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Probability problems with 52 cards deck

  1. Oct 6, 2013 #1
    Hi,

    I am given some questions and I need help.

    (1)You are dealt two cards from a pack of 52 cards. What is the conditional probability that
    (a) both of them are aces if at least one of them is the ace of spades?
    (b) both of them are aces if at least one of them is an ace?

    a) I know that the probability of being dealt 2 aces out of 2 cards is : (4x3)/(52x52)=1/221
    there are 6 possible ways to arrange the aces and 3 to have Ace of spades. So should it be just 1/221/2 ?
    This is the conditional probability of being dealt 2 aces out of 2 cards, right?
    P(A1A2)=P(A1)P(A2|A1)=P(A2|A1)=P(A1A2)/P(A1) , but what should I change to get the condition right?
    b)Isnt it P(A2|A1)/P(A1)?
    And I don't have idea about the 2nd one:
    (2) Consider a pack of 52 cards.
    (a), What is the probability that there is a king right after the first ace?
    (b), What is the probability that there is an ace right after the first ace?
    (c) What is the probability that the 1st ace is the 30-th card?
    (d) If the 1 st ace is the 30-th card, what is the conditional probability the next card is the ace of spades?
    (e) If the 1st ace is the 30-th card, what is the conditional probability the next card is
    the jack of diamonds?
    Note that you have to give these probabilities without looking at any cards from the pack.

    a) if the Ace is the first card it gives 4/52 and the king is right after the ace is 4/51, but what when they are not 1st and 2nd cards
    b) is almost the same 4/52*3/51 ?
    c) I have no idea what is the correct approach, but I think since there are 22 cards left we should have 4/52 - 4/22?
    I have no Idea on the last 2.

    Thanks a lot!
     
  2. jcsd
  3. Oct 6, 2013 #2

    haruspex

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    Not so. How many ways are there in total of drawing two cards from 52 without replacement? How many ways are there of drawing two of the four aces?
     
  4. Oct 7, 2013 #3
    That's what is written in the lecture notes.
     
  5. Oct 7, 2013 #4

    Ray Vickson

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    The probability you give is correct for getting two different aces in two independent draws with replacement; that is, you put back the first card, re-shuffle the deck and then draw the second card. Do you think that is what the question means?
     
  6. Oct 7, 2013 #5

    phinds

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    Despite what others have said, I agree w/ this [(4x3)/(52x52)]. You have properly accounted for the non-replacement issue by using 3 for the second draw. I do NOT agree, however that the result is 1/221, it's 3/676
     
  7. Oct 7, 2013 #6
    It should be (4x3)/(52x51)=1/221. Sorry! Any thoughts about the 2nd question?
    I think you don't have to put the other ace again in the deck, otherwise you will have chance to get 1st time As and second time again As.
     
  8. Oct 7, 2013 #7

    Ray Vickson

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    With replacement: probability = 4×3 /(52×52).
    Without replacement: probability = 4×3/(52×51).

    In fact, I much prefer to look at it sequentially: Probability first Ace = 4/52. With replacement, the probability the second is a different Ace = 3/52. Without replacement, we have 51 cards left, of which 3 are aces, so probability second Ace = 3/51.

    This is all, 100% standard.
     
  9. Oct 7, 2013 #8

    phinds

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    NUTS ... I obviously got that one wrong (the 51 instead of 52)
     
  10. Oct 7, 2013 #9
    what about the conditional probability of this case? P(A1A2)=P(A1)P(A2|A1)=P(A2|A1)=P(A1A2)/P(A1)?
    and for b)
     
  11. Oct 8, 2013 #10

    haruspex

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    I think you mean "P(A1A2)=P(A1)P(A2|A1), therefore P(A2|A1)=P(A1A2)/P(A1)".
    So apply that. You have computed A2, right? But you need P(A1&A2) and P(A1).
     
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