Shuffling a deck of cards












0












$begingroup$


If a card deck was shuffled, what is the probability that the 1st card will be in the same order. The answer I have is (1/52!)...is this correct?
Also, surely the answer is the same for the probability of the last card being in the same order?



Secondly, what is the probability that a KING will be amongst the initial 5 cards?






probability







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  • 1




    $begingroup$
    What do you mean "the first card in same order"? Do you mean that the first card is the same before and after the shuffle?
    $endgroup$
    – Arthur
    Dec 14 '18 at 10:15












  • $begingroup$
    For the second part, I would approach it by considering what is the probability that a king is not in the first 5 cards. That's quite easy to do ...
    $endgroup$
    – Matti P.
    Dec 14 '18 at 10:22










  • $begingroup$
    Who is shuffling?
    $endgroup$
    – Mason
    Dec 14 '18 at 14:19
















0












$begingroup$


If a card deck was shuffled, what is the probability that the 1st card will be in the same order. The answer I have is (1/52!)...is this correct?
Also, surely the answer is the same for the probability of the last card being in the same order?



Secondly, what is the probability that a KING will be amongst the initial 5 cards?






probability







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    What do you mean "the first card in same order"? Do you mean that the first card is the same before and after the shuffle?
    $endgroup$
    – Arthur
    Dec 14 '18 at 10:15












  • $begingroup$
    For the second part, I would approach it by considering what is the probability that a king is not in the first 5 cards. That's quite easy to do ...
    $endgroup$
    – Matti P.
    Dec 14 '18 at 10:22










  • $begingroup$
    Who is shuffling?
    $endgroup$
    – Mason
    Dec 14 '18 at 14:19














0












0








0





$begingroup$


If a card deck was shuffled, what is the probability that the 1st card will be in the same order. The answer I have is (1/52!)...is this correct?
Also, surely the answer is the same for the probability of the last card being in the same order?



Secondly, what is the probability that a KING will be amongst the initial 5 cards?






probability







share|cite|improve this question









$endgroup$




If a card deck was shuffled, what is the probability that the 1st card will be in the same order. The answer I have is (1/52!)...is this correct?
Also, surely the answer is the same for the probability of the last card being in the same order?



Secondly, what is the probability that a KING will be amongst the initial 5 cards?






probability




probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 14 '18 at 10:13









Jasmine078Jasmine078

123




123








  • 1




    $begingroup$
    What do you mean "the first card in same order"? Do you mean that the first card is the same before and after the shuffle?
    $endgroup$
    – Arthur
    Dec 14 '18 at 10:15












  • $begingroup$
    For the second part, I would approach it by considering what is the probability that a king is not in the first 5 cards. That's quite easy to do ...
    $endgroup$
    – Matti P.
    Dec 14 '18 at 10:22










  • $begingroup$
    Who is shuffling?
    $endgroup$
    – Mason
    Dec 14 '18 at 14:19














  • 1




    $begingroup$
    What do you mean "the first card in same order"? Do you mean that the first card is the same before and after the shuffle?
    $endgroup$
    – Arthur
    Dec 14 '18 at 10:15












  • $begingroup$
    For the second part, I would approach it by considering what is the probability that a king is not in the first 5 cards. That's quite easy to do ...
    $endgroup$
    – Matti P.
    Dec 14 '18 at 10:22










  • $begingroup$
    Who is shuffling?
    $endgroup$
    – Mason
    Dec 14 '18 at 14:19








1




1




$begingroup$
What do you mean "the first card in same order"? Do you mean that the first card is the same before and after the shuffle?
$endgroup$
– Arthur
Dec 14 '18 at 10:15






$begingroup$
What do you mean "the first card in same order"? Do you mean that the first card is the same before and after the shuffle?
$endgroup$
– Arthur
Dec 14 '18 at 10:15














$begingroup$
For the second part, I would approach it by considering what is the probability that a king is not in the first 5 cards. That's quite easy to do ...
$endgroup$
– Matti P.
Dec 14 '18 at 10:22




$begingroup$
For the second part, I would approach it by considering what is the probability that a king is not in the first 5 cards. That's quite easy to do ...
$endgroup$
– Matti P.
Dec 14 '18 at 10:22












$begingroup$
Who is shuffling?
$endgroup$
– Mason
Dec 14 '18 at 14:19




$begingroup$
Who is shuffling?
$endgroup$
– Mason
Dec 14 '18 at 14:19










2 Answers
2






active

oldest

votes


















0












$begingroup$

By good shuffling all cards have equal probability to end up as first card of the deck after shuffling.



There are $52$ candidates and only one of them will indeed end up as first card of the deck after shuffling.



That leads to the conclusion that every card (so also the one that was first card before the shuffling) has probability $frac1{52}$ to be the first card after shuffling.



Similar for last card.




Let $E_i$ denote the probability that the $i$-th card is not a king.



Then the probability that among the first $5$ cards there is no king equals:$$P(E_1cap E_2cap E_3cap E_4cap E_5)=$$$$P(E_1)P(E_2mid E_1)P(E_3mid E_1cap E_2)P(E_4mid E_1cap E_2cap E_3)P(E_5mid E_1cap E_2cap E_3cap E_4)=$$$$frac{48}{52}frac{47}{51}frac{46}{50}frac{45}{49}frac{44}{48}$$



So the probability that there is a king is among the first $5$ cards equals:$$1-frac{48}{52}frac{47}{51}frac{46}{50}frac{45}{49}frac{44}{48}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thankyou for your explanation. Could you please clarify further as to why the first fraction is (48/52) - why do you have 48?
    $endgroup$
    – Jasmine078
    Dec 14 '18 at 11:26










  • $begingroup$
    There are $48$ cards that are no king. So the probability that the first card is no king is $frac{48}{52}$.
    $endgroup$
    – drhab
    Dec 14 '18 at 11:28



















0












$begingroup$

In answer to your first question, I believe it would be $frac{51!}{52!}=frac{1}{52}$. If you fix the first card in the deck, there are 51! ways of rearranging the remaining cards, and 52! ways of arranging the entire deck, so that is the probability that a shuffle gives the first card in the same place. This assumes that all possible shuffles are equally likely. The result is true for fixing any card in the deck.






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






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    active

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    0












    $begingroup$

    By good shuffling all cards have equal probability to end up as first card of the deck after shuffling.



    There are $52$ candidates and only one of them will indeed end up as first card of the deck after shuffling.



    That leads to the conclusion that every card (so also the one that was first card before the shuffling) has probability $frac1{52}$ to be the first card after shuffling.



    Similar for last card.




    Let $E_i$ denote the probability that the $i$-th card is not a king.



    Then the probability that among the first $5$ cards there is no king equals:$$P(E_1cap E_2cap E_3cap E_4cap E_5)=$$$$P(E_1)P(E_2mid E_1)P(E_3mid E_1cap E_2)P(E_4mid E_1cap E_2cap E_3)P(E_5mid E_1cap E_2cap E_3cap E_4)=$$$$frac{48}{52}frac{47}{51}frac{46}{50}frac{45}{49}frac{44}{48}$$



    So the probability that there is a king is among the first $5$ cards equals:$$1-frac{48}{52}frac{47}{51}frac{46}{50}frac{45}{49}frac{44}{48}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thankyou for your explanation. Could you please clarify further as to why the first fraction is (48/52) - why do you have 48?
      $endgroup$
      – Jasmine078
      Dec 14 '18 at 11:26










    • $begingroup$
      There are $48$ cards that are no king. So the probability that the first card is no king is $frac{48}{52}$.
      $endgroup$
      – drhab
      Dec 14 '18 at 11:28
















    0












    $begingroup$

    By good shuffling all cards have equal probability to end up as first card of the deck after shuffling.



    There are $52$ candidates and only one of them will indeed end up as first card of the deck after shuffling.



    That leads to the conclusion that every card (so also the one that was first card before the shuffling) has probability $frac1{52}$ to be the first card after shuffling.



    Similar for last card.




    Let $E_i$ denote the probability that the $i$-th card is not a king.



    Then the probability that among the first $5$ cards there is no king equals:$$P(E_1cap E_2cap E_3cap E_4cap E_5)=$$$$P(E_1)P(E_2mid E_1)P(E_3mid E_1cap E_2)P(E_4mid E_1cap E_2cap E_3)P(E_5mid E_1cap E_2cap E_3cap E_4)=$$$$frac{48}{52}frac{47}{51}frac{46}{50}frac{45}{49}frac{44}{48}$$



    So the probability that there is a king is among the first $5$ cards equals:$$1-frac{48}{52}frac{47}{51}frac{46}{50}frac{45}{49}frac{44}{48}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thankyou for your explanation. Could you please clarify further as to why the first fraction is (48/52) - why do you have 48?
      $endgroup$
      – Jasmine078
      Dec 14 '18 at 11:26










    • $begingroup$
      There are $48$ cards that are no king. So the probability that the first card is no king is $frac{48}{52}$.
      $endgroup$
      – drhab
      Dec 14 '18 at 11:28














    0












    0








    0





    $begingroup$

    By good shuffling all cards have equal probability to end up as first card of the deck after shuffling.



    There are $52$ candidates and only one of them will indeed end up as first card of the deck after shuffling.



    That leads to the conclusion that every card (so also the one that was first card before the shuffling) has probability $frac1{52}$ to be the first card after shuffling.



    Similar for last card.




    Let $E_i$ denote the probability that the $i$-th card is not a king.



    Then the probability that among the first $5$ cards there is no king equals:$$P(E_1cap E_2cap E_3cap E_4cap E_5)=$$$$P(E_1)P(E_2mid E_1)P(E_3mid E_1cap E_2)P(E_4mid E_1cap E_2cap E_3)P(E_5mid E_1cap E_2cap E_3cap E_4)=$$$$frac{48}{52}frac{47}{51}frac{46}{50}frac{45}{49}frac{44}{48}$$



    So the probability that there is a king is among the first $5$ cards equals:$$1-frac{48}{52}frac{47}{51}frac{46}{50}frac{45}{49}frac{44}{48}$$






    share|cite|improve this answer











    $endgroup$



    By good shuffling all cards have equal probability to end up as first card of the deck after shuffling.



    There are $52$ candidates and only one of them will indeed end up as first card of the deck after shuffling.



    That leads to the conclusion that every card (so also the one that was first card before the shuffling) has probability $frac1{52}$ to be the first card after shuffling.



    Similar for last card.




    Let $E_i$ denote the probability that the $i$-th card is not a king.



    Then the probability that among the first $5$ cards there is no king equals:$$P(E_1cap E_2cap E_3cap E_4cap E_5)=$$$$P(E_1)P(E_2mid E_1)P(E_3mid E_1cap E_2)P(E_4mid E_1cap E_2cap E_3)P(E_5mid E_1cap E_2cap E_3cap E_4)=$$$$frac{48}{52}frac{47}{51}frac{46}{50}frac{45}{49}frac{44}{48}$$



    So the probability that there is a king is among the first $5$ cards equals:$$1-frac{48}{52}frac{47}{51}frac{46}{50}frac{45}{49}frac{44}{48}$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 14 '18 at 11:41

























    answered Dec 14 '18 at 11:09









    drhabdrhab

    102k545136




    102k545136












    • $begingroup$
      Thankyou for your explanation. Could you please clarify further as to why the first fraction is (48/52) - why do you have 48?
      $endgroup$
      – Jasmine078
      Dec 14 '18 at 11:26










    • $begingroup$
      There are $48$ cards that are no king. So the probability that the first card is no king is $frac{48}{52}$.
      $endgroup$
      – drhab
      Dec 14 '18 at 11:28


















    • $begingroup$
      Thankyou for your explanation. Could you please clarify further as to why the first fraction is (48/52) - why do you have 48?
      $endgroup$
      – Jasmine078
      Dec 14 '18 at 11:26










    • $begingroup$
      There are $48$ cards that are no king. So the probability that the first card is no king is $frac{48}{52}$.
      $endgroup$
      – drhab
      Dec 14 '18 at 11:28
















    $begingroup$
    Thankyou for your explanation. Could you please clarify further as to why the first fraction is (48/52) - why do you have 48?
    $endgroup$
    – Jasmine078
    Dec 14 '18 at 11:26




    $begingroup$
    Thankyou for your explanation. Could you please clarify further as to why the first fraction is (48/52) - why do you have 48?
    $endgroup$
    – Jasmine078
    Dec 14 '18 at 11:26












    $begingroup$
    There are $48$ cards that are no king. So the probability that the first card is no king is $frac{48}{52}$.
    $endgroup$
    – drhab
    Dec 14 '18 at 11:28




    $begingroup$
    There are $48$ cards that are no king. So the probability that the first card is no king is $frac{48}{52}$.
    $endgroup$
    – drhab
    Dec 14 '18 at 11:28











    0












    $begingroup$

    In answer to your first question, I believe it would be $frac{51!}{52!}=frac{1}{52}$. If you fix the first card in the deck, there are 51! ways of rearranging the remaining cards, and 52! ways of arranging the entire deck, so that is the probability that a shuffle gives the first card in the same place. This assumes that all possible shuffles are equally likely. The result is true for fixing any card in the deck.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      In answer to your first question, I believe it would be $frac{51!}{52!}=frac{1}{52}$. If you fix the first card in the deck, there are 51! ways of rearranging the remaining cards, and 52! ways of arranging the entire deck, so that is the probability that a shuffle gives the first card in the same place. This assumes that all possible shuffles are equally likely. The result is true for fixing any card in the deck.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        In answer to your first question, I believe it would be $frac{51!}{52!}=frac{1}{52}$. If you fix the first card in the deck, there are 51! ways of rearranging the remaining cards, and 52! ways of arranging the entire deck, so that is the probability that a shuffle gives the first card in the same place. This assumes that all possible shuffles are equally likely. The result is true for fixing any card in the deck.






        share|cite|improve this answer









        $endgroup$



        In answer to your first question, I believe it would be $frac{51!}{52!}=frac{1}{52}$. If you fix the first card in the deck, there are 51! ways of rearranging the remaining cards, and 52! ways of arranging the entire deck, so that is the probability that a shuffle gives the first card in the same place. This assumes that all possible shuffles are equally likely. The result is true for fixing any card in the deck.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 14 '18 at 10:19









        M.M.M.M.

        637




        637






























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