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Homework Help: Probability of dealing 4 cards from pack of 52 cards

  1. Feb 27, 2010 #1
    1. The problem statement, all variables and given/known data
    The question is:
    Four cards are dealt from a well-shuffled pack of 52 cards.
    What is the probability that all four cards bear the same face value?

    2. Relevant equations

    3. The attempt at a solution
    My guess is that first of all, I pick a card out of random, 1/52. Then after that, I have 12 chances of picking the same face value out of 51 cards, so that is 12/52. Then I continue this for the other cards, so I get:

    1/52 + 12/51 + 11/50 + 10/49 = 0.68

    The thing is, I can't help but think that's a bit high. Have I missed something glaringly obvious, or have I done it right?
  2. jcsd
  3. Feb 27, 2010 #2


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    Welcome to PF!

    Hi something42! Welcome to PF! :smile:
    You want the joint probability (of independent events), so you need to multiply, not add. :wink:
  4. Feb 27, 2010 #3
    It depends on whether you specify the value of the card before you start pulling them from the pack.

    If you don't specify the card then the odds are 1 * 3/51 *2/50 * 1/49 (the 1 at the start is because you must pull a card from the pack)

    If you do specify the value the odds become 4/52 * 3/51 *2/50 *1/49
  5. Feb 27, 2010 #4
    Re: Welcome to PF!

    Hey guys! Thanks for taking the time to answer my question!

    Yay! Thanks! :blushing:

    You know, after I posted this, I knew it was the addition that gave me problems. However, when I saw the results when I multiplied them afterwards, it gave me a shock as to how low the value was, so I didn't think that was appropriate. But wow! That low? :eek:

    Thanks for the tip, though! In my moment of hair pulling, it helps to get some confirmation. :D

    Thanks for taking the time! If you don't mind, I want to "challenge" your post.

    First of all, the 1 at the start completely makes sense. I don't know why I didn't think of this sooner. So I guess that answers your question of the card being specified or not (from what I've seen in the question, it doesn't matter).

    However, I don't get why you have 3/51, 2/50, 1/49. I always thought the definition of

    probability of E = number of sample points in E/total number of sample posts

    and considering that, if the card pack gets less and less for a suit (because that's what I'm reading it as), surely we take into account the number of available suits, and not the amount of cards currently needing to be picked?

    Sorry about the "challenge", but I really want to get this stats stuff right. It seems so simple, yet if I got it wrong, I'd be constantly slapping my head for the rest of my life :(

    But thanks everyone who has posted :D
  6. Feb 27, 2010 #5


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    Hi something42! :smile:
    Jobrag :smile: is right …

    your 12, 11, 10 would be for picking the same suit, because there are 12, 11, 10 left.

    For the same value, there are 3, 2, 1 left. :wink:
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