A deck of cards and the Pigeonhole Principle












0












$begingroup$


I am trying to solve the following problem:




How many cards must be chosen from a standard deck of 52 cards to guarantee that there are at least two cards of each of two different kinds?




It isn't specified, but I am assuming "kind" to mean the rank of the card.



Using the Pigeonhole Principle, I came up with the number 5, but my book says 17. Where am I going wrong?



Thank you!






combinatorics pigeonhole-principle







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    I do not know how you got an answer of $5$ so I cannot tell you "where you are going wrong." Consider a worst case scenario, you draw all thirteen spades. After having drawn all thirteen spades, you draw the remaining three aces. You now currently have one of every rank except ace where you have four of that rank and 16 cards in total in your hand without having at least two cards of each of two different kinds. Drawing one additional card of any kind will have now a second rank that you have at least two cards of.
    $endgroup$
    – JMoravitz
    Dec 15 '18 at 3:38












  • $begingroup$
    @JMoravitz I may be misunderstanding the problem. I got my answer by "drawing" four 2's, then "drawing" the next card of a different rank.
    $endgroup$
    – Lee
    Dec 15 '18 at 3:43






  • 2




    $begingroup$
    I, and the book apparently, interpret it as being that you need at least two ranks each of which have at least two cards of that rank in it. For example $Aspadesuit, Aheartsuit, 2spadesuit, 3heartsuit$ currently has one rank with 2 cards of it's type, another rank with only one card of it's type, and another rank with only one card of it's type. On the other hand $Aspadesuit,Aheartsuit,2spadesuit,2diamondsuit$ has two ranks each of which have at least two cards of it's type. "at least two cards of each of two different kinds", the 'of' there that I put in bold is important.
    $endgroup$
    – JMoravitz
    Dec 15 '18 at 3:45












  • $begingroup$
    @JMoravitz Ahh, I see what you are saying. Thank you for the insight!
    $endgroup$
    – Lee
    Dec 15 '18 at 3:47










  • $begingroup$
    "at least two cards of each of two different kinds" i.e. 'at least two cards of a first kind and at least two cards of a second kind.' This is different than "at least two cards, each of two different kinds" i.e. 'at least two cards, the first of which is of one kind and the second of which is of a different kind.' For the question of needing two cards which disagree on both rank and suit, that would be 14 cards needed there. Take all thirteen spades, then as soon as you take another non-spade it will differ in rank from one of the spades already present @RossMillikan
    $endgroup$
    – JMoravitz
    Dec 15 '18 at 3:52


















0












$begingroup$


I am trying to solve the following problem:




How many cards must be chosen from a standard deck of 52 cards to guarantee that there are at least two cards of each of two different kinds?




It isn't specified, but I am assuming "kind" to mean the rank of the card.



Using the Pigeonhole Principle, I came up with the number 5, but my book says 17. Where am I going wrong?



Thank you!






combinatorics pigeonhole-principle







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    I do not know how you got an answer of $5$ so I cannot tell you "where you are going wrong." Consider a worst case scenario, you draw all thirteen spades. After having drawn all thirteen spades, you draw the remaining three aces. You now currently have one of every rank except ace where you have four of that rank and 16 cards in total in your hand without having at least two cards of each of two different kinds. Drawing one additional card of any kind will have now a second rank that you have at least two cards of.
    $endgroup$
    – JMoravitz
    Dec 15 '18 at 3:38












  • $begingroup$
    @JMoravitz I may be misunderstanding the problem. I got my answer by "drawing" four 2's, then "drawing" the next card of a different rank.
    $endgroup$
    – Lee
    Dec 15 '18 at 3:43






  • 2




    $begingroup$
    I, and the book apparently, interpret it as being that you need at least two ranks each of which have at least two cards of that rank in it. For example $Aspadesuit, Aheartsuit, 2spadesuit, 3heartsuit$ currently has one rank with 2 cards of it's type, another rank with only one card of it's type, and another rank with only one card of it's type. On the other hand $Aspadesuit,Aheartsuit,2spadesuit,2diamondsuit$ has two ranks each of which have at least two cards of it's type. "at least two cards of each of two different kinds", the 'of' there that I put in bold is important.
    $endgroup$
    – JMoravitz
    Dec 15 '18 at 3:45












  • $begingroup$
    @JMoravitz Ahh, I see what you are saying. Thank you for the insight!
    $endgroup$
    – Lee
    Dec 15 '18 at 3:47










  • $begingroup$
    "at least two cards of each of two different kinds" i.e. 'at least two cards of a first kind and at least two cards of a second kind.' This is different than "at least two cards, each of two different kinds" i.e. 'at least two cards, the first of which is of one kind and the second of which is of a different kind.' For the question of needing two cards which disagree on both rank and suit, that would be 14 cards needed there. Take all thirteen spades, then as soon as you take another non-spade it will differ in rank from one of the spades already present @RossMillikan
    $endgroup$
    – JMoravitz
    Dec 15 '18 at 3:52
















0












0








0





$begingroup$


I am trying to solve the following problem:




How many cards must be chosen from a standard deck of 52 cards to guarantee that there are at least two cards of each of two different kinds?




It isn't specified, but I am assuming "kind" to mean the rank of the card.



Using the Pigeonhole Principle, I came up with the number 5, but my book says 17. Where am I going wrong?



Thank you!






combinatorics pigeonhole-principle







share|cite|improve this question









$endgroup$




I am trying to solve the following problem:




How many cards must be chosen from a standard deck of 52 cards to guarantee that there are at least two cards of each of two different kinds?




It isn't specified, but I am assuming "kind" to mean the rank of the card.



Using the Pigeonhole Principle, I came up with the number 5, but my book says 17. Where am I going wrong?



Thank you!






combinatorics pigeonhole-principle




combinatorics pigeonhole-principle






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 15 '18 at 3:34









LeeLee

1015




1015








  • 4




    $begingroup$
    I do not know how you got an answer of $5$ so I cannot tell you "where you are going wrong." Consider a worst case scenario, you draw all thirteen spades. After having drawn all thirteen spades, you draw the remaining three aces. You now currently have one of every rank except ace where you have four of that rank and 16 cards in total in your hand without having at least two cards of each of two different kinds. Drawing one additional card of any kind will have now a second rank that you have at least two cards of.
    $endgroup$
    – JMoravitz
    Dec 15 '18 at 3:38












  • $begingroup$
    @JMoravitz I may be misunderstanding the problem. I got my answer by "drawing" four 2's, then "drawing" the next card of a different rank.
    $endgroup$
    – Lee
    Dec 15 '18 at 3:43






  • 2




    $begingroup$
    I, and the book apparently, interpret it as being that you need at least two ranks each of which have at least two cards of that rank in it. For example $Aspadesuit, Aheartsuit, 2spadesuit, 3heartsuit$ currently has one rank with 2 cards of it's type, another rank with only one card of it's type, and another rank with only one card of it's type. On the other hand $Aspadesuit,Aheartsuit,2spadesuit,2diamondsuit$ has two ranks each of which have at least two cards of it's type. "at least two cards of each of two different kinds", the 'of' there that I put in bold is important.
    $endgroup$
    – JMoravitz
    Dec 15 '18 at 3:45












  • $begingroup$
    @JMoravitz Ahh, I see what you are saying. Thank you for the insight!
    $endgroup$
    – Lee
    Dec 15 '18 at 3:47










  • $begingroup$
    "at least two cards of each of two different kinds" i.e. 'at least two cards of a first kind and at least two cards of a second kind.' This is different than "at least two cards, each of two different kinds" i.e. 'at least two cards, the first of which is of one kind and the second of which is of a different kind.' For the question of needing two cards which disagree on both rank and suit, that would be 14 cards needed there. Take all thirteen spades, then as soon as you take another non-spade it will differ in rank from one of the spades already present @RossMillikan
    $endgroup$
    – JMoravitz
    Dec 15 '18 at 3:52
















  • 4




    $begingroup$
    I do not know how you got an answer of $5$ so I cannot tell you "where you are going wrong." Consider a worst case scenario, you draw all thirteen spades. After having drawn all thirteen spades, you draw the remaining three aces. You now currently have one of every rank except ace where you have four of that rank and 16 cards in total in your hand without having at least two cards of each of two different kinds. Drawing one additional card of any kind will have now a second rank that you have at least two cards of.
    $endgroup$
    – JMoravitz
    Dec 15 '18 at 3:38












  • $begingroup$
    @JMoravitz I may be misunderstanding the problem. I got my answer by "drawing" four 2's, then "drawing" the next card of a different rank.
    $endgroup$
    – Lee
    Dec 15 '18 at 3:43






  • 2




    $begingroup$
    I, and the book apparently, interpret it as being that you need at least two ranks each of which have at least two cards of that rank in it. For example $Aspadesuit, Aheartsuit, 2spadesuit, 3heartsuit$ currently has one rank with 2 cards of it's type, another rank with only one card of it's type, and another rank with only one card of it's type. On the other hand $Aspadesuit,Aheartsuit,2spadesuit,2diamondsuit$ has two ranks each of which have at least two cards of it's type. "at least two cards of each of two different kinds", the 'of' there that I put in bold is important.
    $endgroup$
    – JMoravitz
    Dec 15 '18 at 3:45












  • $begingroup$
    @JMoravitz Ahh, I see what you are saying. Thank you for the insight!
    $endgroup$
    – Lee
    Dec 15 '18 at 3:47










  • $begingroup$
    "at least two cards of each of two different kinds" i.e. 'at least two cards of a first kind and at least two cards of a second kind.' This is different than "at least two cards, each of two different kinds" i.e. 'at least two cards, the first of which is of one kind and the second of which is of a different kind.' For the question of needing two cards which disagree on both rank and suit, that would be 14 cards needed there. Take all thirteen spades, then as soon as you take another non-spade it will differ in rank from one of the spades already present @RossMillikan
    $endgroup$
    – JMoravitz
    Dec 15 '18 at 3:52










4




4




$begingroup$
I do not know how you got an answer of $5$ so I cannot tell you "where you are going wrong." Consider a worst case scenario, you draw all thirteen spades. After having drawn all thirteen spades, you draw the remaining three aces. You now currently have one of every rank except ace where you have four of that rank and 16 cards in total in your hand without having at least two cards of each of two different kinds. Drawing one additional card of any kind will have now a second rank that you have at least two cards of.
$endgroup$
– JMoravitz
Dec 15 '18 at 3:38






$begingroup$
I do not know how you got an answer of $5$ so I cannot tell you "where you are going wrong." Consider a worst case scenario, you draw all thirteen spades. After having drawn all thirteen spades, you draw the remaining three aces. You now currently have one of every rank except ace where you have four of that rank and 16 cards in total in your hand without having at least two cards of each of two different kinds. Drawing one additional card of any kind will have now a second rank that you have at least two cards of.
$endgroup$
– JMoravitz
Dec 15 '18 at 3:38














$begingroup$
@JMoravitz I may be misunderstanding the problem. I got my answer by "drawing" four 2's, then "drawing" the next card of a different rank.
$endgroup$
– Lee
Dec 15 '18 at 3:43




$begingroup$
@JMoravitz I may be misunderstanding the problem. I got my answer by "drawing" four 2's, then "drawing" the next card of a different rank.
$endgroup$
– Lee
Dec 15 '18 at 3:43




2




2




$begingroup$
I, and the book apparently, interpret it as being that you need at least two ranks each of which have at least two cards of that rank in it. For example $Aspadesuit, Aheartsuit, 2spadesuit, 3heartsuit$ currently has one rank with 2 cards of it's type, another rank with only one card of it's type, and another rank with only one card of it's type. On the other hand $Aspadesuit,Aheartsuit,2spadesuit,2diamondsuit$ has two ranks each of which have at least two cards of it's type. "at least two cards of each of two different kinds", the 'of' there that I put in bold is important.
$endgroup$
– JMoravitz
Dec 15 '18 at 3:45






$begingroup$
I, and the book apparently, interpret it as being that you need at least two ranks each of which have at least two cards of that rank in it. For example $Aspadesuit, Aheartsuit, 2spadesuit, 3heartsuit$ currently has one rank with 2 cards of it's type, another rank with only one card of it's type, and another rank with only one card of it's type. On the other hand $Aspadesuit,Aheartsuit,2spadesuit,2diamondsuit$ has two ranks each of which have at least two cards of it's type. "at least two cards of each of two different kinds", the 'of' there that I put in bold is important.
$endgroup$
– JMoravitz
Dec 15 '18 at 3:45














$begingroup$
@JMoravitz Ahh, I see what you are saying. Thank you for the insight!
$endgroup$
– Lee
Dec 15 '18 at 3:47




$begingroup$
@JMoravitz Ahh, I see what you are saying. Thank you for the insight!
$endgroup$
– Lee
Dec 15 '18 at 3:47












$begingroup$
"at least two cards of each of two different kinds" i.e. 'at least two cards of a first kind and at least two cards of a second kind.' This is different than "at least two cards, each of two different kinds" i.e. 'at least two cards, the first of which is of one kind and the second of which is of a different kind.' For the question of needing two cards which disagree on both rank and suit, that would be 14 cards needed there. Take all thirteen spades, then as soon as you take another non-spade it will differ in rank from one of the spades already present @RossMillikan
$endgroup$
– JMoravitz
Dec 15 '18 at 3:52






$begingroup$
"at least two cards of each of two different kinds" i.e. 'at least two cards of a first kind and at least two cards of a second kind.' This is different than "at least two cards, each of two different kinds" i.e. 'at least two cards, the first of which is of one kind and the second of which is of a different kind.' For the question of needing two cards which disagree on both rank and suit, that would be 14 cards needed there. Take all thirteen spades, then as soon as you take another non-spade it will differ in rank from one of the spades already present @RossMillikan
$endgroup$
– JMoravitz
Dec 15 '18 at 3:52












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