve this question









$endgroup$




Given that the hand is drawn from a standard deck of 52 cards, what is the probability that it contains three hearts and two distinct pairs?



I know that there are ${52 choose 5}$ distinct hands of five, and that each pair will contain exactly one heart, with the third heart not being in a pair.



So far, my work is that there are:



${13 choose 3}$ ways to draw three hearts



${3 choose 2}$ ways to select which of the two hearts are part of the two pairs



${3choose2}$ ways to select the suit of the other card for each pair



And so the probability would be $frac{{13 choose 3}{3 choose 2}^3}{{52 choose 5}}$, about 0.003



I can't help but feel like I am leaving something out, though. Have I forgotten to include anything in the numerator?






probability poker




probability poker






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 6 '18 at 18:53









sk13sk13

176




176












  • $begingroup$
    Does a "distinct pair" rule out three of a kind (full house-one pair and then three of another)?
    $endgroup$
    – fleablood
    Dec 6 '18 at 19:00










  • $begingroup$
    "${3 choose 2}$ ways to select the suit of the other card for each pair" That's not correct. You have ${3choose 1}$ way to choose one of the suits and ${3choose 1}$ way to chose the other.
    $endgroup$
    – fleablood
    Dec 6 '18 at 19:15










  • $begingroup$
    Hmm.... no I'm mistaken. Brain fart. YOu DID square it. Sorry.
    $endgroup$
    – fleablood
    Dec 6 '18 at 19:25










  • $begingroup$
    Well, it should be ${3choose 1}$ and not ${3 choose 2}$ for the other suit of each pair but as ${3choose 1} = {3choose 2}$ .....
    $endgroup$
    – fleablood
    Dec 6 '18 at 19:27


















  • $begingroup$
    Does a "distinct pair" rule out three of a kind (full house-one pair and then three of another)?
    $endgroup$
    – fleablood
    Dec 6 '18 at 19:00










  • $begingroup$
    "${3 choose 2}$ ways to select the suit of the other card for each pair" That's not correct. You have ${3choose 1}$ way to choose one of the suits and ${3choose 1}$ way to chose the other.
    $endgroup$
    – fleablood
    Dec 6 '18 at 19:15










  • $begingroup$
    Hmm.... no I'm mistaken. Brain fart. YOu DID square it. Sorry.
    $endgroup$
    – fleablood
    Dec 6 '18 at 19:25










  • $begingroup$
    Well, it should be ${3choose 1}$ and not ${3 choose 2}$ for the other suit of each pair but as ${3choose 1} = {3choose 2}$ .....
    $endgroup$
    – fleablood
    Dec 6 '18 at 19:27
















$begingroup$
Does a "distinct pair" rule out three of a kind (full house-one pair and then three of another)?
$endgroup$
– fleablood
Dec 6 '18 at 19:00




$begingroup$
Does a "distinct pair" rule out three of a kind (full house-one pair and then three of another)?
$endgroup$
– fleablood
Dec 6 '18 at 19:00












$begingroup$
"${3 choose 2}$ ways to select the suit of the other card for each pair" That's not correct. You have ${3choose 1}$ way to choose one of the suits and ${3choose 1}$ way to chose the other.
$endgroup$
– fleablood
Dec 6 '18 at 19:15




$begingroup$
"${3 choose 2}$ ways to select the suit of the other card for each pair" That's not correct. You have ${3choose 1}$ way to choose one of the suits and ${3choose 1}$ way to chose the other.
$endgroup$
– fleablood
Dec 6 '18 at 19:15












$begingroup$
Hmm.... no I'm mistaken. Brain fart. YOu DID square it. Sorry.
$endgroup$
– fleablood
Dec 6 '18 at 19:25




$begingroup$
Hmm.... no I'm mistaken. Brain fart. YOu DID square it. Sorry.
$endgroup$
– fleablood
Dec 6 '18 at 19:25












$begingroup$
Well, it should be ${3choose 1}$ and not ${3 choose 2}$ for the other suit of each pair but as ${3choose 1} = {3choose 2}$ .....
$endgroup$
– fleablood
Dec 6 '18 at 19:27




$begingroup$
Well, it should be ${3choose 1}$ and not ${3 choose 2}$ for the other suit of each pair but as ${3choose 1} = {3choose 2}$ .....
$endgroup$
– fleablood
Dec 6 '18 at 19:27










1 Answer
1






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$begingroup$

This becomes easy once you realize that a pair can't be both hearts. So at most each pair accounts for one heart and the fifth card must be the third. SO the only way you can have two pairs and three hearts is to have a pair with one of the cards a heart, another pair and one of the cards a heart, and the fifth card a heart.



There are there are ${13 choose 2}$ possible options the pairs can be. there are $11$ options the fifth card can be. For each pair there are $3$ options the second card in the pair can be.



So there are ${13choose 2}cdot 3cdot 3cdot 11=$



$frac {13cdot 12}2cdot 99 = 13cdot 6cdot 99 = 7722$ ways to do this.



So probality is $frac {frac {13!}{11!2!}9*11}{frac {52!}{47!5!}} =frac {7722}{2598960} approx 0.003$



Which is the same answer as yours as I did $ {13choose 2}cdot{{13-2} choose 1}=frac {13!}{11!2!}frac {11!}{10!1!}$ whereas you did ${13choose 3}{3choose 1}=frac {13!}{10!3!}frac {3!}{2!1!}$ which are clearly the same value. (there's probably a combinatorial argument why).






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    1 Answer
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    $begingroup$

    This becomes easy once you realize that a pair can't be both hearts. So at most each pair accounts for one heart and the fifth card must be the third. SO the only way you can have two pairs and three hearts is to have a pair with one of the cards a heart, another pair and one of the cards a heart, and the fifth card a heart.



    There are there are ${13 choose 2}$ possible options the pairs can be. there are $11$ options the fifth card can be. For each pair there are $3$ options the second card in the pair can be.



    So there are ${13choose 2}cdot 3cdot 3cdot 11=$



    $frac {13cdot 12}2cdot 99 = 13cdot 6cdot 99 = 7722$ ways to do this.



    So probality is $frac {frac {13!}{11!2!}9*11}{frac {52!}{47!5!}} =frac {7722}{2598960} approx 0.003$



    Which is the same answer as yours as I did $ {13choose 2}cdot{{13-2} choose 1}=frac {13!}{11!2!}frac {11!}{10!1!}$ whereas you did ${13choose 3}{3choose 1}=frac {13!}{10!3!}frac {3!}{2!1!}$ which are clearly the same value. (there's probably a combinatorial argument why).






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      This becomes easy once you realize that a pair can't be both hearts. So at most each pair accounts for one heart and the fifth card must be the third. SO the only way you can have two pairs and three hearts is to have a pair with one of the cards a heart, another pair and one of the cards a heart, and the fifth card a heart.



      There are there are ${13 choose 2}$ possible options the pairs can be. there are $11$ options the fifth card can be. For each pair there are $3$ options the second card in the pair can be.



      So there are ${13choose 2}cdot 3cdot 3cdot 11=$



      $frac {13cdot 12}2cdot 99 = 13cdot 6cdot 99 = 7722$ ways to do this.



      So probality is $frac {frac {13!}{11!2!}9*11}{frac {52!}{47!5!}} =frac {7722}{2598960} approx 0.003$



      Which is the same answer as yours as I did $ {13choose 2}cdot{{13-2} choose 1}=frac {13!}{11!2!}frac {11!}{10!1!}$ whereas you did ${13choose 3}{3choose 1}=frac {13!}{10!3!}frac {3!}{2!1!}$ which are clearly the same value. (there's probably a combinatorial argument why).






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        This becomes easy once you realize that a pair can't be both hearts. So at most each pair accounts for one heart and the fifth card must be the third. SO the only way you can have two pairs and three hearts is to have a pair with one of the cards a heart, another pair and one of the cards a heart, and the fifth card a heart.



        There are there are ${13 choose 2}$ possible options the pairs can be. there are $11$ options the fifth card can be. For each pair there are $3$ options the second card in the pair can be.



        So there are ${13choose 2}cdot 3cdot 3cdot 11=$



        $frac {13cdot 12}2cdot 99 = 13cdot 6cdot 99 = 7722$ ways to do this.



        So probality is $frac {frac {13!}{11!2!}9*11}{frac {52!}{47!5!}} =frac {7722}{2598960} approx 0.003$



        Which is the same answer as yours as I did $ {13choose 2}cdot{{13-2} choose 1}=frac {13!}{11!2!}frac {11!}{10!1!}$ whereas you did ${13choose 3}{3choose 1}=frac {13!}{10!3!}frac {3!}{2!1!}$ which are clearly the same value. (there's probably a combinatorial argument why).






        share|cite|improve this answer











        $endgroup$



        This becomes easy once you realize that a pair can't be both hearts. So at most each pair accounts for one heart and the fifth card must be the third. SO the only way you can have two pairs and three hearts is to have a pair with one of the cards a heart, another pair and one of the cards a heart, and the fifth card a heart.



        There are there are ${13 choose 2}$ possible options the pairs can be. there are $11$ options the fifth card can be. For each pair there are $3$ options the second card in the pair can be.



        So there are ${13choose 2}cdot 3cdot 3cdot 11=$



        $frac {13cdot 12}2cdot 99 = 13cdot 6cdot 99 = 7722$ ways to do this.



        So probality is $frac {frac {13!}{11!2!}9*11}{frac {52!}{47!5!}} =frac {7722}{2598960} approx 0.003$



        Which is the same answer as yours as I did $ {13choose 2}cdot{{13-2} choose 1}=frac {13!}{11!2!}frac {11!}{10!1!}$ whereas you did ${13choose 3}{3choose 1}=frac {13!}{10!3!}frac {3!}{2!1!}$ which are clearly the same value. (there's probably a combinatorial argument why).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 6 '18 at 19:33

























        answered Dec 6 '18 at 19:08









        fleabloodfleablood

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