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Homework Help: Probability of a deck of cards

  1. Feb 27, 2010 #1
    Hi ! I need help with this problem.
    We have a 52 cards deck, and we're drawing at random and without replacement, until cards of one suit only are left. We need to find the probability that the remaining cards are hearts.

    Any suggestions concerning where to start?
     
  2. jcsd
  3. Feb 27, 2010 #2

    sylas

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    Symmetry.
     
  4. Feb 27, 2010 #3
    Umm.. Can you elaborate a bit what you mean ?
    Thanks
     
  5. Feb 27, 2010 #4
    How many suits are there in a standard deck?
     
  6. Feb 27, 2010 #5

    sylas

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    There is no difference between the suits. So whatever probability you have for the last suit being spades will also be the probability for clubs, diamonds and hearts. That is, the suits are symmetrical in this problem.

    Also, consider this. The last suit left is bound to be the suit of the last card in the deck. So you really only need to know the last card to know what suit will be left last.
     
  7. Feb 27, 2010 #6
    Well there are four.
    I tried taking an event An = {the nth card is a spade}, and maybe take the union of the intersections [tex]\bigcup^{52}_{i=39}\bigcap^{52}_{j=i} A_{j}[/tex]. But this is of no use I guess...
     
  8. Feb 27, 2010 #7
    Oh.. So the idea is P(the last suit is hearts) = P(last card is a heart) = 1/4 ?
     
  9. Feb 27, 2010 #8
    I think so, yes.
     
  10. Feb 27, 2010 #9

    sylas

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    Agree, yes. The answer is 0.25.
     
  11. Feb 27, 2010 #10
    hi

    the sample space is 52,

    and the required event is 13 (i.e u r saying that at last there should be all hearts.that means we have 13 hearts in a deck.

    so the event is 13,

    and the probability is n(E)/n(S),(here E=event, S=sample space)
    answer is 13/52,

    i.e 0.25
     
  12. Feb 27, 2010 #11

    Redbelly98

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    (Moderator's note: thread moved from "Set Theory, Logic, Probability, Statistics")

    Please note for the future, homework assignments or any textbook style exercises are to be posted in the appropriate forum in our Homework & Coursework Questions area. This should be done whether the problem is part of one's assigned coursework or just independent study.
     
  13. Mar 11, 2010 #12
    To jinbaw;

    Clearing your thought, we must first understand the range of possibility and the deck.

    *[1] possibility = probability = x; {0 < x < 1}

    Looking at your problem by default, I knew that total number of cards in the deck = sample space = 52 and within the deck there are four suits. {Hearts, Spades, Diamonds, Clubs)

    I will then divide my deck into 4 different group of suits. {52/4 = 13}

    *Always remember that you are to determine the possibility, therefore drawing at random will never be a problem (in this case).

    Now I have confirm 13 cards in deck at random will confirm hearts.

    hence, I can say that the possibility = 13(hearts)/52(sample space) = 0.25

    therefore 0.25 = x where 0 < x < 1, it agree with the condition [1] above.

    To pre-determine your outcome answer: possibility of suit of hearts {13/52 = 0.25}
    possibility of suit of spades {13/52 = 0.25}
    possibility of suit of diamonds {13/52 = 0.25}
    possibility of suit of clubs {13/52 = 0.25}

    The total sum of possibility = probability = 0.25(hearts) + 0.25(spades) + 0.25(diamonds) + 0.25(clubs) = 1 confirm outcome of the deck.
     
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